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1. 1.Computation · 7 min

   Let $V={\mathbb{R}}^2$ as a set, but with the operations

   $$(x_1,y_1)\oplus (x_2,y_2)=(x_1+x_2,y_1+y_2+1),c\odot (x,y)=(cx,cy+c-1).$$

   It is a fact that $V$ is a vector space over $\mathbb{R}$ with these operations.

   1. (a)Find the zero vector of $V$ and the additive inverse of $(x,y)$, working from what the axioms demand. Explain why the answers are not $(0,0)$ and $(-x,-y)$.
   2. (b)Verify axioms 5, 6 and 8 for these operations. (The remaining axioms also hold; you need not check them.)
   3. (c)Find a bijection $T:V\to {\mathbb{R}}^2$ with $T(\mathbf{u}\oplus \mathbf{v})=T(\mathbf{u})+T(\mathbf{v})$ and $T(c\odot \mathbf{v})=cT(\mathbf{v})$, where the right-hand sides use the ordinary operations. Explain why this accounts for all eight axioms at once, and use it to write down a basis of $V$ and $\dim V$.
   Show the solution to question 1

   (a) Solve for the identity. The zero vector is whatever $(a,b)$ satisfies $(x,y)\oplus (a,b)=(x,y)$ for every $(x,y)$. That reads $(x+a,y+b+1)=(x,y)$, so $a=0$ and $b=-1$:

   $$0_V=(0,-1).$$

   The additive inverse of $(x,y)$ is the $(a,b)$ with $(x,y)\oplus (a,b)=0_V$, i.e. $(x+a,y+b+1)=(0,-1)$, so $a=-x$ and $b=-y-2$:

   $$-(x,y)=(-x,-y-2).$$

   Check: $(x,y)\oplus (-x,-y-2)=(0,y-y-2+1)=(0,-1)$ ✓.

   The answers are not the familiar ones because the axioms define $0$ by its behaviour under $\oplus$, not by its appearance. The pair $(0,0)$ is an ordinary nonzero vector of $V$: indeed $(x,y)\oplus (0,0)=(x,y+1)\ne (x,y)$, so it is not an identity. By the elementary-consequences proposition the zero vector is unique, so $(0,-1)$ is the only candidate.

   (b) Axioms 5, 6, 8. Axiom 5: $1\odot (x,y)=(x,y+1-1)=(x,y)$ ✓.

   Axiom 6: $c\odot (d\odot (x,y))=c\odot (dx,dy+d-1)=(cdx,c(dy+d-1)+c-1)=(cdx,cdy+cd-1)$, and $(cd)\odot (x,y)=(cdx,cdy+cd-1)$. Equal ✓.

   Axiom 8: the left side is $(c+d)\odot (x,y)=((c+d)x,(c+d)y+c+d-1)$. The right side is

   $$(cx,cy+c-1)\oplus (dx,dy+d-1)=(cx+dx,(cy+c-1)+(dy+d-1)+1)=((c+d)x,(c+d)y+c+d-1).$$

   Equal ✓.

   (c) The disguise. Take $T(x,y)=(x,y+1)$, a bijection with inverse $(x,y)\mapsto (x,y-1)$. Then

   $$T((x_1,y_1)\oplus (x_2,y_2))=T(x_1+x_2,y_1+y_2+1)=(x_1+x_2,y_1+y_2+2)=(x_1,y_1+1)+(x_2,y_2+1)=T(x_1,y_1)+T(x_2,y_2),$$

   and $T(c\odot (x,y))=T(cx,cy+c-1)=(cx,cy+c)=c(x,y+1)=cT(x,y)$.

   So $T$ converts $\oplus$ into $+$ and $\odot$ into ordinary scaling. Every axiom is an identity between expressions built from the two operations; applying the bijection $T$ to both sides turns it into the same identity in ${\mathbb{R}}^2$ with the usual operations, where it is known to hold, and applying $T^{-1}$ brings it back. That is why all eight hold — exactly as $\ln$ does the work for the positive reals under multiplication. Consistency check: $T(0,-1)=(0,0)$, the zero of ${\mathbb{R}}^2$.

   Because $T$ preserves linear combinations, it carries a basis of $V$ to a basis of ${\mathbb{R}}^2$ and back. The preimages of ${\mathbf{e}}_1$ and ${\mathbf{e}}_2$ are $T^{-1}(1,0)=(1,-1)$ and $T^{-1}(0,1)=(0,0)$, so

   $$((1,-1),(0,0))\text{ is a basis of }V,\dim V=2.$$

   Direct check: $a\odot (1,-1)=(a,-a+a-1)=(a,-1)$ and $b\odot (0,0)=(0,b-1)$, so $a\odot (1,-1)\oplus b\odot (0,0)=(a,-1+b-1+1)=(a,b-1)$, which runs through every vector of $V$ exactly once as $(a,b)$ runs through ${\mathbb{R}}^2$. That is the unique-representation theorem, so it is a basis. Note that the nonzero vector $(0,0)$ is a basis vector.

2. 2.Computation · 8 min

   Decide whether each set is a subspace of the stated space. For a subspace, verify the subspace criterion or the one-step test in full; for a non-subspace, name the condition that fails and exhibit specific elements that break it.

   1. (a)$U_1=\{p\in {\mathcal{P}}_3:p(0)=p(1)\}$ in ${\mathcal{P}}_3$.
   2. (b)$U_2=\{A\in M_{2\times 2}:A^2=A\}$ in $M_{2\times 2}$.
   3. (c)$U_3=\{(x,y,z)\in {\mathbb{R}}^3:x+y+z=0\text{ or }x=y=z\}$ in ${\mathbb{R}}^3$.
   4. (d)$U_4=\{f\in F(\mathbb{R},\mathbb{R}):\text{there is a constant }M\text{ with }|f(x)|\le M\text{ for all }x\}$, the bounded functions, in $F(\mathbb{R},\mathbb{R})$.
   Show the solution to question 2

   (a) Subspace. The zero polynomial has $p(0)=0=p(1)$, so $U_1\ne \varnothing$. One-step test: if $p(0)=p(1)$ and $q(0)=q(1)$, then

   $$(cp+dq)(0)=cp(0)+dq(0)=cp(1)+dq(1)=(cp+dq)(1),$$

   so $cp+dq\in U_1$. Hence $U_1$ is a subspace. In coefficients, $p=a_0+a_{1}x+a_2{x}^2+a_3{x}^3$ lies in $U_1$ exactly when $a_1+a_2+a_3=0$ — a single homogeneous linear equation, which is the shape a subspace condition always has.

   (b) Not a subspace. The zero matrix satisfies $0^2=0$, so the quick zero-vector test does not disqualify $U_2$; closure has to be checked. Now $I^2=I$, so $I\in U_2$, but

   $$(2I{)}^2=4I\ne 2I,$$

   so $2I\notin U_2$. Closure under scalar multiplication fails (and so does closure under addition, since $I+I=2I$). The condition $A^2=A$ is quadratic in the entries, not linear, which is the underlying reason.

   (c) Not a subspace. $U_3=P\cup L$ where $P=\{x+y+z=0\}$ and $L=\{x=y=z\}=\mathrm{span}\{(1,1,1)\}$. Each is a subspace (a homogeneous linear condition, and a span). Neither contains the other: $(1,-1,0)\in P$ but not in $L$, and $(1,1,1)\in L$ but not in $P$ since its coordinates sum to $3$. By the proposition on unions of subspaces, $P\cup L$ is therefore not a subspace.

   Concretely, closure under addition fails:

   $$(1,-1,0)+(1,1,1)=(2,0,1),$$

   whose coordinates sum to $3\ne 0$ and are not all equal, so the sum lies in neither $P$ nor $L$.

   (d) Subspace. The zero function is bounded (by $M=0$). One-step test: suppose $|f(x)|\le M$ and $|g(x)|\le N$ for all $x$. Then for all $x$, by the triangle inequality,

   $$|(cf+dg)(x)|=|cf(x)+dg(x)|\le |c||f(x)|+|d||g(x)|\le |c|M+|d|N,$$

   a constant, so $cf+dg$ is bounded. Hence $U_4$ is a subspace. Note the test only requires that some bound exist; the bound for the combination may depend on $c$ and $d$, and the bounds for $f$ and $g$ need not be the same.

3. 3.Proof · 9 min

   Let $V=F(\mathbb{R},\mathbb{R})$, and let

   $$E=\{f\in V:f(-x)=f(x)\text{ for all }x\},O=\{f\in V:f(-x)=-f(x)\text{ for all }x\}$$

   be the even and the odd functions.

   1. (a)Show that $E$ and $O$ are subspaces of $V$.
   2. (b)Prove that $V=E\oplus O$. Quote the direct-sum criterion when you use it.
   3. (c)Write $f(x)=(x+1{)}^3$ as an even function plus an odd function. Then explain why ${\mathcal{P}}_3=(E\cap {\mathcal{P}}_3)\oplus (O\cap {\mathcal{P}}_3)$, find the dimension of each piece, and check that the dimensions add up correctly.
   Show the solution to question 3

   (a) Subspaces. The zero function satisfies $0(-x)=0=0(x)$ and also $0=-0(x)$, so it lies in both $E$ and $O$. One-step test for $E$: if $f,g\in E$ then

   $$(cf+dg)(-x)=cf(-x)+dg(-x)=cf(x)+dg(x)=(cf+dg)(x)$$

   for every $x$, so $cf+dg\in E$. For $O$ the same computation gives $(cf+dg)(-x)=-cf(x)-dg(x)=-(cf+dg)(x)$. Both are subspaces.

   (b) The sum is all of $V$. Given any $f\in V$ define

   $$f_e(x)=\frac{f(x)+f(-x)}{2},f_o(x)=\frac{f(x)-f(-x)}{2}.$$

   Then $f_e(-x)=\frac{f(-x)+f(x)}{2}=f_e(x)$, so $f_e\in E$; and $f_o(-x)=\frac{f(-x)-f(x)}{2}=-f_o(x)$, so $f_o\in O$; and $f_e+f_o=f$ by adding the two fractions. Hence every $f$ is in $E+O$, i.e. $V=E+O$.

   The sum is direct. Let $f\in E\cap O$. For every $x$, $f(x)=f(-x)$ (even) and $f(-x)=-f(x)$ (odd), so $f(x)=-f(x)$, $2f(x)=0$, $f(x)=0$. Thus $E\cap O=\{0\}$, and by the direct-sum criterion $V=E\oplus O$. $■$

   (The criterion is what saves work: it replaces checking uniqueness of every decomposition by one intersection computation.)

   (c) Expand: $f(x)=(x+1{)}^3=x^3+3x^2+3x+1$ and $f(-x)=(1-x{)}^3=-x^3+3x^2-3x+1$. Then

   $$f_e=\frac{6x^2+2}{2}=3x^2+1,f_o=\frac{2x^3+6x}{2}=x^3+3x,$$

   and indeed $(3x^2+1)+(x^3+3x)=(x+1{)}^3$ ✓. As always, the even part collects the even-degree terms and the odd part the odd-degree ones.

   For ${\mathcal{P}}_3$: if $p\in {\mathcal{P}}_3$ then $p(-x)$ is again a polynomial of degree at most $3$ (replacing $x$ by $-x$ only changes the signs of the odd coefficients), so $p_e$ and $p_o$ lie in ${\mathcal{P}}_3$. Thus ${\mathcal{P}}_3=(E\cap {\mathcal{P}}_3)+(O\cap {\mathcal{P}}_3)$, and the intersection of the two pieces sits inside $E\cap O=\{0\}$, so the sum is direct.

   Dimensions: $p=a+bx+c{x}^2+d{x}^3$ is even exactly when $p(-x)-p(x)=-2bx-2d{x}^3$ is the zero polynomial, and since $\{x,x^3\}$ is independent (a sublist of the monomial basis) that forces $b=d=0$. So $E\cap {\mathcal{P}}_3=\mathrm{span}\{1,x^2\}$, of dimension $2$. Likewise $O\cap {\mathcal{P}}_3=\mathrm{span}\{x,x^3\}$, of dimension $2$. By the dimension formula for a direct sum, $2+2=4=\dim {\mathcal{P}}_3$ ✓.

4. 4.Computation · 7 min

   Every span question below is to be settled by the method of the chapter: write the membership condition as a linear system and row-reduce, or compare coefficients.

   1. (a)Let ${\mathbf{v}}_1=(1,2,-1)$ and ${\mathbf{v}}_2=(2,5,1)$. For which real $k$ is $(1,1,k)\in \mathrm{span}\{{\mathbf{v}}_1,{\mathbf{v}}_2\}$? For that $k$ give the coefficients and check them. Then row-reduce with a general right-hand side $(b_1,b_2,b_3)$ to find the single equation $\alpha x+\beta y+\gamma z=0$ that describes $\mathrm{span}\{{\mathbf{v}}_1,{\mathbf{v}}_2\}$ as a plane through the origin.
   2. (b)Does $\{1+x^2,x+x^2,1+x\}$ span ${\mathcal{P}}_2$? Decide by attempting to hit an arbitrary $a+bx+c{x}^2$, and use your formula to write $x^2$ as a combination of the three polynomials.
   Show the solution to question 4

   (a) Row-reduce the augmented matrix $[{\mathbf{v}}_1{\mathbf{v}}_2\mid \mathbf{b}]$ with $\mathbf{b}=(b_1,b_2,b_3)$ kept general, so that both questions are answered by one reduction:

   $$\left(\begin{array}{ccc}1& 2& b_1\\ 2& 5& b_2\\ -1& 1& b_3\end{array}\right)\to \left(\begin{array}{ccc}1& 2& b_1\\ 0& 1& b_2-2b_1\\ 0& 3& b_3+b_1\end{array}\right)\to \left(\begin{array}{ccc}1& 2& b_1\\ 0& 1& b_2-2b_1\\ 0& 0& 7b_1-3b_2+b_3\end{array}\right),$$

   using $R_2-2R_1$, $R_3+R_1$, then $R_3-3R_2$ (and $b_3+b_1-3(b_2-2b_1)=7b_1-3b_2+b_3$). The system is consistent exactly when the last entry vanishes, so

   $$\mathrm{span}\{{\mathbf{v}}_1,{\mathbf{v}}_2\}=\{(x,y,z):7x-3y+z=0\}.$$

   Sanity check: $7(1)-3(2)+(-1)=0$ and $7(2)-3(5)+1=0$, so both spanning vectors lie on the plane ✓.

   For $\mathbf{b}=(1,1,k)$ the condition is $7-3+k=0$, i.e. $k=-4$. Back-substituting with $b=(1,1,-4)$: $c_2=b_2-2b_1=-1$ and $c_1=b_1-2c_2=3$. Check: $3(1,2,-1)-(2,5,1)=(1,1,-4)$ ✓. For every other $k$ the reduced matrix has a row $[00\mid k+4]$ with $k+4\ne 0$ and the vector is not in the span.

   (b) Compare coefficients. A combination is

   $$\alpha (1+x^2)+\beta (x+x^2)+\gamma (1+x)=(\alpha +\gamma )+(\beta +\gamma )x+(\alpha +\beta )x^2.$$

   Two polynomials are equal exactly when their coefficients agree, so hitting $a+bx+c{x}^2$ means solving

   $$\alpha +\gamma =a,\beta +\gamma =b,\alpha +\beta =c.$$

   Add all three: $2(\alpha +\beta +\gamma )=a+b+c$, so $\alpha +\beta +\gamma =\frac{1}{2}(a+b+c)$. Subtracting each original equation from this gives

   $$\beta =\frac{1}{2}(-a+b+c),\alpha =\frac{1}{2}(a-b+c),\gamma =\frac{1}{2}(a+b-c).$$

   The system is solvable for every $(a,b,c)$, so the three polynomials span ${\mathcal{P}}_2$. (Since $\dim {\mathcal{P}}_2=3$, the counting corollary then says they are also a basis.)

   For $x^2$ take $a=b=0$, $c=1$: $\alpha =\frac{1}{2}$, $\beta =\frac{1}{2}$, $\gamma =-\frac{1}{2}$. Check:

   $$\frac{1}{2}(1+x^2)+\frac{1}{2}(x+x^2)-\frac{1}{2}(1+x)=\frac{1}{2}-\frac{1}{2}+(\frac{1}{2}-\frac{1}{2})x+(\frac{1}{2}+\frac{1}{2})x^2=x^2✓.$$

5. 5.Computation · 8 min

   Independence is a statement about the only solution of a linear equation. Exhibit the relation whenever you claim dependence, and check it.

   1. (a)For which real $k$ are the vectors $(1,1,k)$, $(1,k,1)$, $(k,1,1)$ linearly dependent? For each such $k$ write down an explicit dependence relation and verify it.
   2. (b)Show that $\{{\sin }^{2}x,{\cos }^{2}x,1\}$ is linearly dependent in $F(\mathbb{R},\mathbb{R})$, and then show that the sublist $\{{\sin }^{2}x,{\cos }^{2}x\}$ is linearly independent.
   3. (c)Show that $\{x,|x|\}$ is linearly independent in $F(\mathbb{R},\mathbb{R})$ even though the two functions agree at every $x\ge 0$. Then decide whether the same two functions are independent as elements of $F([0,\infty ),\mathbb{R})$.
   Show the solution to question 5

   (a) Use the determinant test. Put the vectors as the columns of a square matrix; by the independence recipe they are independent exactly when the determinant is nonzero. Expanding along the first row,

   $$\det \left(\begin{array}{ccc}1& 1& k\\ 1& k& 1\\ k& 1& 1\end{array}\right)=1(k\cdot 1-1\cdot 1)-1(1\cdot 1-1\cdot k)+k(1\cdot 1-k\cdot k)=(k-1)+(k-1)+k-k^3=-k^3+3k-2.$$

   Now $k=1$ is a root, and dividing out gives $-k^3+3k-2=-(k-1)(k^2+k-2)=-(k-1{)}^2(k+2)$. (Check by expanding: $(k-1{)}^2(k+2)=(k^2-2k+1)(k+2)=k^3-3k+2$ ✓.) So the vectors are dependent exactly for $k=1$ and $k=-2$.

   $k=1$: all three vectors equal $(1,1,1)$, and $1\cdot {\mathbf{v}}_1-1\cdot {\mathbf{v}}_2+0\cdot {\mathbf{v}}_3=0$ is a relation with a nonzero coefficient.

   $k=-2$: the vectors are $(1,1,-2)$, $(1,-2,1)$, $(-2,1,1)$, and

   $${\mathbf{v}}_1+{\mathbf{v}}_2+{\mathbf{v}}_3=(1+1-2,1-2+1,-2+1+1)=(0,0,0)✓.$$

   Sanity check on a value that should be independent: at $k=0$ the determinant is $-2\ne 0$, and indeed $(1,1,0),(1,0,1),(0,1,1)$ are independent.

   (b) Dependent triple. For every real $x$, ${\sin }^{2}x+{\cos }^{2}x=1$, so

   $$1\cdot {\sin }^{2}x+1\cdot {\cos }^{2}x+(-1)\cdot 1=0\text{for all }x,$$

   a dependence relation with coefficients $(1,1,-1)$, not all zero.

   Independent pair. Suppose $a{\sin }^{2}x+b{\cos }^{2}x=0$ for all $x$. Evaluate at $x=0$: $a\cdot 0+b\cdot 1=0$, so $b=0$. Evaluate at $x=\frac{\pi }{2}$: $a\cdot 1+b\cdot 0=0$, so $a=0$. The only relation is trivial, so the pair is independent. This is the linear dependence lemma in action: in the list $1,{\sin }^{2}x,{\cos }^{2}x$ the last vector lies in the span of its predecessors (${\cos }^{2}x=1-{\sin }^{2}x$), and removing it leaves an independent list.

   (c) Suppose $ax+b|x|=0$ for every real $x$. At $x=1$: $a+b=0$. At $x=-1$: $-a+b=0$. Adding gives $2b=0$, so $b=0$ and then $a=0$. Independent.

   Agreement on $x\ge 0$ is irrelevant: it shows only that the particular combination $x-|x|$ vanishes on half the domain, whereas a dependence relation must vanish at every point of the domain, and $x-|x|=2x\ne 0$ for $x<0$.

   On the domain $[0,\infty )$ the situation reverses. There $|x|=x$ at every point of the domain, so $x$ and $|x|$ are the same element of $F([0,\infty ),\mathbb{R})$, and $1\cdot x+(-1)\cdot |x|=0$ is a nontrivial relation: the two functions are dependent. The domain is part of the data of a function space, and changing it can change independence.

6. 6.Computation · 9 min

   Both parts use the left-to-right procedure: run through the list, keeping a vector if it lies outside the span of the vectors kept so far and deleting it otherwise.

   1. (a)Let $p_1=1+x$, $p_2=1-x$, $p_3=2$, $p_4=x+x^2$, $p_5=1+x^2$ in ${\mathcal{P}}_2$. Reduce the list to a basis of its span, saying at each step whether the vector is kept or deleted and why, and writing every deleted vector explicitly as a combination of the vectors kept before it. Then explain why the result is in fact a basis of all of ${\mathcal{P}}_2$, and confirm your answer by row-reducing the matrix whose columns are the monomial coordinate vectors.
   2. (b)Extend $\{I,J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\}$ to a basis of $M_{2\times 2}$ by appending the matrix units $E_{11},E_{12},E_{21},E_{22}$ in that order and running the procedure. Say at which point the counting corollary lets you stop, and express each discarded matrix unit as a combination of your basis.
   Show the solution to question 6

   (a) Run the procedure.

   $p_1=1+x$: nonzero, so not in $\mathrm{span}\varnothing =\{0\}$. Keep.

   $p_2=1-x$: in $\mathrm{span}\{p_1\}$ only if $1-x=c(1+x)$, which needs $c=1$ from the constants and $c=-1$ from the $x$-coefficients. Impossible. Keep.

   $p_3=2$: $p_3=(1+x)+(1-x)=p_1+p_2$. Delete.

   $p_4=x+x^2$: every combination of $p_1,p_2$ has degree at most $1$, and $p_4$ has degree $2$. Not in the span. Keep.

   $p_5=1+x^2$: the kept list $p_1,p_2,p_4$ is independent (each kept vector was outside the span of the earlier ones, so by the linear dependence lemma no relation is possible), and it has $3$ members in the $3$-dimensional space ${\mathcal{P}}_2$. By the counting corollary it is a basis of ${\mathcal{P}}_2$, so $p_5$ is certainly in its span: explicitly $p_5=(1-x)+(x+x^2)=p_2+p_4$. Delete.

   Basis of the span: $(1+x,1-x,x+x^2)$. Since a deletion never changes the span (redundancy proposition), the span of the five polynomials equals the span of these three, which is ${\mathcal{P}}_2$ by the counting argument just made. So the original list did span ${\mathcal{P}}_2$, and the three kept vectors are a basis of it.

   Confirmation by columns. In the monomial basis the coordinate vectors are $(1,1,0)$, $(1,-1,0)$, $(2,0,0)$, $(0,1,1)$, $(1,0,1)$. Row-reduce:

   $$\left(\begin{array}{ccccc}1& 1& 2& 0& 1\\ 1& -1& 0& 1& 0\\ 0& 0& 0& 1& 1\end{array}\right)\stackrel{R_2-R_1}{\to }\left(\begin{array}{ccccc}1& 1& 2& 0& 1\\ 0& -2& -2& 1& -1\\ 0& 0& 0& 1& 1\end{array}\right).$$

   This is already in echelon form with pivots in columns $1$, $2$ and $4$ — exactly the vectors $p_1,p_2,p_4$ that were kept. ✓

   (b) In coordinates $(a_{11},a_{12},a_{21},a_{22})$ the two given matrices are $I\mapsto (1,0,0,1)$ and $J\mapsto (0,1,1,0)$; neither is a multiple of the other, so the pair is independent and both are kept. A general combination is $aI+bJ\mapsto (a,b,b,a)$.

   $E_{11}\mapsto (1,0,0,0)$: equal to $(a,b,b,a)$ would need $a=1$ (first slot) and $a=0$ (last slot). Not in the span. Keep.

   $E_{12}\mapsto (0,1,0,0)$: a combination $aI+bJ+c{E}_{11}\mapsto (a+c,b,b,a)$ would need $b=1$ (second slot) and $b=0$ (third slot). Not in the span. Keep.

   Now the kept list $I,J,E_{11},E_{12}$ is independent with $4$ members, and $\dim M_{2\times 2}=4$. By the counting corollary it is already a basis, so the procedure can stop: $E_{21}$ and $E_{22}$ must lie in the span and would be deleted. Explicitly,

   $$E_{21}=J-E_{12},E_{22}=I-E_{11},$$

   both immediate from the entries. Basis: $(I,J,E_{11},E_{12})$. Sanity check: the extension theorem promised a basis containing $I$ and $J$, and this one does.

7. 7.Critique · 8 min

   A student is asked for the dimension of

   $$U=\{p\in {\mathcal{P}}_4:p(1)=0\text{ and }p(-1)=0\}$$

   and submits the following.

   Step 1. ${\mathcal{P}}_4$ is the space of polynomials of degree at most $4$, so $\dim {\mathcal{P}}_4=4$.

   Step 2. Each of $p(1)=0$ and $p(-1)=0$ is one linear equation in the coefficients of $p$, and each equation costs one dimension. So $\dim U=4-2=2$.

   Step 3. Check: $x^2-1$ and $x^3-x$ both vanish at $1$ and at $-1$, and neither is a multiple of the other, so they are linearly independent. Two independent vectors in $U$ form a basis of $U$, confirming $\dim U=2$.

   1. (a)Find every error. For each one quote the sentence, say why it is wrong, and name the definition or theorem being misused.
   2. (b)Find $\dim U$ correctly: give a basis of $U$ and prove that it is one.
   3. (c)State precisely what the two polynomials of Step 3 do prove about $U$, and exhibit a vector of $U$ that is not in their span.
   Show the solution to question 7

   (a) There are two errors, and the second is the one that let the first go unnoticed.

   Step 1: “$\dim {\mathcal{P}}_4=4$.” False: the monomial basis $1,x,x^2,x^3,x^4$ has five members, so $\dim {\mathcal{P}}_4=5$. The constants count. This is the standard off-by-one, and it feeds a wrong input into Step 2, whose arithmetic then gives $5-2=3$ rather than $2$ once corrected. (Step 2 itself is a heuristic rather than a proof — two conditions cost two dimensions only if neither is a consequence of the other — but it is not the source of the wrong answer; part (b) shows the count $3$ is right.)

   Step 3: “Two independent vectors in $U$ form a basis of $U$.” This misuses the counting corollary. That corollary says an independent list of exactly $\dim U$ vectors is a basis — so it can only be applied once $\dim U$ is known, and here it is being used to establish $\dim U$. The argument is circular. What the definition of basis actually requires is independence and spanning, and spanning was never checked. Independence alone gives only a lower bound: by the exchange lemma, an independent list in $U$ is no longer than a basis of $U$, so $\dim U\ge 2$. The “confirmation” confirmed nothing, because it silently assumed the conclusion of Step 2.

   (b) Claim: $U=\{(x^2-1)q(x):q\in {\mathcal{P}}_2\}$.

   If $p\in U$, then $p(1)=0$, so by the factor theorem $p(x)=(x-1)r(x)$ with $\mathrm{deg}r\le 3$. Then $0=p(-1)=(-2)r(-1)$ forces $r(-1)=0$, so $r(x)=(x+1)q(x)$ with $\mathrm{deg}q\le 2$, and $p=(x^2-1)q$. Conversely any $(x^2-1)q$ with $\mathrm{deg}q\le 2$ has degree at most $4$ and vanishes at $\pm 1$. This proves the claim.

   Writing $q=a+bx+c{x}^2$,

   $$p=a(x^2-1)+b(x^3-x)+c(x^4-x^2),$$

   so the three polynomials $x^2-1,x^3-x,x^4-x^2$ span $U$. They are independent: their degrees are $2,3,4$, so in a relation $a(x^2-1)+b(x^3-x)+c(x^4-x^2)=0$ the coefficient of $x^4$ is $c$, forcing $c=0$; then the coefficient of $x^3$ is $b$, forcing $b=0$; then $a=0$. Independent and spanning, so a basis, and

   $$\dim U=3.$$

   Sanity check against the corrected heuristic: $5-2=3$ ✓. The two conditions really are independent of each other — $x-1$ satisfies the first and not the second — which is why the heuristic count happens to be right.

   (c) The pair $x^2-1$, $x^3-x$ is an independent list in $U$, which proves exactly that $\dim U\ge 2$ — nothing more. It does not span: $x^4-x^2=(x^2-1)x^2$ vanishes at $\pm 1$, so it lies in $U$, but every combination $a(x^2-1)+b(x^3-x)$ has degree at most $3$, so $x^4-x^2$ is outside their span. Adjoining it gives the basis of part (b) — the extension theorem carried out by hand.

8. 8.Computation · 9 min

   Let $\mathcal{M}=(1,x,x^2)$ be the monomial basis of ${\mathcal{P}}_2$ and let

   $$\mathcal{B}=(1,x-1,(x-1{)}^2).$$

   1. (a)Prove that $\mathcal{B}$ is a basis of ${\mathcal{P}}_2$ using the counting corollary, so that you have to check only one of the two basis properties.
   2. (b)Write down $P_{\mathcal{M}\leftarrow \mathcal{B}}$ and $P_{\mathcal{B}\leftarrow \mathcal{M}}$ — the second by expressing each monomial in $\mathcal{B}$, not by an inversion algorithm — and verify that their product is $I_3$.
   3. (c)Find $[p{]}_{\mathcal{B}}$ for $p(x)=2x^2-3x+5$ in two independent ways: by multiplying by the appropriate matrix, and by comparing coefficients directly. Say which matrix you used and why that one.
   4. (d)Show that $p\in {\mathcal{P}}_2$ satisfies $p(1)=0$ exactly when the first entry of $[p{]}_{\mathcal{B}}$ is $0$. Deduce a basis and the dimension of $U=\{p\in {\mathcal{P}}_2:p(1)=0\}$.
   Show the solution to question 8

   (a) $\dim {\mathcal{P}}_2=3$ and $\mathcal{B}$ has three members, so by the counting corollary it suffices to show independence. Suppose $a\cdot 1+b(x-1)+c(x-1{)}^2$ is the zero polynomial. Expanding,

   $$a+b(x-1)+c(x^2-2x+1)=(a-b+c)+(b-2c)x+c{x}^2,$$

   and a polynomial is zero exactly when all its coefficients are: $c=0$, then $b-2c=0$ gives $b=0$, then $a-b+c=0$ gives $a=0$. Independent, hence a basis. (Equivalently: the three have distinct degrees $0,1,2$, so no nontrivial relation can cancel the top-degree term.)

   (b) The columns of $P_{\mathcal{M}\leftarrow \mathcal{B}}$ are the $\mathcal{M}$-coordinates of the $\mathcal{B}$-vectors: $[1{]}_{\mathcal{M}}=(1,0,0)$, $[x-1{]}_{\mathcal{M}}=(-1,1,0)$, $[(x-1{)}^2{]}_{\mathcal{M}}=[x^2-2x+1{]}_{\mathcal{M}}=(1,-2,1)$. So

   $$P_{\mathcal{M}\leftarrow \mathcal{B}}=\left(\begin{array}{ccc}1& -1& 1\\ 0& 1& -2\\ 0& 0& 1\end{array}\right).$$

   For the other direction express each monomial in $\mathcal{B}$: $1=1$; $x=1+(x-1)$; $x^2=((x-1)+1{)}^2=(x-1{)}^2+2(x-1)+1$. The columns are therefore $(1,0,0)$, $(1,1,0)$, $(1,2,1)$:

   $$P_{\mathcal{B}\leftarrow \mathcal{M}}=\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 2\\ 0& 0& 1\end{array}\right).$$

   Product, row by row: row $1$ of the first against the columns of the second gives $(1,1-1,1-2+1)=(1,0,0)$; row $2$ gives $(0,1,2-2)=(0,1,0)$; row $3$ gives $(0,0,1)$. So $P_{\mathcal{M}\leftarrow \mathcal{B}}{P}_{\mathcal{B}\leftarrow \mathcal{M}}=I_3$ ✓, as the change-of-coordinates theorem requires.

   (c) $[p{]}_{\mathcal{M}}=(5,-3,2)$. We have $\mathcal{M}$-coordinates and want $\mathcal{B}$-coordinates, so the matrix is $P_{\mathcal{B}\leftarrow \mathcal{M}}$ — read the arrow: $\mathcal{M}$ in, $\mathcal{B}$ out:

   $$[p{]}_{\mathcal{B}}=\left(\begin{array}{ccc}1& 1& 1\\ 0& 1& 2\\ 0& 0& 1\end{array}\right)\left(\begin{array}{c}5\\ -3\\ 2\end{array}\right)=\left(\begin{array}{c}5-3+2\\ -3+4\\ 2\end{array}\right)=\left(\begin{array}{c}4\\ 1\\ 2\end{array}\right).$$

   Directly: by the expansion in (a), $a+b(x-1)+c(x-1{)}^2=(a-b+c)+(b-2c)x+c{x}^2$ must equal $5-3x+2x^2$. Top down: $c=2$; $b-4=-3$ so $b=1$; $a-1+2=5$ so $a=4$. Both routes give $[p{]}_{\mathcal{B}}=(4,1,2)$. Check: $4+(x-1)+2(x-1{)}^2=4+x-1+2x^2-4x+2=2x^2-3x+5$ ✓.

   (d) If $[p{]}_{\mathcal{B}}=(a,b,c)$ then $p(x)=a+b(x-1)+c(x-1{)}^2$, and evaluating at $x=1$ kills every term but the first: $p(1)=a$. Hence $p(1)=0⟺a=0$.

   So $U=\{b(x-1)+c(x-1{)}^2\}=\mathrm{span}\{x-1,(x-1{)}^2\}$. Those two are independent, being a sublist of the basis $\mathcal{B}$, so they form a basis of $U$ and $\dim U=2$. Sanity check: one homogeneous linear condition on a $3$-dimensional space costs one dimension, and $p=2x^2-3x+5$ has $p(1)=4\ne 0$, matching its first $\mathcal{B}$-coordinate $4$. This is the point of choosing a basis adapted to the problem: in $\mathcal{B}$ the subspace $U$ is visible as a coordinate hyperplane, which it is not in $\mathcal{M}$.

9. 9.Sketch · 9 min

   Let $A=\left(\begin{array}{cc}1& 2\\ 3& 6\end{array}\right)$. Every line you draw must be labelled with the subspace it is and justified by a computation.

   1. (a)Find $\mathrm{Row}(A)$, $\mathrm{Null}(A)$, $\mathrm{Col}(A)$ and $\mathrm{Null}(A^T)$, each as a line through the origin with a direction vector.
   2. (b)On one pair of axes (the input plane) sketch $\mathrm{Row}(A)$, $\mathrm{Null}(A)$ and the solution set of $A\mathbf{x}=(2,6)$. On a second pair (the output plane) sketch $\mathrm{Col}(A)$ and $\mathrm{Null}(A^T)$, and mark the points $(2,6)$ and $(1,1)$. Say which of $A\mathbf{x}=(2,6)$ and $A\mathbf{x}=(1,1)$ is consistent and how the picture shows it.
   3. (c)Explain why ${\mathbb{R}}^2=\mathrm{Row}(A)\oplus \mathrm{Null}(A)$, decompose $(1,0)$ accordingly, and verify that $A$ applied to the row-space part alone already gives $A(1,0)$.
   Show the solution to question 9

   (a) Row $2$ is $3$ times row $1$, so $A$ row-reduces to $\left(\begin{array}{cc}1& 2\\ 0& 0\end{array}\right)$: one pivot, $\mathrm{rank}A=1$.

   $\mathrm{Row}(A)$: the nonzero row of the reduced matrix, $\mathrm{span}\{(1,2)\}$ — the line $y=2x$.

   $\mathrm{Null}(A)$: solve $x+2y=0$; with $y=t$ free, $\mathbf{x}=t(-2,1)$, so $\mathrm{span}\{(-2,1)\}$ — the line $y=-\frac{1}{2}x$. (Rank Theorem check: $1+1=2$ columns ✓.)

   $\mathrm{Col}(A)$: the pivot column of the original matrix, $\mathrm{span}\{(1,3)\}$ — the line $y=3x$. (The second column $(2,6)=2(1,3)$ adds nothing.)

   $\mathrm{Null}(A^T)$: $A^T=\left(\begin{array}{cc}1& 3\\ 2& 6\end{array}\right)$ reduces to $\left(\begin{array}{cc}1& 3\\ 0& 0\end{array}\right)$, so solve $x+3y=0$: $\mathrm{span}\{(-3,1)\}$ — the line $y=-\frac{1}{3}x$. Dimension $m-r=2-1=1$ ✓.

   (b) Input plane. Draw the line through the origin and $(1,2)$, labelled $\mathrm{Row}(A)$, and the line through the origin and $(-2,1)$, labelled $\mathrm{Null}(A)$. They are perpendicular, as the fundamental theorem predicts: $(1,2)\cdot (-2,1)=-2+2=0$.

   The system $A\mathbf{x}=(2,6)$ is the two equations $x+2y=2$ and $3x+6y=6$, which are the same equation. Its solution set is the line $x+2y=2$, through $(2,0)$ and $(0,1)$; in parametric form $(2,0)+t(-2,1)$. Draw it dashed: it is parallel to $\mathrm{Null}(A)$ (same direction vector) but does not pass through the origin, so it is a translate of the null space and not a subspace — exactly “particular solution plus null space”.

   Output plane. Draw the line through the origin and $(1,3)$, labelled $\mathrm{Col}(A)$, and the line through the origin and $(-3,1)$, labelled $\mathrm{Null}(A^T)$; again perpendicular, since $(1,3)\cdot (-3,1)=0$. Mark $(2,6)$: it equals $2(1,3)$, so it lies on $\mathrm{Col}(A)$. Mark $(1,1)$: it would need $(1,1)=c(1,3)$, impossible ($c=1$ and $c=\frac{1}{3}$), so it lies off the line.

   Consistency: $A\mathbf{x}=\mathbf{b}$ is consistent exactly when $\mathbf{b}\in \mathrm{Col}(A)$. So $A\mathbf{x}=(2,6)$ is consistent (its solutions are the dashed line in the input plane) and $A\mathbf{x}=(1,1)$ is not. The left null space gives the same verdict by a dot product: $(2,6)\cdot (-3,1)=-6+6=0$ while $(1,1)\cdot (-3,1)=-2\ne 0$, and a vector of $\mathrm{Col}(A)$ must be orthogonal to $\mathrm{Null}(A^T)$.

   (c) The two lines meet only at the origin: a common point satisfies $s(1,2)=t(-2,1)$, so $s=-2t$ and $2s=t$, giving $-4t=t$, $t=0=s$. So $\mathrm{Row}(A)\cap \mathrm{Null}(A)=\{0\}$ and the sum is direct by the direct-sum criterion. Its dimension is $1+1=2=\dim {\mathbb{R}}^2$ by the direct-sum dimension formula, and a subspace of full dimension is the whole space (subspace-dimension theorem). Hence ${\mathbb{R}}^2=\mathrm{Row}(A)\oplus \mathrm{Null}(A)$.

   Decompose $(1,0)=\alpha (1,2)+\beta (-2,1)$: the equations are $\alpha -2\beta =1$ and $2\alpha +\beta =0$. From the second $\beta =-2\alpha$, so $\alpha +4\alpha =1$, $\alpha =\frac{1}{5}$, $\beta =-\frac{2}{5}$. Check: $\frac{1}{5}(1,2)-\frac{2}{5}(-2,1)=(\frac{1}{5}+\frac{4}{5},\frac{2}{5}-\frac{2}{5})=(1,0)$ ✓.

   Now $A(1,0)=(1,3)$, the first column. The row-space part alone gives

   $$A\cdot \frac{1}{5}(1,2)=\frac{1}{5}(1+4,3+12)=\frac{1}{5}(5,15)=(1,3),$$

   the same answer, because the null-space part $-\frac{2}{5}(-2,1)$ is sent to $0$. The matrix sees only the row-space component of its input: a rank-$1$ machine in a $2\times 2$ costume.

10. 10.Proof · 10 min

    Throughout, $V$ is a finite-dimensional vector space with $\dim V=n$. Name each theorem as you use it.

    1. (a)Let $U$ and $W$ be subspaces of $V$ with $\dim U+\dim W>n$. Prove that $U\cap W$ contains a nonzero vector.
    2. (b)Prove that every subspace $U$ of $V$ has a complement: a subspace $W$ with $V=U\oplus W$. Then show that every complement of $U$ has the same dimension, and say what it is.
    3. (c)Show by an explicit example in ${\mathbb{R}}^2$ that a complement is not unique, verifying directly that each of your two candidates is a complement.
    Show the solution to question 10

    (a) $U+W$ is a subspace of $V$ (the sum proposition), so by the subspace-dimension theorem $\dim (U+W)\le \dim V=n$. Now apply the dimension formula for a sum:

    $$\dim (U\cap W)=\dim U+\dim W-\dim (U+W)\ge \dim U+\dim W-n>0.$$

    So $U\cap W$ is a subspace of dimension at least $1$. It therefore has a basis with at least one vector, and a basis vector is nonzero (a list containing $0$ is dependent). Hence $U\cap W\ne \{0\}$. $■$

    This is the “two planes in ${\mathbb{R}}^3$ must share a line” fact in general form: $2+2>3$. The bound is sharp — with $\dim U+\dim W=n$ the intersection can be trivial, as two distinct lines through the origin in ${\mathbb{R}}^2$ show.

    (b) Existence. Let $k=\dim U$. By the subspace-dimension theorem $U$ is finite-dimensional, so it has a basis ${\mathbf{u}}_1,\dots ,{\mathbf{u}}_k$. This list is independent in $V$, so by the extension theorem there are ${\mathbf{w}}_1,\dots ,{\mathbf{w}}_r$ such that

    $${\mathbf{u}}_1,\dots ,{\mathbf{u}}_k,{\mathbf{w}}_1,\dots ,{\mathbf{w}}_r$$

    is a basis of $V$; by the dimension theorem $k+r=n$. Put $W=\mathrm{span}\{{\mathbf{w}}_1,\dots ,{\mathbf{w}}_r\}$, a subspace by the span theorem.

    $V=U+W$: every $\mathbf{v}\in V$ is a combination $\sum a_i{\mathbf{u}}_i+\sum b_j{\mathbf{w}}_j$ of the basis; the first sum lies in $U$ and the second in $W$.

    The sum is direct: let $\mathbf{v}\in U\cap W$. Then $\mathbf{v}=\sum a_i{\mathbf{u}}_i$ for some scalars (as $\mathbf{v}\in U$) and $\mathbf{v}=\sum b_j{\mathbf{w}}_j$ for some scalars (as $\mathbf{v}\in W$). Subtracting,

    $$\sum a_i{\mathbf{u}}_i-\sum b_j{\mathbf{w}}_j=0$$

    is a relation among the members of a basis of $V$, so every $a_i$ and $b_j$ is zero and $\mathbf{v}=0$. Thus $U\cap W=\{0\}$, and by the direct-sum criterion $V=U\oplus W$. $■$

    Dimension of any complement. If $W^{\prime }$ is any subspace with $V=U\oplus W^{\prime }$, the dimension formula for a direct sum gives $n=\dim V=\dim U+\dim W^{\prime }$, so

    $$\dim W^{\prime }=n-k,$$

    independently of which complement was chosen. (For the $W$ constructed above this is $r=n-k$, as it should be.)

    (c) Take $U=\mathrm{span}\{(1,0)\}$, the $x$-axis, and the two candidates $W_1=\mathrm{span}\{(0,1)\}$ and $W_2=\mathrm{span}\{(1,1)\}$.

    $U\cap W_1$: a common vector is $(s,0)=(0,t)$, forcing $s=t=0$. $U\cap W_2$: $(s,0)=(t,t)$ forces $t=0$ and then $s=0$. Both intersections are trivial, so both sums are direct by the criterion. Each direct sum has dimension $1+1=2=\dim {\mathbb{R}}^2$, so by the subspace-dimension theorem each equals ${\mathbb{R}}^2$: ${\mathbb{R}}^2=U\oplus W_1=U\oplus W_2$.

    The complements differ, since $(1,1)\notin W_1$. In fact every line through the origin other than $U$ itself is a complement of $U$. The decomposition of a vector depends on the complement: $(3,1)=(3,0)+(0,1)$ using $W_1$ but $(3,1)=(2,0)+(1,1)$ using $W_2$ — same vector, different filing system.

11. 11.Computation · 12 min

    Let

    $$A=\left(\begin{array}{cccc}1& 2& 0& 1\\ 2& 4& 1& 3\\ -1& -2& 1& k\end{array}\right),$$

    where $k$ is a real parameter.

    1. (a)Row-reduce $A$ and find $\mathrm{rank}A$ and $\mathrm{nullity}A$ as functions of $k$. For the values of $k$ with $\mathrm{rank}A=3$, explain without further computation why $A\mathbf{x}=\mathbf{b}$ is consistent for every $\mathbf{b}\in {\mathbb{R}}^3$.
    2. (b)Now set $k=0$. Find a basis for each of $\mathrm{Col}(A)$, $\mathrm{Row}(A)$, $\mathrm{Null}(A)$ and $\mathrm{Null}(A^T)$, stating for each which recipe you are applying and to which matrix.
    3. (c)Still with $k=0$: verify by dot products that every basis vector of $\mathrm{Null}(A)$ is orthogonal to every basis vector of $\mathrm{Row}(A)$, and likewise for $\mathrm{Null}(A^T)$ against $\mathrm{Col}(A)$. Check that the four dimensions satisfy the counts of the four-fundamental-subspaces corollary.
    4. (d)Still with $k=0$: show that $\mathrm{Col}(A)$ is exactly the set of $\mathbf{b}=(b_1,b_2,b_3)$ with $3b_1-b_2+b_3=0$, using a dimension argument. Then decide, from that equation alone, whether $A\mathbf{x}=(1,3,0)$ and $A\mathbf{x}=(1,3,1)$ are consistent, and confirm the consistent one by finding a solution.
    Show the solution to question 11

    (a) Row-reduce. $R_2-2R_1$ gives $(0,0,1,1)$; $R_3+R_1$ gives $(0,0,1,k+1)$; then $R_3-R_2$ gives $(0,0,0,k)$:

    $$A\to \left(\begin{array}{cccc}1& 2& 0& 1\\ 0& 0& 1& 1\\ 0& 0& 0& k\end{array}\right).$$

    If $k\ne 0$ there are pivots in columns $1,3,4$: $\mathrm{rank}A=3$ and, by the Rank Theorem with $n=4$ columns, $\mathrm{nullity}A=1$. If $k=0$ the last row vanishes, pivots are in columns $1$ and $3$: $\mathrm{rank}A=2$ and $\mathrm{nullity}A=2$.

    When $\mathrm{rank}A=3$, $\mathrm{Col}(A)$ is a $3$-dimensional subspace of ${\mathbb{R}}^3$, hence equal to ${\mathbb{R}}^3$ by the subspace-dimension theorem. Every $\mathbf{b}$ lies in $\mathrm{Col}(A)$, and $A\mathbf{x}=\mathbf{b}$ is consistent exactly when $\mathbf{b}\in \mathrm{Col}(A)$.

    (b) The four subspaces for $k=0$. The echelon form is

    $$R=\left(\begin{array}{cccc}1& 2& 0& 1\\ 0& 0& 1& 1\\ 0& 0& 0& 0\end{array}\right),$$

    which is already reduced (the entry above the second pivot is $0$). Pivot columns $1$ and $3$; free columns $2$ and $4$.

    $\mathrm{Col}(A)$: the pivot-position columns of the original $A$, never of $R$:

    $$\{(1,2,-1),(0,1,1)\}.$$

    $\mathrm{Row}(A)$: the nonzero rows of $R$ (row operations preserve the row space):

    $$\{(1,2,0,1),(0,0,1,1)\}.$$

    $\mathrm{Null}(A)$: solve $R\mathbf{x}=0$ with $x_2=s$, $x_4=t$ free. The equations $x_1+2x_2+x_4=0$ and $x_3+x_4=0$ give $x_1=-2s-t$, $x_3=-t$, so $\mathbf{x}=s(-2,1,0,0)+t(-1,0,-1,1)$ and the basis is

    $$\{(-2,1,0,0),(-1,0,-1,1)\}.$$

    Check against the original $A$: $A(-1,0,-1,1)$ has entries $-1+0+0+1$, $-2+0-1+3$, $1+0-1+0$, all $0$ ✓; and $A(-2,1,0,0)$ has entries $-2+2$, $-4+4$, $2-2$, all $0$ ✓.

    $\mathrm{Null}(A^T)$: apply the null-space recipe to $A^T$, whose rows are the columns of $A$:

    $$A^T=\left(\begin{array}{ccc}1& 2& -1\\ 2& 4& -2\\ 0& 1& 1\\ 1& 3& 0\end{array}\right)\to \left(\begin{array}{ccc}1& 2& -1\\ 0& 1& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),$$

    via $R_2-2R_1$ (which zeroes row $2$), $R_4-R_1$ (giving $(0,1,1)$), then $R_4-R_3$. With $y_3=t$ free: $y_2=-t$ and $y_1=-2y_2+y_3=3t$, so the basis is

    $$\{(3,-1,1)\}.$$

    (c) Orthogonality. Rows against null vectors: $(1,2,0,1)\cdot (-2,1,0,0)=-2+2=0$; $(1,2,0,1)\cdot (-1,0,-1,1)=-1+1=0$; $(0,0,1,1)\cdot (-2,1,0,0)=0$; $(0,0,1,1)\cdot (-1,0,-1,1)=-1+1=0$ ✓. Columns against the left null vector: $(1,2,-1)\cdot (3,-1,1)=3-2-1=0$; $(0,1,1)\cdot (3,-1,1)=-1+1=0$ ✓.

    Dimensions: $\dim \mathrm{Col}(A)=\dim \mathrm{Row}(A)=2=r$; $\dim \mathrm{Null}(A)=2=n-r=4-2$; $\dim \mathrm{Null}(A^T)=1=m-r=3-2$. Row space and null space live in ${\mathbb{R}}^4$ with $2+2=4$; column space and left null space live in ${\mathbb{R}}^3$ with $2+1=3$ ✓.

    (d) Column space as an equation. Let $H=\{\mathbf{b}\in {\mathbb{R}}^3:3b_1-b_2+b_3=0\}$. This is the null space of the $1\times 3$ matrix $(3-11)$, which has rank $1$, so by the Rank Theorem $\dim H=3-1=2$. By (c) both basis vectors of $\mathrm{Col}(A)$ satisfy the equation, so $\mathrm{Col}(A)\subseteq H$. Both are $2$-dimensional, and a subspace of the same dimension is the whole space (subspace-dimension theorem), so $\mathrm{Col}(A)=H$.

    Hence $A\mathbf{x}=\mathbf{b}$ is consistent exactly when $3b_1-b_2+b_3=0$. For $(1,3,0)$: $3-3+0=0$, consistent. For $(1,3,1)$: $3-3+1=1\ne 0$, inconsistent.

    Confirmation: we need $c_1(1,2,-1)+c_2(0,1,1)=(1,3,0)$. The first coordinate gives $c_1=1$, the second gives $2+c_2=3$ so $c_2=1$, and the third checks: $-1+1=0$ ✓. Since those are columns $1$ and $3$ of $A$, one solution is $\mathbf{x}=(1,0,1,0)$; the full solution set is $(1,0,1,0)+\mathrm{Null}(A)$, a $2$-dimensional translate.

12. 12.Proof · 11 min

    All matrices are over $\mathbb{R}$. Write each argument in terms of the column space and null space; the Rank Theorem and the subspace-dimension theorem are the tools.

    1. (a)Let $A$ be $m\times n$ and $B$ be $n\times p$. Prove that $\mathrm{Col}(AB)\subseteq \mathrm{Col}(A)$ and $\mathrm{Null}(B)\subseteq \mathrm{Null}(AB)$, and deduce that $\mathrm{rank}(AB)\le \min (\mathrm{rank}A,\mathrm{rank}B)$.
    2. (b)Let $A$ be $n\times n$ with $A^2=0$. Prove that $\mathrm{Col}(A)\subseteq \mathrm{Null}(A)$, and deduce that $\mathrm{rank}A\le \frac{n}{2}$.
    3. (c)Give a nonzero $2\times 2$ matrix with $A^2=0$ and check the bound of (b). Then show that for every even $n$ there is an $n\times n$ matrix with $A^2=0$ and $\mathrm{rank}A=\frac{n}{2}$ exactly.
    4. (d)Use (a) to prove: if $A$ is $m\times n$ and $B$ is $n\times m$ with $m>n$, then $AB$ is not invertible. Illustrate with $m=2$, $n=1$.
    Show the solution to question 12

    (a) The inclusions. Take $\mathbf{y}\in \mathrm{Col}(AB)$, so $\mathbf{y}=(AB)\mathbf{x}$ for some $\mathbf{x}\in {\mathbb{R}}^p$. Then $\mathbf{y}=A(B\mathbf{x})$ is $A$ applied to the vector $B\mathbf{x}\in {\mathbb{R}}^n$, so $\mathbf{y}\in \mathrm{Col}(A)$. Hence $\mathrm{Col}(AB)\subseteq \mathrm{Col}(A)$.

    Take $\mathbf{x}\in \mathrm{Null}(B)$, so $B\mathbf{x}=0$. Then $(AB)\mathbf{x}=A(B\mathbf{x})=A0=0$, so $\mathbf{x}\in \mathrm{Null}(AB)$. Hence $\mathrm{Null}(B)\subseteq \mathrm{Null}(AB)$.

    The rank bound. By the subspace-dimension theorem a subspace has dimension at most that of any space containing it. From the first inclusion,

    $$\mathrm{rank}(AB)=\dim \mathrm{Col}(AB)\le \dim \mathrm{Col}(A)=\mathrm{rank}A.$$

    From the second, $\mathrm{nullity}B\le \mathrm{nullity}(AB)$. Both $B$ and $AB$ have $p$ columns, so the Rank Theorem applied to each gives

    $$\mathrm{rank}(AB)=p-\mathrm{nullity}(AB)\le p-\mathrm{nullity}B=\mathrm{rank}B.$$

    Together, $\mathrm{rank}(AB)\le \min (\mathrm{rank}A,\mathrm{rank}B)$. $■$

    (b) Take $\mathbf{y}\in \mathrm{Col}(A)$, so $\mathbf{y}=A\mathbf{x}$ for some $\mathbf{x}$. Then $A\mathbf{y}=A(A\mathbf{x})=A^2\mathbf{x}=0$, so $\mathbf{y}\in \mathrm{Null}(A)$. Hence $\mathrm{Col}(A)\subseteq \mathrm{Null}(A)$ (both are subspaces of ${\mathbb{R}}^n$, since $A$ is square).

    Write $r=\mathrm{rank}A$. By the subspace-dimension theorem $r=\dim \mathrm{Col}(A)\le \dim \mathrm{Null}(A)=\mathrm{nullity}A$, and by the Rank Theorem $\mathrm{nullity}A=n-r$. So $r\le n-r$, i.e. $2r\le n$, i.e. $\mathrm{rank}A\le \frac{n}{2}$. $■$

    In words: a matrix that squares to zero must send everything it produces into what it destroys, so its output cannot occupy more than half the space.

    (c) Take $A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. Then $A^2=\left(\begin{array}{cc}0\cdot 0+1\cdot 0& 0\cdot 1+1\cdot 0\\ 0& 0\end{array}\right)=0$, and $\mathrm{rank}A=1$ (one nonzero column). The bound says $1\le \frac{2}{2}=1$ — attained.

    For even $n$ write $n=2q$ and define $A$ on the standard basis by

    $$A{\mathbf{e}}_j=0(1\le j\le q),A{\mathbf{e}}_{q+j}={\mathbf{e}}_j(1\le j\le q);$$

    concretely $A=E_{1,q+1}+E_{2,q+2}+\cdots +E_{q,2q}$, the matrix whose only nonzero entries are $q$ ones on a diagonal shifted $q$ places to the right (for $n=4$: $A=E_{13}+E_{24}$, and for $n=2$ it is the matrix above). Then $A^2{\mathbf{e}}_j=A0=0$ and $A^2{\mathbf{e}}_{q+j}=A{\mathbf{e}}_j=0$, so $A^2$ kills every basis vector, and a matrix that sends every standard basis vector to $0$ has all columns zero: $A^2=0$. Its columns are $q$ zero columns followed by ${\mathbf{e}}_1,\dots ,{\mathbf{e}}_q$, which are independent, so $\mathrm{Col}(A)=\mathrm{span}\{{\mathbf{e}}_1,\dots ,{\mathbf{e}}_q\}$ and $\mathrm{rank}A=q=\frac{n}{2}$. The bound in (b) is sharp for every even $n$.

    (d) $AB$ is $m\times m$. By (a), $\mathrm{rank}(AB)\le \mathrm{rank}B$, and $\mathrm{rank}B=\dim \mathrm{Col}(B)\le \dim {\mathbb{R}}^n=n$ because $\mathrm{Col}(B)$ is a subspace of ${\mathbb{R}}^n$ (subspace-dimension theorem). So

    $$\mathrm{rank}(AB)\le n<m.$$

    By the Invertible Matrix Theorem, an $m\times m$ matrix is invertible if and only if its rank is $m$. The rank of $AB$ is less than $m$, so $AB$ is not invertible. $■$

    Illustration with $m=2$, $n=1$: $A=\left(\begin{array}{c}1\\ 1\end{array}\right)$, $B=\left(\begin{array}{cc}1& 1\end{array}\right)$, and

    $$AB=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right),$$

    whose rank is $1<2$ (the columns coincide) and whose determinant is $0$. The other product $BA=(2)$ is a perfectly good invertible $1\times 1$ matrix — the bound only constrains the larger product, and it is the shape $m>n$ that dooms it.