Ch.5: Geometrical aspects of linear algebra

Chapter 5 of the book "No Bullshit Guide to Linear Algebra" by Ivan Savov Lines and planes Projections Coordinate projections Vector spaces Vector space techniques

111 flashcards · Mathematics · by lakesare

Part of Textbooks › No Bullshit Guide to Linear Algebra (Ivan Savov) · 10 courses

1. [Geometrical aspects of linear algebra] Geometrically, what is the solution set of a system of three linear equations in three variables x,y,zx, y, zx,y,z ?

   The intersection of three planes in R3\mathbb{R}^3R3 . Each equation Aix+Biy+Ciz=DiA_ix + B_iy + C_iz = D_iAi​x+Bi​y+Ci​z=Di​ describes a plane, and a solution (x,y,z)(x, y, z)(x,y,z) lies in all three. Example: the planes x=1x = 1x=1 , y=2y = 2y=2 , z=3z = 3z=3 intersect in the single point (1,2,3)(1, 2, 3)(1,2,3) .

2. [Lines and planes] ppp and qqq are points in R3\mathbb{R}^3R3 . Which vector takes ppp to qqq ?

   The displacement vector d⃗=q−p\vec{d} = q - pd=q−p . A point ppp and the vector p⃗\vec{p}p​ from the origin to ppp have the same coordinates, so points can be added and subtracted like vectors.Example: from p=(1,0,0)p = (1, 0, 0)p=(1,0,0) to q=(2,3,4)q = (2, 3, 4)q=(2,3,4) : d⃗=(1,3,4)\vec{d} = (1, 3, 4)d=(1,3,4) .

3. [Lines and planes] When do two lines in R2\mathbb{R}^2R2 intersect in exactly one point, and how do you find that point?

   When the lines are not parallel; solve their two equations simultaneously. Example: x−y=−1x - y = -1x−y=−1 and 3x+y=93x + y = 93x+y=9 ; adding them gives 4x=84x = 84x=8 , so p=(2,3)p = (2, 3)p=(2,3) .

4. [Lines and planes] Where do the lines x+2y=5x + 2y = 5x+2y=5 and 3x+9y=213x + 9y = 213x+9y=21 intersect?

   At p=(1,2)p = (1, 2)p=(1,2) . From the first equation x=5−2yx = 5 - 2yx=5−2y ; then 15−6y+9y=2115 - 6y + 9y = 2115−6y+9y=21 , so y=2y = 2y=2 and x=1x = 1x=1 .

5. [Lines and planes] What is the parametric equation of the line ℓ\ellℓ through the point pop_{\text{o}}po​ with direction vector v⃗\vec{v}v ?

   ℓ:{po+t v⃗, t∈R}\boldsymbol{\ell : \{ p_{\text{o}} + t\,\vec{v}, \ t \in \mathbb{R} \}}ℓ:{po​+tv, t∈R} The points reached from pop_{\text{o}}po​ by adding a multiple of v⃗\vec{v}v . The line is parametrized by ttt . Example: {(1,0,0)+t(0,1,1)}\{(1, 0, 0) + t(0, 1, 1)\}{(1,0,0)+t(0,1,1)} contains (1,0,0)(1, 0, 0)(1,0,0) ( t=0t = 0t=0 ) and (1,2,2)(1, 2, 2)(1,2,2) ( t=2t = 2t=2 ).

6. [Lines and planes] What is the symmetric equation of the line through po=(pox,poy,poz)p_{\text{o}} = (p_{\text{o}x}, p_{\text{o}y}, p_{\text{o}z})po​=(pox​,poy​,poz​) with direction vector v⃗=(vx,vy,vz)\vec{v} = (v_x, v_y, v_z)v=(vx​,vy​,vz​) ?

   x−poxvx=y−poyvy=z−pozvz\boldsymbol{\dfrac{x - p_{\text{o}x}}{v_x} = \dfrac{y - p_{\text{o}y}}{v_y} = \dfrac{z - p_{\text{o}z}}{v_z}}vx​x−pox​​=vy​y−poy​​=vz​z−poz​​ It describes the line by relations between xxx , yyy , zzz , without a parameter (here vx,vy,vz≠0v_x, v_y, v_z \neq 0vx​,vy​,vz​=0 ).Example: po=(1,1,1)p_{\text{o}} = (1, 1, 1)po​=(1,1,1) , v⃗=(1,2,3)\vec{v} = (1, 2, 3)v=(1,2,3) : x−1=y−12=z−13\ x - 1 = \frac{y - 1}{2} = \frac{z - 1}{3} x−1=2y−1​=3z−1​ .

7. [Lines and planes] Parametric equation of a line: (x,y,z)=po+t v⃗(x, y, z) = p_{\text{o}} + t\,\vec{v}(x,y,z)=po​+tv , t∈Rt \in \mathbb{R}t∈R .How do you get the symmetric equation of the line from the parametric one?

   Solve each coordinate equation for ttt and set the results equal. x=pox+tvxx = p_{\text{o}x} + t v_xx=pox​+tvx​ gives t=x−poxvxt = \frac{x - p_{\text{o}x}}{v_x}t=vx​x−pox​​ , and likewise for yyy and zzz .

8. [Lines and planes] A non-vertical line in R2\mathbb{R}^2R2 passes through pop_{\text{o}}po​ with direction v⃗=(vx,vy)\vec{v} = (v_x, v_y)v=(vx​,vy​) , vx≠0v_x \neq 0vx​=0 . Write it as y=mx+by = mx + by=mx+b : what are mmm and bbb ?

   m=vyvx\boldsymbol{m = \frac{v_y}{v_x}}m=vx​vy​​ , b=poy−vyvxpox\ \boldsymbol{b = p_{\text{o}y} - \frac{v_y}{v_x} p_{\text{o}x}} b=poy​−vx​vy​​pox​ The slope mmm is the change in yyy divided by the change in xxx along the line.Example: po=(1,1)p_{\text{o}} = (1, 1)po​=(1,1) , v⃗=(2,4)\vec{v} = (2, 4)v=(2,4) : m=2m = 2m=2 , b=1−2=−1b = 1 - 2 = -1b=1−2=−1 , so y=2x−1y = 2x - 1y=2x−1 .

9. [Lines and planes] What is a parametric equation of the line through two points ppp and qqq ?

   ℓ:{p+t (q−p), t∈R}\boldsymbol{\ell : \{ p + t\,(q - p), \ t \in \mathbb{R} \}}ℓ:{p+t(q−p), t∈R} q−pq - pq−p serves as the direction vector. Any nonzero vector parallel to q−pq - pq−p is also a valid direction vector.

10. [Lines and planes] Find a parametric equation and the symmetric equation of the line through p=(1,1,1)p = (1, 1, 1)p=(1,1,1) and q=(2,3,4)q = (2, 3, 4)q=(2,3,4) .

    {(1,1,1)+t(1,2,3), t∈R}\boldsymbol{\{(1, 1, 1) + t(1, 2, 3), \ t \in \mathbb{R}\}}{(1,1,1)+t(1,2,3), t∈R} and x−11=y−12=z−13\boldsymbol{\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 1}{3}}1x−1​=2y−1​=3z−1​ Direction v⃗=q−p=(1,2,3)\vec{v} = q - p = (1, 2, 3)v=q−p=(1,2,3) . The direction (−1,−2,−3)(-1, -2, -3)(−1,−2,−3) gives an equally valid parametric equation.

11. [Lines and planes] What is the intersection of two non-parallel planes in R3\mathbb{R}^3R3 , and how do you compute it?

    A line; solve the system of the two plane equations with Gauss–Jordan elimination. A1x+B1y+C1z=D1A_1x + B_1y + C_1z = D_1A1​x+B1​y+C1​z=D1​ , A2x+B2y+C2z=D2\ A_2x + B_2y + C_2z = D_2 A2​x+B2​y+C2​z=D2​ : two equations in three unknowns, so one free variable, which is the line's parameter. Example: the xyxyxy -plane z=0z = 0z=0 and the xzxzxz -plane y=0y = 0y=0 intersect in the xxx -axis {(0,0,0)+t(1,0,0)}\{(0, 0, 0) + t(1, 0, 0)\}{(0,0,0)+t(1,0,0)} .

12. [Lines and planes] Find the intersection of the planes 0x+0y+1z=00x + 0y + 1z = 00x+0y+1z=0 and 0x+1y+1z=00x + 1y + 1z = 00x+1y+1z=0 .

    The xxx -axis: {(0,0,0)+t(1,0,0), t∈R}\{(0, 0, 0) + t(1, 0, 0), \ t \in \mathbb{R}\}{(0,0,0)+t(1,0,0), t∈R} . [00100110]∼[01100010]∼[01000010]\left[\begin{array}{ccc|c} 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right][00​01​11​00​]∼[00​10​11​00​]∼[00​10​01​00​] The first column has no pivot, so x=tx = tx=t is free; y=0y = 0y=0 , z=0…

13. [Lines and planes] What is the general equation of a plane PPP in R3\mathbb{R}^3R3 ?

    P:{(x,y,z)∈R3∣Ax+By+Cz=D}\boldsymbol{P : \{(x, y, z) \in \mathbb{R}^3 \mid Ax + By + Cz = D\}}P:{(x,y,z)∈R3∣Ax+By+Cz=D} The algebraic view of a plane, useful for calculations.Example: x+y+z=1x + y + z = 1x+y+z=1 is the plane through (1,0,0)(1, 0, 0)(1,0,0) , (0,1,0)(0, 1, 0)(0,1,0) and (0,0,1)(0, 0, 1)(0,0,1) .

14. [Lines and planes] What is the geometric equation of the plane through the point pop_{\text{o}}po​ with normal vector n⃗\vec{n}n ?

    P: n⃗⋅[(x,y,z)−po]=0\boldsymbol{P : \ \vec{n} \cdot [(x, y, z) - p_{\text{o}}] = 0}P: n⋅[(x,y,z)−po​]=0 A normal vector is perpendicular to the plane.

15. [Lines and planes] Geometric equation of a plane: n⃗⋅[(x,y,z)−po]=0\vec{n} \cdot [(x, y, z) - p_{\text{o}}] = 0n⋅[(x,y,z)−po​]=0 , with pop_{\text{o}}po​ a point in the plane and n⃗\vec{n}n a normal vector.Why does this equation describe exactly the points of the plane?

    (x,y,z)−po(x, y, z) - p_{\text{o}}(x,y,z)−po​ is the vector from pop_{\text{o}}po​ to (x,y,z)(x, y, z)(x,y,z) , and the equation keeps exactly the points for which this vector is perpendicular to n⃗\vec{n}n . The dot product of two vectors is zero if and only if they are orthogonal.

16. [Lines and planes] How do you convert the geometric equation n⃗⋅[(x,y,z)−po]=0\vec{n} \cdot [(x, y, z) - p_{\text{o}}] = 0n⋅[(x,y,z)−po​]=0 of a plane into its general equation Ax+By+Cz=DAx + By + Cz = DAx+By+Cz=D ?

    A=nx, B=ny, C=nz, D=n⃗⋅po\boldsymbol{A = n_x, \ B = n_y, \ C = n_z, \ D = \vec{n} \cdot p_{\text{o}}}A=nx​, B=ny​, C=nz​, D=n⋅po​ Expand the dot product and move n⃗⋅po\vec{n} \cdot p_{\text{o}}n⋅po​ to the right-hand side.Example: n⃗=(1,2,3)\vec{n} = (1, 2, 3)n=(1,2,3) , po=(1,0,1)p_{\text{o}} = (1, 0, 1)po​=(1,0,1) : x+2y+3z=4\ x + 2y + 3z = 4 x+2y+3z=4 .

17. [Lines and planes] Do Ax+By+Cz=DAx + By + Cz = DAx+By+Cz=D and αAx+αBy+αCz=αD\alpha Ax + \alpha By + \alpha Cz = \alpha DαAx+αBy+αCz=αD ( α≠0\alpha \neq 0α=0 ) describe the same plane? What does this say about the normal vector?

    Yes, the same plane; only the direction of n⃗\vec{n}n matters, not its length. Likewise n⃗⋅[(x,y,z)−po]=0\vec{n} \cdot [(x, y, z) - p_{\text{o}}] = 0n⋅[(x,y,z)−po​]=0 and αn⃗⋅[(x,y,z)−po]=0\alpha\vec{n} \cdot [(x, y, z) - p_{\text{o}}] = 0αn⋅[(x,y,z)−po​]=0 define the same plane.Example: x+y+z=1x + y + z = 1x+y+z=1 and 2x+2y+2z=22x + 2y + 2z = 22x+2y+2z=2 ; normals (1,1,1)(1, 1, 1)(1,1,1) and (2,2,2)(2, 2, 2)(2,2,2) .

18. [Lines and planes] What is the parametric equation of a plane PPP through the point pop_{\text{o}}po​ ?

    P:{po+s v⃗+t w⃗, s,t∈R}\boldsymbol{P : \{ p_{\text{o}} + s\,\vec{v} + t\,\vec{w}, \ s, t \in \mathbb{R} \}}P:{po​+sv+tw, s,t∈R} with v⃗,w⃗\vec{v}, \vec{w}v,w linearly independent vectors parallel to the plane. A plane is two-dimensional, so it needs two parameters sss and ttt .Example: {(0,0,0)+s(1,0,0)+t(0,1,0)}\{(0, 0, 0) + s(1, 0, 0) + t(0, 1, 0)\}{(0,0,0)+s(1,0,0)+t(0,1,0)} is the xyxyxy -plane.

19. [Lines and planes] The points ppp , qqq , rrr lie in a plane (not all on one line). How do you find a normal vector n⃗\vec{n}n of the plane?

    n⃗=(q−p)×(r−p)\boldsymbol{\vec{n} = (q - p) \times (r - p)}n=(q−p)×(r−p) q−pq - pq−p and r−pr - pr−p are parallel to the plane, and their cross product is perpendicular to both. The geometric equation is then n⃗⋅[(x,y,z)−p]=0\vec{n} \cdot [(x, y, z) - p] = 0n⋅[(x,y,z)−p]=0 .

20. [Lines and planes] Find the geometric, general and parametric equations of the plane through p=(1,0,0)p = (1, 0, 0)p=(1,0,0) , q=(0,1,0)q = (0, 1, 0)q=(0,1,0) , r=(0,0,1)r = (0, 0, 1)r=(0,0,1) .

    (1,1,1)⋅[(x,y,z)−(1,0,0)]=0\boldsymbol{(1, 1, 1) \cdot [(x, y, z) - (1, 0, 0)] = 0}(1,1,1)⋅[(x,y,z)−(1,0,0)]=0 , x+y+z=1\ \boldsymbol{x + y + z = 1} x+y+z=1 , {(1,0,0)+s(−1,1,0)+t(−1,0,1)}\ \boldsymbol{\{(1, 0, 0) + s(-1, 1, 0) + t(-1, 0, 1)\}} {(1,0,0)+s(−1,1,0)+t(−1,0,1)} v⃗=q−p=(−1,1,0)\vec{v} = q - p = (-1, 1, 0)v=q−p=(−1,1,0) , w⃗=r−p=(−1,0,1)\vec{w} = r - p = (-1, 0, 1)w=r−p=(−1,0,1) , n⃗=v⃗×w⃗=(1,1,1)\vec{n} = \vec{v} \times \vec{w} = (1, 1, 1)n=v×w=(1,1,1) , and D=n⃗⋅p=1D = \vec{n} \cdo…

21. [Lines and planes] What is the distance d(p,q)d(p, q)d(p,q) between points ppp and qqq in R3\mathbb{R}^3R3 ?

    d(p,q)=∥q−p∥=(qx−px)2+(qy−py)2+(qz−pz)2\boldsymbol{d(p, q) = \|q - p\| = \sqrt{(q_x - p_x)^2 + (q_y - p_y)^2 + (q_z - p_z)^2}}d(p,q)=∥q−p∥=(qx​−px​)2+(qy​−py​)2+(qz​−pz​)2​ Example: d((1,2,3),(4,6,3))=9+16+0=5d\big((1, 2, 3), (4, 6, 3)\big) = \sqrt{9 + 16 + 0} = 5d((1,2,3),(4,6,3))=9+16+0​=5 .

22. [Lines and planes] What is the closest distance between the line ℓ:{po+t v⃗, t∈R}\ell : \{p_{\text{o}} + t\,\vec{v}, \ t \in \mathbb{R}\}ℓ:{po​+tv, t∈R} and the origin OOO ?

    d(ℓ,O)=∥po−po⋅v⃗∥v⃗∥2 v⃗∥\boldsymbol{d(\ell, O) = \left\| p_{\text{o}} - \frac{p_{\text{o}} \cdot \vec{v}}{\|\vec{v}\|^2}\,\vec{v} \right\|}d(ℓ,O)=∥∥​po​−∥v∥2po​⋅v​v∥∥​ From pop_{\text{o}}po​ , subtract its component along v⃗\vec{v}v ; what remains is perpendicular to the line.

23. [Lines and planes] d(ℓ,O)=∥po−po⋅v⃗∥v⃗∥2 v⃗∥d(\ell, O) = \left\| p_{\text{o}} - \frac{p_{\text{o}} \cdot \vec{v}}{\|\vec{v}\|^2}\,\vec{v} \right\|d(ℓ,O)=∥∥​po​−∥v∥2po​⋅v​v∥∥​ for the line ℓ:{po+t v⃗}\ell : \{p_{\text{o}} + t\,\vec{v}\}ℓ:{po​+tv} and the origin OOO .Compute the closest distance between ℓ:{(4,5,6)+t(1,0,1), t∈R}\ell : \{(4, 5, 6) + t(1, 0, 1), \ t \in \mathbb{R}\}ℓ:{(4,5,6)+t(1,0,1), t∈R} and the origin.

    33\boldsymbol{3\sqrt{3}}33​ (4,5,6)⋅(1,0,1)12+02+12=102=5\frac{(4, 5, 6) \cdot (1, 0, 1)}{1^2 + 0^2 + 1^2} = \frac{10}{2} = 512+02+12(4,5,6)⋅(1,0,1)​=210​=5 , so ∥(4,5,6)−5(1,0,1)∥=∥(−1,5,1)∥=27=33\|(4, 5, 6) - 5(1, 0, 1)\| = \|(-1, 5, 1)\| = \sqrt{27} = 3\sqrt{3}∥(4,5,6)−5(1,0,1)∥=∥(−1,5,1)∥=27​=33​ .

24. [Lines and planes] What is the closest distance between the plane P:n⃗⋅[(x,y,z)−po]=0P : \vec{n} \cdot [(x, y, z) - p_{\text{o}}] = 0P:n⋅[(x,y,z)−po​]=0 and the origin OOO ?

    d(P,O)=∣n⃗⋅po∣∥n⃗∥\boldsymbol{d(P, O) = \dfrac{|\vec{n} \cdot p_{\text{o}}|}{\|\vec{n}\|}}d(P,O)=∥n∥∣n⋅po​∣​ Example: n⃗=(−3,0,−4)\vec{n} = (-3, 0, -4)n=(−3,0,−4) , po=(1,2,3)p_{\text{o}} = (1, 2, 3)po​=(1,2,3) : ∣−3−12∣5=155=3\frac{|-3 - 12|}{5} = \frac{15}{5} = 35∣−3−12∣​=515​=3 .

25. [Projections] According to Savov, what should you learn about each projection formula instead of memorizing it?

    Its vector diagram, the picture of the sequence of operations that computes it. From the diagram you can reproduce the formula using only vector subtraction, vector length and the dot product.

26. [Projections] Which sets are the vector subspaces of R3\mathbb{R}^3R3 ?

    Lines and planes through the origin (and also {0⃗}\{\vec{0}\}{0} and R3\mathbb{R}^3R3 itself). Example: {t(1,2,3)}\{t(1, 2, 3)\}{t(1,2,3)} is a subspace; {(1,0,0)+t(0,1,0)}\{(1, 0, 0) + t(0, 1, 0)\}{(1,0,0)+t(0,1,0)} is not, since it does not contain 0⃗\vec{0}0 .

27. [Projections] What is the orthogonal space S⊥S^\perpS⊥ of a subspace S⊆R3S \subseteq \mathbb{R}^3S⊆R3 ?

    S⊥={w⃗∈R3∣w⃗⋅s⃗=0 for all s⃗∈S}\boldsymbol{S^\perp = \{\vec{w} \in \mathbb{R}^3 \mid \vec{w} \cdot \vec{s} = 0 \ \text{for all} \ \vec{s} \in S\}}S⊥={w∈R3∣w⋅s=0 for all s∈S} The vectors perpendicular to every vector of SSS ; ⊥\perp⊥ stands for "perpendicular to".Example: for SSS = the xxx -axis, S⊥S^\perpS⊥ is the yzyzyz -plane.

28. [Projections] ΠS:R3→S\Pi_S : \mathbb{R}^3 \to SΠS​:R3→S is the projection onto the subspace SSS . What is ΠS(v⃗)\Pi_S(\vec{v})ΠS​(v) for v⃗∈S\vec{v} \in Sv∈S , and ΠS(w⃗)\Pi_S(\vec{w})ΠS​(w) for w⃗∈S⊥\vec{w} \in S^\perpw∈S⊥ ?

    ΠS(v⃗)=v⃗\boldsymbol{\Pi_S(\vec{v}) = \vec{v}}ΠS​(v)=v and ΠS(w⃗)=0⃗\boldsymbol{\Pi_S(\vec{w}) = \vec{0}}ΠS​(w)=0 With linearity: ΠS(αv⃗+βw⃗)=αv⃗\Pi_S(\alpha\vec{v} + \beta\vec{w}) = \alpha\vec{v}ΠS​(αv+βw)=αv . So ΠS\Pi_SΠS​ removes the part of the input that lies in S⊥S^\perpS⊥ .Example: SSS = xxx -axis: ΠS(2,0,0)=(2,0,0)\Pi_S(2, 0, 0) = (2, 0, 0)ΠS​(2,0,0)=(2,0,0) , ΠS(0,5,6)=(0,0,0)\Pi_S(0, 5, 6) = (0, 0, 0)ΠS​(0,5,6)=(0,0,0) .

29. [Projections] What does it mean that R3\mathbb{R}^3R3 is the direct sum S⊕S⊥S \oplus S^\perpS⊕S⊥ of a subspace SSS and its orthogonal space?

    Every u⃗∈R3\vec{u} \in \mathbb{R}^3u∈R3 splits as u⃗=v⃗+w⃗\vec{u} = \vec{v} + \vec{w}u=v+w with v⃗=ΠS(u⃗)∈S\vec{v} = \Pi_S(\vec{u}) \in Sv=ΠS​(u)∈S and w⃗=ΠS⊥(u⃗)∈S⊥\vec{w} = \Pi_{S^\perp}(\vec{u}) \in S^\perpw=ΠS⊥​(u)∈S⊥ . Example: SSS = xxx -axis: (4,5,6)=(4,0,0)+(0,5,6)(4, 5, 6) = (4, 0, 0) + (0, 5, 6)(4,5,6)=(4,0,0)+(0,5,6) .

30. [Projections] Projections are idempotent: ΠS(ΠS(u⃗))=ΠS(u⃗)\Pi_S(\Pi_S(\vec{u})) = \Pi_S(\vec{u})ΠS​(ΠS​(u))=ΠS​(u) for every u⃗\vec{u}u .

    ΠS(u⃗)\Pi_S(\vec{u})ΠS​(u) already lies in SSS , and ΠS\Pi_SΠS​ leaves vectors of SSS unchanged; projecting a second time has no effect.

31. [Projections] What is the projection of u⃗\vec{u}u onto the line ℓ:{t v⃗, t∈R}\ell : \{t\,\vec{v}, \ t \in \mathbb{R}\}ℓ:{tv, t∈R} through the origin?

    Πℓ(u⃗)=u⃗⋅v⃗∥v⃗∥2 v⃗\boldsymbol{\Pi_\ell(\vec{u}) = \dfrac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2}\,\vec{v}}Πℓ​(u)=∥v∥2u⋅v​v Example: u⃗=(3,4,0)\vec{u} = (3, 4, 0)u=(3,4,0) , v⃗=(2,0,0)\vec{v} = (2, 0, 0)v=(2,0,0) : 64(2,0,0)=(3,0,0)\frac{6}{4}(2, 0, 0) = (3, 0, 0)46​(2,0,0)=(3,0,0) .

32. [Projections] Why is the projection of u⃗\vec{u}u onto the line with direction v⃗\vec{v}v equal to u⃗⋅v⃗∥v⃗∥2 v⃗\dfrac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2}\,\vec{v}∥v∥2u⋅v​v ?

    Its length is ∥u⃗∥cos⁡θ=u⃗⋅v⃗∥v⃗∥\|\vec{u}\| \cos\theta = \frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|}∥u∥cosθ=∥v∥u⋅v​ and its direction is the unit vector v⃗∥v⃗∥\frac{\vec{v}}{\|\vec{v}\|}∥v∥v​ . θ\thetaθ is the angle between u⃗\vec{u}u and v⃗\vec{v}v , and u⃗⋅v⃗=∥u⃗∥∥v⃗∥cos⁡θ\vec{u} \cdot \vec{v} = \|\vec{u}\|\|\vec{v}\|\cos\thetau⋅v=∥u∥∥v∥cosθ . Multiplying the length by the unit vector gives the formula.

33. [Projections] What is the orthogonal space ℓ⊥\ell^\perpℓ⊥ of the line ℓ\ellℓ through the origin with direction vector v⃗\vec{v}v ?

    The plane through the origin with normal vector v⃗\vec{v}v : ℓ⊥:{(x,y,z)∣(x,y,z)⋅v⃗=0}\ \ell^\perp : \{(x, y, z) \mid (x, y, z) \cdot \vec{v} = 0\} ℓ⊥:{(x,y,z)∣(x,y,z)⋅v=0} Example: for v⃗=(0,0,1)\vec{v} = (0, 0, 1)v=(0,0,1) ( zzz -axis), ℓ⊥\ell^\perpℓ⊥ is the xyxyxy -plane z=0z = 0z=0 .

34. [Projections] Πℓ(u⃗)=u⃗⋅v⃗∥v⃗∥2 v⃗\Pi_\ell(\vec{u}) = \frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2}\,\vec{v}Πℓ​(u)=∥v∥2u⋅v​v is the projection onto the line ℓ\ellℓ through the origin with direction v⃗\vec{v}v .What is Πℓ⊥(u⃗)\Pi_{\ell^\perp}(\vec{u})Πℓ⊥​(u) , and why?

    Πℓ⊥(u⃗)=u⃗−u⃗⋅v⃗∥v⃗∥2 v⃗\boldsymbol{\Pi_{\ell^\perp}(\vec{u}) = \vec{u} - \dfrac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2}\,\vec{v}}Πℓ⊥​(u)=u−∥v∥2u⋅v​v u⃗=Πℓ(u⃗)+Πℓ⊥(u⃗)\vec{u} = \Pi_\ell(\vec{u}) + \Pi_{\ell^\perp}(\vec{u})u=Πℓ​(u)+Πℓ⊥​(u) , so the ℓ⊥\ell^\perpℓ⊥ part is what remains of u⃗\vec{u}u after removing the ℓ\ellℓ part.

35. [Projections] Πℓ(u⃗)=u⃗⋅v⃗∥v⃗∥2 v⃗\Pi_\ell(\vec{u}) = \frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2}\,\vec{v}Πℓ​(u)=∥v∥2u⋅v​v and Πℓ⊥(u⃗)=u⃗−Πℓ(u⃗)\Pi_{\ell^\perp}(\vec{u}) = \vec{u} - \Pi_\ell(\vec{u})Πℓ⊥​(u)=u−Πℓ​(u) for the line ℓ\ellℓ through the origin with direction v⃗\vec{v}v .For ℓ:{t(1,2,3)}\ell : \{t(1, 2, 3)\}ℓ:{t(1,2,3)} and u⃗=(4,5,6)\vec{u} = (4, 5, 6)u=(4,5,6) , compute Πℓ(u⃗)\Pi_\ell(\vec{u})Πℓ​(u) and Πℓ⊥(u⃗)\Pi_{\ell^\perp}(\vec{u})Πℓ⊥​(u) .

    Πℓ(u⃗)=(167,327,487)\boldsymbol{\Pi_\ell(\vec{u}) = (\frac{16}{7}, \frac{32}{7}, \frac{48}{7})}Πℓ​(u)=(716​,732​,748​) , Πℓ⊥(u⃗)=(127,37,−67)\ \boldsymbol{\Pi_{\ell^\perp}(\vec{u}) = (\frac{12}{7}, \frac{3}{7}, \frac{-6}{7})} Πℓ⊥​(u)=(712​,73​,7−6​) u⃗⋅v⃗=4+10+18=32\vec{u} \cdot \vec{v} = 4 + 10 + 18 = 32u⋅v=4+10+18=32 , ∥v⃗∥2=14\|\vec{v}\|^2 = 14∥v∥2=14 , so Πℓ(u⃗)=167(1,2,3)\Pi_\ell(\vec{u}) = \frac{16}{7}(1, 2, 3)Πℓ​(u)=716​(1,2,3) .Checks: (127,37,−67)⋅(1,2,3)=12+6−187=0(\frac{12}{7}, \frac…

36. [Projections] What is the projection of u⃗\vec{u}u onto the plane P:{n⃗⋅(x,y,z)=0}P : \{\vec{n} \cdot (x, y, z) = 0\}P:{n⋅(x,y,z)=0} through the origin?

    ΠP(u⃗)=u⃗−u⃗⋅n⃗∥n⃗∥2 n⃗\boldsymbol{\Pi_P(\vec{u}) = \vec{u} - \dfrac{\vec{u} \cdot \vec{n}}{\|\vec{n}\|^2}\,\vec{n}}ΠP​(u)=u−∥n∥2u⋅n​n P⊥P^\perpP⊥ is the line with direction n⃗\vec{n}n , so ΠP⊥(u⃗)=u⃗⋅n⃗∥n⃗∥2 n⃗\Pi_{P^\perp}(\vec{u}) = \frac{\vec{u} \cdot \vec{n}}{\|\vec{n}\|^2}\,\vec{n}ΠP⊥​(u)=∥n∥2u⋅n​n , and ΠP(u⃗)=u⃗−ΠP⊥(u⃗)\Pi_P(\vec{u}) = \vec{u} - \Pi_{P^\perp}(\vec{u})ΠP​(u)=u−ΠP⊥​(u) .

37. [Projections] To project u⃗\vec{u}u onto a plane PPP through the origin, why compute ΠP⊥(u⃗)\Pi_{P^\perp}(\vec{u})ΠP⊥​(u) first?

    P⊥P^\perpP⊥ is a line (with direction the normal n⃗\vec{n}n ), and projecting onto a line is one formula; then ΠP(u⃗)=u⃗−ΠP⊥(u⃗)\Pi_P(\vec{u}) = \vec{u} - \Pi_{P^\perp}(\vec{u})ΠP​(u)=u−ΠP⊥​(u) . This uses R3=P⊕P⊥\mathbb{R}^3 = P \oplus P^\perpR3=P⊕P⊥ .

38. [Projections] ΠP⊥(u⃗)=u⃗⋅n⃗∥n⃗∥2 n⃗\Pi_{P^\perp}(\vec{u}) = \frac{\vec{u} \cdot \vec{n}}{\|\vec{n}\|^2}\,\vec{n}ΠP⊥​(u)=∥n∥2u⋅n​n and ΠP(u⃗)=u⃗−ΠP⊥(u⃗)\Pi_P(\vec{u}) = \vec{u} - \Pi_{P^\perp}(\vec{u})ΠP​(u)=u−ΠP⊥​(u) for the plane PPP through the origin with normal n⃗\vec{n}n .For P:(1,1,1)⋅(x,y,z)=0P : (1, 1, 1) \cdot (x, y, z) = 0P:(1,1,1)⋅(x,y,z)=0 and u⃗=(4,5,6)\vec{u} = (4, 5, 6)u=(4,5,6) , compute ΠP(u⃗)\Pi_P(\vec{u})ΠP​(u) and ΠP⊥(u⃗)\Pi_{P^\perp}(\vec{u})ΠP⊥​(u) .

    ΠP(u⃗)=(−1,0,1)\boldsymbol{\Pi_P(\vec{u}) = (-1, 0, 1)}ΠP​(u)=(−1,0,1) , ΠP⊥(u⃗)=(5,5,5)\ \boldsymbol{\Pi_{P^\perp}(\vec{u}) = (5, 5, 5)} ΠP⊥​(u)=(5,5,5) u⃗⋅n⃗∥n⃗∥2=153=5\frac{\vec{u} \cdot \vec{n}}{\|\vec{n}\|^2} = \frac{15}{3} = 5∥n∥2u⋅n​=315​=5 . Check: (−1,0,1)+(5,5,5)=(4,5,6)(-1, 0, 1) + (5, 5, 5) = (4, 5, 6)(−1,0,1)+(5,5,5)=(4,5,6) .

39. [Projections] Πℓ⊥(u⃗)=u⃗−u⃗⋅v⃗∥v⃗∥2 v⃗\Pi_{\ell^\perp}(\vec{u}) = \vec{u} - \frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2}\,\vec{v}Πℓ⊥​(u)=u−∥v∥2u⋅v​v projects onto the plane perpendicular to the line ℓ\ellℓ through the origin with direction v⃗\vec{v}v .Why is the closest distance between the line {po+t v⃗}\{p_{\text{o}} + t\,\vec{v}\}{po​+tv} and the origin equal to ∥Πℓ⊥(po)∥\|\Pi_{\ell^\perp}(p_{\text{o}})\|∥Πℓ⊥​(po​)∥ , with ℓ={t v⃗}\ell = \{t\,\vec{v}\}ℓ={tv} ?

    The distance equals the distance between the point pop_{\text{o}}po​ and the line ℓ\ellℓ , and the shortest vector from ℓ\ellℓ to pop_{\text{o}}po​ is Πℓ⊥(po)\Pi_{\ell^\perp}(p_{\text{o}})Πℓ⊥​(po​) . Shifting by −po-p_{\text{o}}−po​ moves the line to ℓ\ellℓ and OOO to −po-p_{\text{o}}−po​ , which has the same distance to ℓ\ellℓ as pop_{\text{o}}po​ . The part of pop_{\text{o}}po​ along ℓ\ellℓ is removed; the perpendicular part is the shortest connection.So d=∥po−po⋅v⃗∥v⃗∥2 v⃗∥d = \left\| p_{\te…

40. [Projections] Why is the closest distance between the plane n⃗⋅[(x,y,z)−po]=0\vec{n} \cdot [(x, y, z) - p_{\text{o}}] = 0n⋅[(x,y,z)−po​]=0 and the origin equal to ∣n⃗⋅po∣∥n⃗∥\dfrac{|\vec{n} \cdot p_{\text{o}}|}{\|\vec{n}\|}∥n∥∣n⋅po​∣​ ?

    It is the length of the projection of pop_{\text{o}}po​ onto the normal direction n⃗\vec{n}n : ∥po⋅n⃗∥n⃗∥2 n⃗∥=∣n⃗⋅po∣∥n⃗∥\ \left\| \frac{p_{\text{o}} \cdot \vec{n}}{\|\vec{n}\|^2}\,\vec{n} \right\| = \frac{|\vec{n} \cdot p_{\text{o}}|}{\|\vec{n}\|} ∥∥​∥n∥2po​⋅n​n∥∥​=∥n∥∣n⋅po​∣​ . It equals the distance between the parallel plane P′:n⃗⋅(x,y,z)=0P' : \vec{n} \cdot (x, y, z) = 0P′:n⋅(x,y,z)=0 through the origin and the point pop_{\text{o}}po​ , which is measured along P′⊥P'^\perpP′⊥ , t…

41. [Projections] Why can every projection Π\PiΠ be computed as a matrix-vector product MΠu⃗M_\Pi \vec{u}MΠ​u ? What is the matrix of the projection onto the xyxyxy -plane in R3\mathbb{R}^3R3 ?

    Projections are linear transformations; onto the xyxyxy -plane: MΠ=[100010000]\boldsymbol{M_\Pi = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}}MΠ​=⎣⎡​100​010​000​⎦⎤​ MΠM_\PiMΠ​ keeps the xxx - and yyy -components and sets the zzz -component to 0.Example: MΠ(4,5,6)T=(4,5,0)TM_\Pi (4, 5, 6)^{\mathsf{T}} = (4, 5, 0)^{\mathsf{T}}MΠ​(4,5,6)T=(4,5,0)T .

42. [Projections] What is the coordinate projection Πx(v⃗)\Pi_x(\vec{v})Πx​(v) , and how is it computed?

    The projection onto the xxx -axis: Πx(v⃗)=v⃗⋅ı^∥ı^∥2 ı^=(v⃗⋅ı^) ı^=vxı^\ \boldsymbol{\Pi_x(\vec{v}) = \frac{\vec{v} \cdot \hat{\imath}}{\|\hat{\imath}\|^2}\,\hat{\imath} = (\vec{v} \cdot \hat{\imath})\,\hat{\imath} = v_x \hat{\imath}} Πx​(v)=∥^∥2v⋅^​^=(v⋅^)^=vx​^ Likewise Πy(v⃗)=(v⃗⋅ȷ^)ȷ^\Pi_y(\vec{v}) = (\vec{v} \cdot \hat{\jmath})\hat{\jmath}Πy​(v)=(v⋅^​)^​ and Πz(v⃗)=(v⃗⋅k^)k^\Pi_z(\vec{v}) = (\vec{v} \cdot \hat{k})\hat{k}Πz​(v)=(v⋅k^)k^ . Coordinate projections give the components o…

43. [Coordinate projections] Why does writing a vector as components (vx,vy,vz)(v_x, v_y, v_z)(vx​,vy​,vz​) not determine it on its own?

    Components are always taken with respect to a basis, which can be written as a subscript: v⃗=(vx,vy,vz)ı^ȷ^k^\vec{v} = (v_x, v_y, v_z)_{\hat{\imath}\hat{\jmath}\hat{k}}v=(vx​,vy​,vz​)^^​k^​ . The same vector has different components in a different basis. Arithmetic like u⃗+v⃗\vec{u} + \vec{v}u+v works on components only if both use the same basis.

44. [Coordinate projections] A force F⃗\vec{F}F in the plane makes the angle θ\thetaθ with the xxx -axis. What are its components FxF_xFx​ , FyF_yFy​ in the basis {ı^,ȷ^}\{\hat{\imath}, \hat{\jmath}\}{^,^​} ?

    Fx=∥F⃗∥cos⁡θ\boldsymbol{F_x = \|\vec{F}\| \cos\theta}Fx​=∥F∥cosθ , Fy=∥F⃗∥sin⁡θ\ \boldsymbol{F_y = \|\vec{F}\| \sin\theta} Fy​=∥F∥sinθ FxF_xFx​ is the length of the projection of F⃗\vec{F}F onto the xxx -axis: Fxı^=F⃗⋅ı^∥ı^∥2ı^=[1000][FxFy]F_x\hat{\imath} = \frac{\vec{F} \cdot \hat{\imath}}{\|\hat{\imath}\|^2}\hat{\imath} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} F_x \\ F_y \end{bmatrix}Fx​^=∥^∥2F⋅^​^=[10​00​][Fx​Fy​​] .Example: ∥F⃗∥=2\|\vec{F}\| = 2∥F∥=2 , θ=60°\theta…

45. [Coordinate projections] A vector's components are also called by two other names. Which, and in which context?

    Coordinates (in the context of a coordinate system) and coefficients (in the context of a linear combination). All three mean the numbers viv_ivi​ in v⃗=v1e⃗1+⋯+vne⃗n\vec{v} = v_1\vec{e}_1 + \dots + v_n\vec{e}_nv=v1​e1​+⋯+vn​en​ .

46. [Coordinate projections] What two properties define a basis B={e⃗1,…,e⃗n}B = \{\vec{e}_1, \dots, \vec{e}_n\}B={e1​,…,en​} of a vector space VVV ?

    Spanning property Every v⃗∈V\vec{v} \in Vv∈V is a linear combination v1e⃗1+⋯+vne⃗nv_1\vec{e}_1 + \dots + v_n\vec{e}_nv1​e1​+⋯+vn​en​ ; the basis is sufficient to represent every vector. Example: (1,0)(1, 0)(1,0) and (1,1)(1, 1)(1,1) span R2\mathbb{R}^2R2 : (a,b)=(a−b)(1,0)+b(1,1)(a, b) = (a - b)(1, 0) + b(1, 1)(a,b)=(a−b)(1,0)+b(1,1) . Linear independence property The vectors e⃗1,…,e⃗n\vec{e}_1, \dots, \vec{e}_ne1​,…,en​ are linearly independent; none of them is redundant. Example: {(1,0),(1,1)…

47. [Coordinate projections] What are the three types of bases of an nnn -dimensional vector space VVV ?

    Generic basis Bf={f⃗1,…,f⃗n}B_f = \{\vec{f}_1, \dots, \vec{f}_n\}Bf​={f​1​,…,f​n​} Any nnn linearly independent vectors in VVV . Example: {(1,1),(3,0)}\{(1, 1), (3, 0)\}{(1,1),(3,0)} in R2\mathbb{R}^2R2 . Orthogonal basis Be={e⃗1,…,e⃗n}B_e = \{\vec{e}_1, \dots, \vec{e}_n\}Be​={e1​,…,en​} Mutually orthogonal vectors: e⃗i⋅e⃗j=0\vec{e}_i \cdot \vec{e}_j = 0ei​⋅ej​=0 for all i≠ji \neq ji=j . Example: {(1,1),(1,−1)}\{(1, 1), (1, -1)\}{(1,1),(1,−1)} in R2\mathbb{R}^2R2 . Orthonormal basis Be^={e^1,……

48. [Coordinate projections] How do you compute the coordinates (a1,…,an)(a_1, \dots, a_n)(a1​,…,an​) of a⃗\vec{a}a with respect to an orthonormal basis {e^1,…,e^n}\{\hat{e}_1, \dots, \hat{e}_n\}{e^1​,…,e^n​} ?

    ai=a⃗⋅e^i\boldsymbol{a_i = \vec{a} \cdot \hat{e}_i}ai​=a⋅e^i​ So a⃗=(a⃗⋅e^1)e^1+⋯+(a⃗⋅e^n)e^n\vec{a} = (\vec{a} \cdot \hat{e}_1)\hat{e}_1 + \dots + (\vec{a} \cdot \hat{e}_n)\hat{e}_na=(a⋅e^1​)e^1​+⋯+(a⋅e^n​)e^n​ . For {ı^,ȷ^,k^}\{\hat{\imath}, \hat{\jmath}, \hat{k}\}{^,^​,k^} : vx=v⃗⋅ı^v_x = \vec{v} \cdot \hat{\imath}vx​=v⋅^ , etc.Example: a⃗=(3,1)\vec{a} = (3, 1)a=(3,1) , e^1=12(1,1)\hat{e}_1 = \frac{1}{\sqrt{2}}(1, 1)e^1​=2​1​(1,1) , e^2=12(1,−1)\hat{e}_2 = \frac{1}{\sqrt{2}}(1, -1)e^2​=2​…

49. [Coordinate projections] How do you compute the coordinates (b1,…,bn)(b_1, \dots, b_n)(b1​,…,bn​) of b⃗\vec{b}b with respect to an orthogonal (not orthonormal) basis {e⃗1,…,e⃗n}\{\vec{e}_1, \dots, \vec{e}_n\}{e1​,…,en​} ?

    bi=b⃗⋅e⃗i∥e⃗i∥2\boldsymbol{b_i = \dfrac{\vec{b} \cdot \vec{e}_i}{\|\vec{e}_i\|^2}}bi​=∥ei​∥2b⋅ei​​ bie⃗ib_i\vec{e}_ibi​ei​ is the projection of b⃗\vec{b}b onto the line with direction e⃗i\vec{e}_iei​ .Example: b⃗=(3,1)\vec{b} = (3, 1)b=(3,1) , e⃗1=(1,1)\vec{e}_1 = (1, 1)e1​=(1,1) , e⃗2=(1,−1)\vec{e}_2 = (1, -1)e2​=(1,−1) : b1=42=2b_1 = \frac{4}{2} = 2b1​=24​=2 , b2=22=1b_2 = \frac{2}{2} = 1b2​=22​=1 ; indeed 2(1,1)+1(1,−1)=(3,1)2(1, 1) + 1(1, -1) = (3, 1)2(1,1)+1(1,−1)=(3,1) .

50. [Coordinate projections] In an orthogonal basis {e⃗1,…,e⃗n}\{\vec{e}_1, \dots, \vec{e}_n\}{e1​,…,en​} , why can each coefficient bib_ibi​ of b⃗\vec{b}b be computed independently of the others?

    Taking the dot product of b⃗=b1e⃗1+⋯+bne⃗n\vec{b} = b_1\vec{e}_1 + \dots + b_n\vec{e}_nb=b1​e1​+⋯+bn​en​ with e⃗i\vec{e}_iei​ gives b⃗⋅e⃗i=bi∥e⃗i∥2\vec{b} \cdot \vec{e}_i = b_i\|\vec{e}_i\|^2b⋅ei​=bi​∥ei​∥2 , since e⃗j⋅e⃗i=0\vec{e}_j \cdot \vec{e}_i = 0ej​⋅ei​=0 for j≠ij \neq ij=i . So bib_ibi​ needs only b⃗\vec{b}b and e⃗i\vec{e}_iei​ : nnn independent orthogonal projections.

51. [Coordinate projections] How do you find the coordinates (c1,…,cn)(c_1, \dots, c_n)(c1​,…,cn​) of c⃗\vec{c}c with respect to a generic basis {f⃗1,…,f⃗n}\{\vec{f}_1, \dots, \vec{f}_n\}{f​1​,…,f​n​} , and why is this more work than for an orthogonal basis?

    Solve the system c1f⃗1+⋯+cnf⃗n=c⃗c_1\vec{f}_1 + \dots + c_n\vec{f}_n = \vec{c}c1​f​1​+⋯+cn​f​n​=c of nnn equations in nnn unknowns. The f⃗i\vec{f}_if​i​ are not orthogonal, so the coefficients must be found simultaneously. In orthogonal and orthonormal bases each coefficient is a single projection.

52. [Coordinate projections] Express v⃗=(5,6)ı^ȷ^\vec{v} = (5, 6)_{\hat{\imath}\hat{\jmath}}v=(5,6)^^​​ in the basis Bf={f⃗1,f⃗2}B_f = \{\vec{f}_1, \vec{f}_2\}Bf​={f​1​,f​2​} with f⃗1=(1,1)ı^ȷ^\vec{f}_1 = (1, 1)_{\hat{\imath}\hat{\jmath}}f​1​=(1,1)^^​​ and f⃗2=(3,0)ı^ȷ^\vec{f}_2 = (3, 0)_{\hat{\imath}\hat{\jmath}}f​2​=(3,0)^^​​ .

    v⃗=(6,−13)Bf\boldsymbol{\vec{v} = (6, -\tfrac{1}{3})_{B_f}}v=(6,−31​)Bf​​ v1f⃗1+v2f⃗2=(5,6)v_1\vec{f}_1 + v_2\vec{f}_2 = (5, 6)v1​f​1​+v2​f​2​=(5,6) gives v1+3v2=5v_1 + 3v_2 = 5v1​+3v2​=5 and v1=6v_1 = 6v1​=6 , so v2=−13v_2 = -\frac{1}{3}v2​=−31​ .Check: 6(1,1)−13(3,0)=(5,6)6(1, 1) - \frac{1}{3}(3, 0) = (5, 6)6(1,1)−31​(3,0)=(5,6) .

53. [Coordinate projections] What does the change-of-basis matrix B′[1 ⁣ ⁣1]B\,{}_{B'}[\mathrm{1\!\!1}]_BB′​[11]B​ do?

    It converts the coordinates of a vector in the basis BBB into its coordinates in the basis B′B'B′ : [v⃗]B′=B′[1 ⁣ ⁣1]B [v⃗]B\ [\vec{v}]_{B'} = {}_{B'}[\mathrm{1\!\!1}]_B \, [\vec{v}]_B [v]B′​=B′​[11]B​[v]B​ [v⃗]B=(v1,…,vn)B[\vec{v}]_B = (v_1, \dots, v_n)_B[v]B​=(v1​,…,vn​)B​ is v⃗\vec{v}v expressed in BBB ; the vector itself does not change, only its representation.

54. [Coordinate projections] Why is the change-of-basis matrix written B′[1 ⁣ ⁣1]B\,{}_{B'}[\mathrm{1\!\!1}]_BB′​[11]B​ , with the identity symbol 1 ⁣ ⁣1\mathrm{1\!\!1}11 ?

    It acts as the identity transformation: the vector v⃗\vec{v}v stays the same, only the basis in which it is expressed changes. (v1′,v2′,v3′)B′=v1′e^1′+v2′e^2′+v3′e^3′=v⃗=v1e^1+v2e^2+v3e^3=(v1,v2,v3)B(v_1', v_2', v_3')_{B'} = v_1'\hat{e}_1' + v_2'\hat{e}_2' + v_3'\hat{e}_3' = \vec{v} = v_1\hat{e}_1 + v_2\hat{e}_2 + v_3\hat{e}_3 = (v_1, v_2, v_3)_B(v1′​,v2′​,v3′​)B′​=v1′​e^1′​+v2′​e^2′​+v3′​e^3′​=v=v1​e^1​+v2​e^2​+v3​e^3​=(v1​,v2​,v3​)B​

55. [Coordinate projections] B={e^1,…,e^n}B = \{\hat{e}_1, \dots, \hat{e}_n\}B={e^1​,…,e^n​} and B′={e^1′,…,e^n′}B' = \{\hat{e}_1', \dots, \hat{e}_n'\}B′={e^1′​,…,e^n′​} are orthonormal bases. What are the entries of B′[1 ⁣ ⁣1]B\,{}_{B'}[\mathrm{1\!\!1}]_BB′​[11]B​ , and what are its columns?

    Entry (i,j)(i, j)(i,j) is e^i′⋅e^j\hat{e}_i' \cdot \hat{e}_je^i′​⋅e^j​ ; column jjj is the old basis vector e^j\hat{e}_je^j​ expressed in the new basis B′B'B′ . e^1=(e^1′⋅e^1)e^1′+(e^2′⋅e^1)e^2′+(e^3′⋅e^1)e^3′\hat{e}_1 = (\hat{e}_1' \cdot \hat{e}_1)\hat{e}_1' + (\hat{e}_2' \cdot \hat{e}_1)\hat{e}_2' + (\hat{e}_3' \cdot \hat{e}_1)\hat{e}_3'e^1​=(e^1′​⋅e^1​)e^1′​+(e^2′​⋅e^1​)e^2′​+(e^3′​⋅e^1​)e^3′​ , the orthonormal coordinate formula.Example: B={ı^,ȷ^}B =…

56. [Coordinate projections] B={e⃗1,…,e⃗n}B = \{\vec{e}_1, \dots, \vec{e}_n\}B={e1​,…,en​} and B′={e⃗1′,…,e⃗n′}B' = \{\vec{e}_1', \dots, \vec{e}_n'\}B′={e1′​,…,en′​} are orthogonal but not necessarily orthonormal. What is entry (i,j)(i, j)(i,j) of B′[1 ⁣ ⁣1]B\,{}_{B'}[\mathrm{1\!\!1}]_BB′​[11]B​ ?

    (B′[1 ⁣ ⁣1]B)ij=e⃗i′⋅e⃗j∥e⃗i′∥2\boldsymbol{\left({}_{B'}[\mathrm{1\!\!1}]_B\right)_{ij} = \dfrac{\vec{e}_i' \cdot \vec{e}_j}{\|\vec{e}_i'\|^2}}(B′​[11]B​)ij​=∥ei′​∥2ei′​⋅ej​​ Column jjj holds the B′B'B′ -coordinates of e⃗j\vec{e}_jej​ , and in the orthogonal basis B′B'B′ the iii -th coordinate of a vector x⃗\vec{x}x is x⃗⋅e⃗i′∥e⃗i′∥2\frac{\vec{x} \cdot \vec{e}_i'}{\|\vec{e}_i'\|^2}∥ei′​∥2x⋅ei′​​ . For non-orthogonal bases the computation is more complicated.Ex…

57. [Vector spaces] What is a vector space VVV (in Savov's description)?

    A set of vectors together with all linear combinations of these vectors. So if v⃗1,v⃗2∈V\vec{v}_1, \vec{v}_2 \in Vv1​,v2​∈V , then αv⃗1+βv⃗2∈V\alpha\vec{v}_1 + \beta\vec{v}_2 \in Vαv1​+βv2​∈V for all α,β∈R\alpha, \beta \in \mathbb{R}α,β∈R .Example: R3\mathbb{R}^3R3 , the set of all three-dimensional vectors, is a vector space.

58. [Vector spaces] Which two closure properties follow from a vector space VVV containing all linear combinations of its vectors?

    Closed under addition ∀v⃗1,v⃗2∈V, v⃗1+v⃗2∈V\forall \vec{v}_1, \vec{v}_2 \in V, \ \vec{v}_1 + \vec{v}_2 \in V∀v1​,v2​∈V, v1​+v2​∈V Example: (1,0,2),(0,3,1)∈R3(1, 0, 2), (0, 3, 1) \in \mathbb{R}^3(1,0,2),(0,3,1)∈R3 , and their sum (1,3,3)∈R3(1, 3, 3) \in \mathbb{R}^3(1,3,3)∈R3 . Closed under scalar multiplication ∀α∈R, ∀v⃗∈V, αv⃗∈V\forall \alpha \in \mathbb{R}, \ \forall \vec{v} \in V, \ \alpha\vec{v} \in V∀α∈R, ∀v∈V, αv∈V Example: (1,0,2)∈R3(1, 0, 2) \in \mathbb{R}^3(1,0,2)∈R3 , and −2(1,0,2)=(−…

59. [Vector spaces] What is the span of vectors v⃗1,…,v⃗n∈V\vec{v}_1, \dots, \vec{v}_n \in Vv1​,…,vn​∈V ?

    The set of all linear combinations of them: span⁡{v⃗1,…,v⃗n}≡{v⃗∈V∣v⃗=α1v⃗1+⋯+αnv⃗n, αi∈R}\operatorname{span}\{\vec{v}_1, \dots, \vec{v}_n\} \equiv \{\vec{v} \in V \mid \vec{v} = \alpha_1\vec{v}_1 + \dots + \alpha_n\vec{v}_n, \ \alpha_i \in \mathbb{R}\}span{v1​,…,vn​}≡{v∈V∣v=α1​v1​+⋯+αn​vn​, αi​∈R} Example: span⁡{(1,1)}\operatorname{span}\{(1, 1)\}span{(1,1)} is the line y=xy = xy=x in R2\mathbb{R}^2R2 .

60. [Vector spaces] What shape is span⁡{v⃗1}\operatorname{span}\{\vec{v}_1\}span{v1​} for a nonzero v⃗1∈R3\vec{v}_1 \in \mathbb{R}^3v1​∈R3 , and span⁡{v⃗1,v⃗2}\operatorname{span}\{\vec{v}_1, \vec{v}_2\}span{v1​,v2​} for v⃗1,v⃗2\vec{v}_1, \vec{v}_2v1​,v2​ not multiples of each other?

    span⁡{v⃗1}\operatorname{span}\{\vec{v}_1\}span{v1​} : an infinite line through the origin. span⁡{v⃗1,v⃗2}\operatorname{span}\{\vec{v}_1, \vec{v}_2\}span{v1​,v2​} : an infinite plane through the origin. The line is all multiples αv⃗1\alpha\vec{v}_1αv1​ ; the plane is all αv⃗1+βv⃗2\alpha\vec{v}_1 + \beta\vec{v}_2αv1​+βv2​ .Example: span⁡{(1,0,0),(0,1,0)}\operatorname{span}\{(1, 0, 0), (0, 1, 0)\}span{(1,0,0),(0,1,0)} is the xyxyxy -plane.

61. [Vector spaces] v⃗3=v⃗1+v⃗2\vec{v}_3 = \vec{v}_1 + \vec{v}_2v3​=v1​+v2​ . Why is span⁡{v⃗1,v⃗2,v⃗3}=span⁡{v⃗1,v⃗2}\operatorname{span}\{\vec{v}_1, \vec{v}_2, \vec{v}_3\} = \operatorname{span}\{\vec{v}_1, \vec{v}_2\}span{v1​,v2​,v3​}=span{v1​,v2​} ?

    v⃗3\vec{v}_3v3​ is already a linear combination of v⃗1\vec{v}_1v1​ and v⃗2\vec{v}_2v2​ , so adding it creates no new linear combinations. Any av⃗1+bv⃗2+cv⃗3=(a+c)v⃗1+(b+c)v⃗2a\vec{v}_1 + b\vec{v}_2 + c\vec{v}_3 = (a + c)\vec{v}_1 + (b + c)\vec{v}_2av1​+bv2​+cv3​=(a+c)v1​+(b+c)v2​ . Geometrically, v⃗3\vec{v}_3v3​ lies in the plane of v⃗1\vec{v}_1v1​ and v⃗2\vec{v}_2v2​ .

62. [Vector spaces] v⃗1≠0⃗\vec{v}_1 \neq \vec{0}v1​=0 and v⃗2′=γv⃗1\vec{v}_2' = \gamma\vec{v}_1v2′​=γv1​ for some γ∈R\gamma \in \mathbb{R}γ∈R . What is the dimension of span⁡{v⃗1,v⃗2′}\operatorname{span}\{\vec{v}_1, \vec{v}_2'\}span{v1​,v2′​} ?

    1: span⁡{v⃗1,v⃗2′}=span⁡{v⃗1}\operatorname{span}\{\vec{v}_1, \vec{v}_2'\} = \operatorname{span}\{\vec{v}_1\}span{v1​,v2′​}=span{v1​} , a line. αv⃗1+βv⃗2′=(α+βγ)v⃗1\alpha\vec{v}_1 + \beta\vec{v}_2' = (\alpha + \beta\gamma)\vec{v}_1αv1​+βv2′​=(α+βγ)v1​ is always a multiple of v⃗1\vec{v}_1v1​ .

63. [Vector spaces] What three conditions make WWW a subspace of the vector space VVV (written W⊆VW \subseteq VW⊆V )?

    WWW is contained in VVV If w⃗∈W\vec{w} \in Ww∈W , then w⃗∈V\vec{w} \in Vw∈V . Example: the xyxyxy -plane {(x,y,0)}\{(x, y, 0)\}{(x,y,0)} consists of vectors of R3\mathbb{R}^3R3 . Closed under addition For all w⃗1,w⃗2∈W\vec{w}_1, \vec{w}_2 \in Ww1​,w2​∈W : w⃗1+w⃗2∈W\vec{w}_1 + \vec{w}_2 \in Ww1​+w2​∈W . Example: (1,2,0)+(3,−1,0)=(4,1,0)(1, 2, 0) + (3, -1, 0) = (4, 1, 0)(1,2,0)+(3,−1,0)=(4,1,0) is again in the xyxyxy -plane. Closed under scalar multiplication For all w⃗∈W\vec{w} \in Ww∈W and α∈R\…

64. [Vector spaces] Why does every subspace WWW contain the zero vector?

    WWW is closed under scalar multiplication, and 0w⃗=0⃗0\vec{w} = \vec{0}0w=0 for any w⃗∈W\vec{w} \in Ww∈W . Example: the line {(t,2t,1)∣t∈R}\{(t, 2t, 1) \mid t \in \mathbb{R}\}{(t,2t,1)∣t∈R} does not contain 0⃗\vec{0}0 , so it is not a subspace of R3\mathbb{R}^3R3 .

65. [Vector spaces] Which subsets of R3\mathbb{R}^3R3 are its vector subspaces (besides {0⃗}\{\vec{0}\}{0} and R3\mathbb{R}^3R3 itself)?

    Lines and planes that pass through the origin. A line or plane not through the origin does not contain 0⃗\vec{0}0 .

66. [Vector spaces] How can the xyxyxy -plane Pxy⊆R3P_{xy} \subseteq \mathbb{R}^3Pxy​⊆R3 be specified by a constraint?

    Pxy={(x,y,z)∈R3∣(0,0,1)⋅(x,y,z)=0}\boldsymbol{P_{xy} = \{(x, y, z) \in \mathbb{R}^3 \mid (0, 0, 1) \cdot (x, y, z) = 0\}}Pxy​={(x,y,z)∈R3∣(0,0,1)⋅(x,y,z)=0} The set of all vectors perpendicular to k^=(0,0,1)\hat{k} = (0, 0, 1)k^=(0,0,1) .Example: (3,−2,0)⋅(0,0,1)=0(3, -2, 0) \cdot (0, 0, 1) = 0(3,−2,0)⋅(0,0,1)=0 , so (3,−2,0)∈Pxy(3, -2, 0) \in P_{xy}(3,−2,0)∈Pxy​ ; (0,0,1)⋅(1,1,1)=1(0, 0, 1) \cdot (1, 1, 1) = 1(0,0,1)⋅(1,1,1)=1 , so (1,1,1)∉Pxy(1, 1, 1) \notin P_{xy}(1,1,1)∈/Pxy​ .

67. [Vector spaces] How can the xyxyxy -plane Pxy⊆R3P_{xy} \subseteq \mathbb{R}^3Pxy​⊆R3 be specified as a span, and what are the parameters of this description?

    Pxy=span⁡{(1,0,0),(0,1,0)}\boldsymbol{P_{xy} = \operatorname{span}\{(1, 0, 0), (0, 1, 0)\}}Pxy​=span{(1,0,0),(0,1,0)} , with parameters α,β\alpha, \betaα,β in v⃗=α(1,0,0)+β(0,1,0)\vec{v} = \alpha(1, 0, 0) + \beta(0, 1, 0)v=α(1,0,0)+β(0,1,0) . Each point of the plane corresponds to exactly one pair (α,β)(\alpha, \beta)(α,β) . A parametrization of an mmm -dimensional vector space needs mmm parameters.Example: (4,−1,0)(4, -1, 0)(4,−1,0) has α=4\alpha = 4α=4 , β=−1\beta = -1β=−1 .

68. [Vector spaces] What is the difference between a subset and a subspace of a vector space VVV ?

    A subset is any set {v⃗∈V∣conditions}\{\vec{v} \in V \mid \text{conditions}\}{v∈V∣conditions} ; a subspace is a subset that is also closed under addition and scalar multiplication. Calling a set a subspace says that adding and scaling its elements gives elements of the same set. Both relations are written ⊆\subseteq⊆ .Example: the unit circle in R2\mathbb{R}^2R2 is a subset but not a subspace: 2(1,0)=(2,0)2(1, 0) = (2, 0)2(1,0)=(2,0) is not on the circle.

69. [Vector spaces] A system of linear equations Ax⃗=0⃗A\vec{x} = \vec{0}Ax=0 , whose right-hand side constants are all zero, is called homogeneous.

    Ax⃗=b⃗A\vec{x} = \vec{b}Ax=b with b⃗≠0⃗\vec{b} \neq \vec{0}b=0 is non-homogeneous. (Savov spells it "homogenous".)

70. [Vector spaces] Ax⃗=b⃗A\vec{x} = \vec{b}Ax=b has a particular solution c⃗\vec{c}c , and v⃗n\vec{v}_nvn​ spans the null space of AAA (the solutions of Ax⃗=0⃗A\vec{x} = \vec{0}Ax=0 ). What is the full solution set of Ax⃗=b⃗A\vec{x} = \vec{b}Ax=b ?

    {c⃗+sv⃗n∣s∈R}\boldsymbol{\{\vec{c} + s\vec{v}_n \mid s \in \mathbb{R}\}}{c+svn​∣s∈R} The solution set of the homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0}Ax=0 is {sv⃗n∣s∈R}=span⁡{v⃗n}\{s\vec{v}_n \mid s \in \mathbb{R}\} = \operatorname{span}\{\vec{v}_n\}{svn​∣s∈R}=span{vn​} .

71. [Vector spaces] Is the solution set of Ax⃗=0⃗A\vec{x} = \vec{0}Ax=0 a vector space? Is the solution set of Ax⃗=b⃗A\vec{x} = \vec{b}Ax=b with b⃗≠0⃗\vec{b} \neq \vec{0}b=0 ?

    Ax⃗=0⃗A\vec{x} = \vec{0}Ax=0 : yes (the null space of AAA ). Ax⃗=b⃗A\vec{x} = \vec{b}Ax=b : no. If Ax⃗1=Ax⃗2=0⃗A\vec{x}_1 = A\vec{x}_2 = \vec{0}Ax1​=Ax2​=0 , then A(αx⃗1+βx⃗2)=0⃗A(\alpha\vec{x}_1 + \beta\vec{x}_2) = \vec{0}A(αx1​+βx2​)=0 . But if Ax⃗1=Ax⃗2=b⃗A\vec{x}_1 = A\vec{x}_2 = \vec{b}Ax1​=Ax2​=b , then A(x⃗1+x⃗2)=2b⃗≠b⃗A(\vec{x}_1 + \vec{x}_2) = 2\vec{b} \neq \vec{b}A(x1​+x2​)=2b=b , and 0⃗\vec{0}0 is not a solution.

72. [Vector spaces] What is the column space C(M)\mathcal{C}(M)C(M) of a matrix M∈Rm×nM \in \mathbb{R}^{m \times n}M∈Rm×n ?

    The span of the columns of MMM , i.e. all outputs Mv⃗M\vec{v}Mv : C(M)≡{w⃗∈Rm∣w⃗=Mv⃗ for some v⃗∈Rn}⊆Rm\mathcal{C}(M) \equiv \{\vec{w} \in \mathbb{R}^m \mid \vec{w} = M\vec{v} \text{ for some } \vec{v} \in \mathbb{R}^n\} \subseteq \mathbb{R}^mC(M)≡{w∈Rm∣w=Mv for some v∈Rn}⊆Rm Example: M=[1224]M = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}M=[12​24​] has C(M)=span⁡{(1,2)}\mathcal{C}(M) = \operatorname{span}\{(1, 2)\}C(M)=span{(1,2)} .

73. [Vector spaces] What is the null space N(M)\mathcal{N}(M)N(M) of a matrix M∈Rm×nM \in \mathbb{R}^{m \times n}M∈Rm×n ?

    All vectors v⃗∈Rn\vec{v} \in \mathbb{R}^nv∈Rn with Mv⃗=0⃗M\vec{v} = \vec{0}Mv=0 : N(M)≡{v⃗∈Rn∣Mv⃗=0⃗}⊆Rn\mathcal{N}(M) \equiv \{\vec{v} \in \mathbb{R}^n \mid M\vec{v} = \vec{0}\} \subseteq \mathbb{R}^nN(M)≡{v∈Rn∣Mv=0}⊆Rn Example: M=[1224]M = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}M=[12​24​] has N(M)=span⁡{(2,−1)}\mathcal{N}(M) = \operatorname{span}\{(2, -1)\}N(M)=span{(2,−1)} .

74. [Vector spaces] The null space of a matrix is also called its kernel.

75. [Vector spaces] What is the row space R(M)\mathcal{R}(M)R(M) of a matrix M∈Rm×nM \in \mathbb{R}^{m \times n}M∈Rm×n ?

    The span of the rows of MMM , i.e. all products w⃗TM\vec{w}^{\mathsf{T}}MwTM : R(M)≡{v⃗∈Rn∣v⃗=w⃗TM for some w⃗∈Rm}⊆Rn\mathcal{R}(M) \equiv \{\vec{v} \in \mathbb{R}^n \mid \vec{v} = \vec{w}^{\mathsf{T}}M \text{ for some } \vec{w} \in \mathbb{R}^m\} \subseteq \mathbb{R}^nR(M)≡{v∈Rn∣v=wTM for some w∈Rm}⊆Rn The transpose makes w⃗\vec{w}w a row vector. Example: M=[1224]M = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}M=[12​24​] has R(M)=span⁡{(1,2)}\mathcal{R}(M) = \operatorname{span}\{(1, 2)\}R(M)=s…

76. [Vector spaces] What is the left null space N(MT)\mathcal{N}(M^{\mathsf{T}})N(MT) of a matrix M∈Rm×nM \in \mathbb{R}^{m \times n}M∈Rm×n ?

    All vectors w⃗∈Rm\vec{w} \in \mathbb{R}^mw∈Rm with w⃗TM=0⃗T\vec{w}^{\mathsf{T}}M = \vec{0}^{\mathsf{T}}wTM=0T : N(MT)≡{w⃗∈Rm∣w⃗TM=0⃗T}⊆Rm\mathcal{N}(M^{\mathsf{T}}) \equiv \{\vec{w} \in \mathbb{R}^m \mid \vec{w}^{\mathsf{T}}M = \vec{0}^{\mathsf{T}}\} \subseteq \mathbb{R}^mN(MT)≡{w∈Rm∣wTM=0T}⊆Rm Example: M=[1224]M = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}M=[12​24​] has N(MT)=span⁡{(2,−1)}\mathcal{N}(M^{\mathsf{T}}) = \operatorname{span}\{(2, -1)\}N(MT)=span{(2,−1)} : (2,−1)M=(0,0)(2, -1)M =…

77. [Vector spaces] Why is the left null space of MMM written N(MT)\mathcal{N}(M^{\mathsf{T}})N(MT) ?

    w⃗TM=0⃗T\vec{w}^{\mathsf{T}}M = \vec{0}^{\mathsf{T}}wTM=0T is equivalent to MTw⃗=0⃗M^{\mathsf{T}}\vec{w} = \vec{0}MTw=0 , so the left null space of MMM is the null space of MTM^{\mathsf{T}}MT . Transpose both sides: (w⃗TM)T=MTw⃗(\vec{w}^{\mathsf{T}}M)^{\mathsf{T}} = M^{\mathsf{T}}\vec{w}(wTM)T=MTw . In general, multiplying AAA from the left is multiplying ATA^{\mathsf{T}}AT from the right: a⃗TA=b⃗T⇔ATa⃗=b⃗\vec{a}^{\mathsf{T}}A = \vec{b}^{\mathsf{T}} \Leftrightarrow A^{\mathsf{T}}\vec{a} = \vec{…

78. [Vector spaces] Mx⃗=b⃗M\vec{x} = \vec{b}Mx=b has a particular solution c⃗\vec{c}c , and N(M)=span⁡{v⃗1,…,v⃗k}\mathcal{N}(M) = \operatorname{span}\{\vec{v}_1, \dots, \vec{v}_k\}N(M)=span{v1​,…,vk​} . What is the full solution set, and why is every vector in it a solution?

    x⃗=c⃗+span⁡{v⃗1,…,v⃗k}\boldsymbol{\vec{x} = \vec{c} + \operatorname{span}\{\vec{v}_1, \dots, \vec{v}_k\}}x=c+span{v1​,…,vk​} M(c⃗+α1v⃗1+⋯+αkv⃗k)=Mc⃗+α1Mv⃗1+⋯+αkMv⃗k=b⃗+0⃗=b⃗M(\vec{c} + \alpha_1\vec{v}_1 + \dots + \alpha_k\vec{v}_k) = M\vec{c} + \alpha_1 M\vec{v}_1 + \dots + \alpha_k M\vec{v}_k = \vec{b} + \vec{0} = \vec{b}M(c+α1​v1​+⋯+αk​vk​)=Mc+α1​Mv1​+⋯+αk​Mvk​=b+0=b .So if N(M)≠{0⃗}\mathcal{N}(M) \neq \{\vec{0}\}N(M)={0} and a solution exists, there are infinitely many solutions; if N(M)={0…

79. [Vector spaces] Which two linear transformations does one matrix M∈Rm×nM \in \mathbb{R}^{m \times n}M∈Rm×n define?

    TM:Rn→RmT_M : \mathbb{R}^n \to \mathbb{R}^mTM​:Rn→Rm , TM(x⃗)=Mx⃗T_M(\vec{x}) = M\vec{x}TM​(x)=Mx (multiplication from the right), and TMT:Rm→RnT_{M^{\mathsf{T}}} : \mathbb{R}^m \to \mathbb{R}^nTMT​:Rm→Rn , given by a⃗TM=b⃗T\vec{a}^{\mathsf{T}}M = \vec{b}^{\mathsf{T}}aTM=bT (multiplication from the left). TMTT_{M^{\mathsf{T}}}TMT​ is called the adjoint of TMT_MTM​ . It is written with MTM^{\mathsf{T}}MT because a⃗TM=b⃗T\vec{a}^{\mathsf{T}}M = \vec{b}^{\mathsf{T}}aTM=bT means MTa⃗=b⃗M^{\mathsf{T…

80. [Vector spaces] For M∈Rm×nM \in \mathbb{R}^{m \times n}M∈Rm×n , what are the left space and the right space of MMM ?

    Left space: Rm\boldsymbol{\mathbb{R}^m}Rm (vectors that can multiply MMM from the left). Right space: Rn\boldsymbol{\mathbb{R}^n}Rn (vectors that can multiply MMM from the right). R(M),N(M)⊆Rn\mathcal{R}(M), \mathcal{N}(M) \subseteq \mathbb{R}^nR(M),N(M)⊆Rn and C(M),N(MT)⊆Rm\mathcal{C}(M), \mathcal{N}(M^{\mathsf{T}}) \subseteq \mathbb{R}^mC(M),N(MT)⊆Rm .

81. [Vector spaces] M∈Rm×nM \in \mathbb{R}^{m \times n}M∈Rm×n . How does the right space Rn\mathbb{R}^nRn split into fundamental spaces of MMM ?

    R(M)⊕N(M)=Rn\boldsymbol{\mathcal{R}(M) \oplus \mathcal{N}(M) = \mathbb{R}^n}R(M)⊕N(M)=Rn Every v⃗∈Rn\vec{v} \in \mathbb{R}^nv∈Rn is a sum v⃗=v⃗r+v⃗n\vec{v} = \vec{v}_r + \vec{v}_nv=vr​+vn​ with v⃗r∈R(M)\vec{v}_r \in \mathcal{R}(M)vr​∈R(M) , v⃗n∈N(M)\vec{v}_n \in \mathcal{N}(M)vn​∈N(M) , and v⃗r⋅v⃗n=0\vec{v}_r \cdot \vec{v}_n = 0vr​⋅vn​=0 ( ⊕\oplus⊕ is the orthogonal sum). Hence dim⁡R(M)+dim⁡N(M)=n\dim\mathcal{R}(M) + \dim\mathcal{N}(M) = ndimR(M)+dimN(M)=n .Example: M=[11]M = \begin{bmatrix} 1…

82. [Vector spaces] M∈Rm×nM \in \mathbb{R}^{m \times n}M∈Rm×n . How does the left space Rm\mathbb{R}^mRm split into fundamental spaces of MMM ?

    C(M)⊕N(MT)=Rm\boldsymbol{\mathcal{C}(M) \oplus \mathcal{N}(M^{\mathsf{T}}) = \mathbb{R}^m}C(M)⊕N(MT)=Rm An orthogonal sum, so dim⁡C(M)+dim⁡N(MT)=m\dim\mathcal{C}(M) + \dim\mathcal{N}(M^{\mathsf{T}}) = mdimC(M)+dimN(MT)=m .Example: M=[1224]M = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}M=[12​24​] : C(M)=span⁡{(1,2)}\mathcal{C}(M) = \operatorname{span}\{(1, 2)\}C(M)=span{(1,2)} and N(MT)=span⁡{(2,−1)}\mathcal{N}(M^{\mathsf{T}}) = \operatorname{span}\{(2, -1)\}N(MT)=span{(2,−1)} , dimensions 1+1=…

83. [Vector spaces] What is the rank of a matrix MMM ?

    rank⁡(M)≡dim⁡R(M)=dim⁡C(M)\boldsymbol{\operatorname{rank}(M) \equiv \dim\mathcal{R}(M) = \dim\mathcal{C}(M)}rank(M)≡dimR(M)=dimC(M) The number of linearly independent rows of MMM , which equals the number of linearly independent columns.Example: [1224]\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}[12​24​] has rank 1: the second row is twice the first.

84. [Vector spaces] What is the nullity of a matrix MMM ?

    nullity⁡(M)≡dim⁡N(M)\boldsymbol{\operatorname{nullity}(M) \equiv \dim\mathcal{N}(M)}nullity(M)≡dimN(M) Example: [1224]\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}[12​24​] has N=span⁡{(2,−1)}\mathcal{N} = \operatorname{span}\{(2, -1)\}N=span{(2,−1)} , so nullity 1.

85. [Vector spaces] Rank–nullity theorem For any matrix M∈Rm×nM \in \mathbb{R}^{m \times n}M∈Rm×n : rank⁡(M)+nullity⁡(M)= ?\operatorname{rank}(M) + \operatorname{nullity}(M) = \; ?rank(M)+nullity(M)=?

    n\boldsymbol{n}n (the number of columns, dim⁡Rn\dim \mathbb{R}^ndimRn ) The right space Rn\mathbb{R}^nRn splits into row-space dimensions and null-space dimensions.Example: [1224]\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}[12​24​] : rank 1, nullity 1, 1+1=21 + 1 = 21+1=2 .

86. [Vector spaces] Rank–nullity theorem: for M∈Rm×nM \in \mathbb{R}^{m \times n}M∈Rm×n , rank⁡(M)+nullity⁡(M)=n\operatorname{rank}(M) + \operatorname{nullity}(M) = nrank(M)+nullity(M)=n .What is the key idea of the proof?

    A basis BrB_rBr​ of R(M)\mathcal{R}(M)R(M) ( kkk vectors) together with a basis BnB_nBn​ of N(M)\mathcal{N}(M)N(M) ( ℓ\ellℓ vectors) is a basis of Rn\mathbb{R}^nRn , so k+ℓ=nk + \ell = nk+ℓ=n . Independent: every vector of N(M)\mathcal{N}(M)N(M) is orthogonal to every row of MMM , so Br∪BnB_r \cup B_nBr​∪Bn​ is linearly independent, and k+ℓ>nk + \ell > nk+ℓ>n is impossible.Spanning: for v⃗∈Rn\vec{v} \in \mathbb{R}^nv∈Rn , let v⃗r\vec{v}_rvr​ be its projection onto R(M)\mathcal{R}(M)R(M) . Then…

87. [Vector spaces] Rank–nullity theorem: for M∈Rm×nM \in \mathbb{R}^{m \times n}M∈Rm×n , rank⁡(M)+nullity⁡(M)=n\operatorname{rank}(M) + \operatorname{nullity}(M) = nrank(M)+nullity(M)=n . A∈R3×6A \in \mathbb{R}^{3 \times 6}A∈R3×6 , and one row of AAA is a linear combination of the other two, which are independent. What are rank⁡(A)\operatorname{rank}(A)rank(A) and nullity⁡(A)\operatorname{nullity}(A)nullity(A) ?

    rank⁡(A)=2\operatorname{rank}(A) = 2rank(A)=2 , nullity⁡(A)=4\operatorname{nullity}(A) = 4nullity(A)=4 The row space is two-dimensional, and 6−2=46 - 2 = 46−2=4 .Likewise, if N(M)={0⃗}\mathcal{N}(M) = \{\vec{0}\}N(M)={0} for M∈Rm×nM \in \mathbb{R}^{m \times n}M∈Rm×n , then rank⁡(M)=n−0=n\operatorname{rank}(M) = n - 0 = nrank(M)=n−0=n .

88. [Vector spaces] M∈Rm×nM \in \mathbb{R}^{m \times n}M∈Rm×n . What does multiplication by MMM do to each of its four fundamental spaces?

    R(M)undefinedMC(M)\mathcal{R}(M) \xrightarrow{M} \mathcal{C}(M)R(M)M​C(M) and back, N(M)undefinedM0⃗∈Rm\ \mathcal{N}(M) \xrightarrow{M} \vec{0} \in \mathbb{R}^m N(M)M​0∈Rm , N(MT)undefinedM0⃗∈Rn\ \mathcal{N}(M^{\mathsf{T}}) \xrightarrow{M} \vec{0} \in \mathbb{R}^n N(MT)M​0∈Rn Right multiplication Mv⃗M\vec{v}Mv sends R(M)\mathcal{R}(M)R(M) one-to-one onto C(M)\mathcal{C}(M)C(M) and N(M)\mathcal{N}(M)N(M) to 0⃗\vec{0}0 . Left multiplication w⃗TM\vec{w}^{\mathsf{T}}MwTM sends C(M)\mathcal{C}(M)C(M…

89. [Vector spaces] When is a set of vectors {v⃗1,…,v⃗n}\{\vec{v}_1, \dots, \vec{v}_n\}{v1​,…,vn​} linearly independent?

    When the only solution of α1v⃗1+⋯+αnv⃗n=0⃗\alpha_1\vec{v}_1 + \dots + \alpha_n\vec{v}_n = \vec{0}α1​v1​+⋯+αn​vn​=0 is αi=0\alpha_i = 0αi​=0 for all iii . Then no v⃗i\vec{v}_ivi​ is a linear combination of the others.Example: (1,2),(2,4)(1, 2), (2, 4)(1,2),(2,4) are not independent: 2(1,2)−1(2,4)=0⃗2(1, 2) - 1(2, 4) = \vec{0}2(1,2)−1(2,4)=0 .

90. [Vector spaces] Vectors satisfy α1v⃗1+α2v⃗2+α3v⃗3=0⃗\alpha_1\vec{v}_1 + \alpha_2\vec{v}_2 + \alpha_3\vec{v}_3 = \vec{0}α1​v1​+α2​v2​+α3​v3​=0 with α1=−1,α2=1,α3=2\alpha_1 = -1, \alpha_2 = 1, \alpha_3 = 2α1​=−1,α2​=1,α3​=2 . What follows?

    v⃗1=v⃗2+2v⃗3\vec{v}_1 = \vec{v}_2 + 2\vec{v}_3v1​=v2​+2v3​ , so the vectors are linearly dependent. Any nonzero αi\alpha_iαi​ lets you solve for v⃗i\vec{v}_ivi​ in terms of the others.

91. [Vector spaces] Why is linear independence defined by "the only solution of α1v⃗1+⋯+αnv⃗n=0⃗\alpha_1\vec{v}_1 + \dots + \alpha_n\vec{v}_n = \vec{0}α1​v1​+⋯+αn​vn​=0 is all αi=0\alpha_i = 0αi​=0 ", rather than "no vector is a combination of the others"?

    The equation treats all vectors symmetrically: one condition, without singling out any v⃗i\vec{v}_ivi​ . It is equivalent: a solution with some αi≠0\alpha_i \neq 0αi​=0 expresses v⃗i\vec{v}_ivi​ as a combination of the others, and vice versa.

92. [Vector spaces] What is a basis of an nnn -dimensional vector space VVV ?

    A set of nnn linearly independent vectors in VVV that span VVV . A basis serves as a coordinate system for VVV : every v⃗∈V\vec{v} \in Vv∈V is v1e⃗1+⋯+vne⃗nv_1\vec{e}_1 + \dots + v_n\vec{e}_nv1​e1​+⋯+vn​en​ .Example: {(1,0),(1,1)}\{(1, 0), (1, 1)\}{(1,0),(1,1)} is a basis of R2\mathbb{R}^2R2 .

93. [Vector spaces] What is the dimension dim⁡(V)\dim(V)dim(V) of a vector space VVV , and why does every basis of VVV have the same size?

    The number of vectors in a basis of VVV . In an nnn -dimensional space, fewer than nnn vectors cannot span it, and more than nnn vectors cannot be linearly independent.Example: dim⁡R3=3\dim \mathbb{R}^3 = 3dimR3=3 ; dim⁡span⁡{(1,0,0),(0,1,0)}=2\dim\operatorname{span}\{(1, 0, 0), (0, 1, 0)\} = 2dimspan{(1,0,0),(0,1,0)}=2 .

94. [Vector spaces] A set of more than nnn vectors that spans an nnn -dimensional vector space is called over-complete.

    Its vectors cannot all be linearly independent, so some must be removed to get a basis. Example: {(1,0),(0,1),(1,1)}\{(1, 0), (0, 1), (1, 1)\}{(1,0),(0,1),(1,1)} spans R2\mathbb{R}^2R2 and is over-complete.

95. [Vector space techniques] V=span⁡{u⃗1,…,u⃗m}V = \operatorname{span}\{\vec{u}_1, \dots, \vec{u}_m\}V=span{u1​,…,um​} , where the u⃗i\vec{u}_iui​ need not be linearly independent. How do you distill a basis of VVV ?

    Write the u⃗i\vec{u}_iui​ as the rows of a matrix MMM , compute rref⁡(M)\operatorname{rref}(M)rref(M) ; its nonzero rows form a basis of VVV . The dependent vectors turn into rows of zeros.Example: rows (1,2),(2,4),(0,1)(1, 2), (2, 4), (0, 1)(1,2),(2,4),(0,1) have RREF rows (1,0),(0,1),(0,0)(1, 0), (0, 1), (0, 0)(1,0),(0,1),(0,0) , so the span is R2\mathbb{R}^2R2 with basis {(1,0),(0,1)}\{(1, 0), (0, 1)\}{(1,0),(0,1)} .

96. [Vector space techniques] When distilling a basis of span⁡{u⃗1,…,u⃗m}\operatorname{span}\{\vec{u}_1, \dots, \vec{u}_m\}span{u1​,…,um​} with Gauss–Jordan elimination, why do the nonzero rows of rref⁡(M)\operatorname{rref}(M)rref(M) span the same space as the u⃗i\vec{u}_iui​ (the rows of MMM )?

    Row operations do not change the row space: R(M)=R(rref⁡(M))\mathcal{R}(M) = \mathcal{R}(\operatorname{rref}(M))R(M)=R(rref(M)) . Each row operation replaces rows by linear combinations of rows, and can be undone. The nonzero RREF rows are independent: each has a leading 1 where the others have 0.

97. [Vector space techniques] V=span⁡{u⃗1,…,u⃗m}V = \operatorname{span}\{\vec{u}_1, \dots, \vec{u}_m\}V=span{u1​,…,um​} with dependent u⃗i\vec{u}_iui​ . Why describe VVV by a basis {e⃗1,…,e⃗n}\{\vec{e}_1, \dots, \vec{e}_n\}{e1​,…,en​} instead?

    Each v⃗∈V\vec{v} \in Vv∈V has exactly one list of coefficients (v1,…,vn)(v_1, \dots, v_n)(v1​,…,vn​) with v⃗=v1e⃗1+⋯+vne⃗n\vec{v} = v_1\vec{e}_1 + \dots + v_n\vec{e}_nv=v1​e1​+⋯+vn​en​ , and n=dim⁡Vn = \dim Vn=dimV is visible. With dependent u⃗i\vec{u}_iui​ , the same v⃗\vec{v}v is many different combinations v1′u⃗1+⋯+vm′u⃗mv_1'\vec{u}_1 + \dots + v_m'\vec{u}_mv1′​u1​+⋯+vm′​um​ .Example: with u⃗1=(1,0),u⃗2=(0,1),u⃗3=(1,1)\vec{u}_1 = (1, 0), \vec{u}_2 = (0, 1), \vec{u}_3 = (1, 1)u1​=(1…

98. [Vector space techniques] Pxy=span⁡{(1,0,0),(0,1,0),(1,1,0)}P_{xy} = \operatorname{span}\{(1, 0, 0), (0, 1, 0), (1, 1, 0)\}Pxy​=span{(1,0,0),(0,1,0),(1,1,0)} , so every p∈Pxyp \in P_{xy}p∈Pxy​ is a(1,0,0)+b(0,1,0)+c(1,1,0)a(1, 0, 0) + b(0, 1, 0) + c(1, 1, 0)a(1,0,0)+b(0,1,0)+c(1,1,0) . Why is this description misleading?

    Three coefficients suggest PxyP_{xy}Pxy​ is three-dimensional, but (1,1,0)(1, 1, 0)(1,1,0) is redundant and PxyP_{xy}Pxy​ is two-dimensional. p=(a+c)(1,0,0)+(b+c)(0,1,0)p = (a + c)(1, 0, 0) + (b + c)(0, 1, 0)p=(a+c)(1,0,0)+(b+c)(0,1,0) , so the basis {(1,0,0),(0,1,0)}\{(1, 0, 0), (0, 1, 0)\}{(1,0,0),(0,1,0)} and two coefficients suffice.

99. [Vector space techniques] How do you find a basis for the row space R(A)\mathcal{R}(A)R(A) of a matrix AAA ?

    Compute rref⁡(A)\operatorname{rref}(A)rref(A) ; its nonzero rows form a basis of R(A)\mathcal{R}(A)R(A) . Example: A=[1224]A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}A=[12​24​] , rref⁡(A)=[1200]\operatorname{rref}(A) = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}rref(A)=[10​20​] , basis {(1,2)}\{(1, 2)\}{(1,2)} .

100. [Vector space techniques] How do you find a basis for the column space C(A)\mathcal{C}(A)C(A) of a matrix AAA ?

     Compute rref⁡(A)\operatorname{rref}(A)rref(A) , find the columns with pivots; the same columns of the original AAA form a basis of C(A)\mathcal{C}(A)C(A) . Alternatively: find a basis of the row space of ATA^{\mathsf{T}}AT , since C(A)=R(AT)\mathcal{C}(A) = \mathcal{R}(A^{\mathsf{T}})C(A)=R(AT) .Example: A=[1326]A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}A=[12​36​] , rref⁡(A)=[1300]\operatorname{rref}(A) = \begin{bmatrix} 1 & 3 \\ 0 & 0 \end{bmatrix}rref(A)=[10​30​] : pivot in column 1, so…