Print this page and work through it on paper. Revealed answers are never printed, so you can check one question and still print a clean sheet.

1. 1.Computation · 6 min

   For each map decide whether it is linear. If it is, verify the one-line linearity test $T(c\mathbf{u}+\mathbf{v})=cT(\mathbf{u})+T(\mathbf{v})$ for general inputs. If it is not, give one explicit numerical counterexample and name the axiom it breaks (additivity, or homogeneity with a stated scalar). In every case say first whether the origin test $T(0)=0$ settles the question.

   1. (a)$T:P_2\to {\mathbb{R}}^2$, $T(p)=(p(1),p^{\prime }(0))$.
   2. (b)$T:M_{2\times 2}\to \mathbb{R}$, $T(A)=\det A$.
   3. (c)$T:{\mathbb{R}}^2\to \mathbb{R}$, $T(x,y)=\max (x,y)$.
   Show the solution to question 1

   (a) Linear. The origin test passes — the zero polynomial has $p(1)=0$ and $p^{\prime }(0)=0$ — so it decides nothing; the test can only ever refute linearity. Run the one-line test. For polynomials $p,q$ and a scalar $c$, the definition of adding and scaling functions gives $(cp+q)(1)=cp(1)+q(1)$, and the sum and constant-multiple rules of differentiation give $(cp+q{)}^{\prime }=c{p}^{\prime }+q^{\prime }$, hence $(cp+q{)}^{\prime }(0)=c{p}^{\prime }(0)+q^{\prime }(0)$. Therefore

   $$T(cp+q)=(cp(1)+q(1),c{p}^{\prime }(0)+q^{\prime }(0))=c(p(1),p^{\prime }(0))+(q(1),q^{\prime }(0))=cT(p)+T(q).$$

   Sanity check with numbers: $p=t^2$ gives $T(p)=(1,0)$ and $q=1+t$ gives $T(q)=(2,1)$; linearity predicts $T(2p+q)=2(1,0)+(2,1)=(4,1)$, and directly $2p+q=1+t+2t^2$ has value $4$ at $1$ and derivative $1+4t$ equal to $1$ at $0$. ✓

   (b) Not linear. $\det$ of the zero matrix is $0$, so the origin test passes and says nothing. Homogeneity fails with $c=2$: $\det (2I)=\det \left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)=4$, whereas $2\det I=2$. (Additivity fails too: $\det \left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)+\det \left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)=0$, but the two matrices add to $I$ with determinant $1$.) One counterexample is enough. The determinant of a $2\times 2$ matrix scales like $c^2$, not like $c$.

   (c) Not linear. $\max (0,0)=0$, so again the origin test passes. Additivity fails: $T((1,0)+(0,1))=T(1,1)=1$, but $T(1,0)+T(0,1)=1+1=2$. Homogeneity fails as well, but only for negative scalars: $T(-1,0)=\max (-1,0)=0$ while $-T(1,0)=-1$. For $c>0$ we do have $\max (cx,cy)=c\max (x,y)$, so a check with positive scalars alone would have passed — the same trap as the absolute value.

2. 2.Sketch · 7 min

   Let $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ be the map that first applies the horizontal shear $H_{-1}$ (so ${\mathbf{e}}_2\mapsto (-1,1)$) and then reflects across the $x$-axis. A drawing with no computation beside it earns nothing: every vertex you plot must be justified.

   1. (a)Find the images of the four corners $0$, ${\mathbf{e}}_1$, ${\mathbf{e}}_1+{\mathbf{e}}_2$, ${\mathbf{e}}_2$ of the unit square, and write down the standard matrix of $T$.
   2. (b)On one pair of axes draw the unit square and its image, labelling each image vertex with the corner it came from. Walking the corners in the order $0,{\mathbf{e}}_1,{\mathbf{e}}_1+{\mathbf{e}}_2,{\mathbf{e}}_2$, does the image go round the same way as the square?
   3. (c)Compute $\det T$ and explain how its size and its sign are both visible in your drawing.
   4. (d)Compute the standard matrix of $T\circ T$. Which vertices of your drawing does $T$ swap, and which does it fix?
   Show the solution to question 2

   (a) Track the basis vectors through both steps. The shear $H_{-1}$ fixes ${\mathbf{e}}_1$ and sends ${\mathbf{e}}_2$ to $(-1,1)$; the reflection $F_x$ fixes $(1,0)$ and sends $(-1,1)$ to $(-1,-1)$. So $T({\mathbf{e}}_1)=(1,0)$ and $T({\mathbf{e}}_2)=(-1,-1)$, and stacking these as columns gives the standard matrix

   $$A=\left(\begin{array}{cc}1& -1\\ 0& -1\end{array}\right).$$

   As a product this is $F_x{H}_{-1}=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)$, the reflection on the left because it is applied second. ✓

   The origin goes to itself, and by linearity $T({\mathbf{e}}_1+{\mathbf{e}}_2)=T({\mathbf{e}}_1)+T({\mathbf{e}}_2)=(1,0)+(-1,-1)=(0,-1)$. The fourth corner was never free to choose. So the corners go

   $$(0,0)\mapsto (0,0),(1,0)\mapsto (1,0),(1,1)\mapsto (0,-1),(0,1)\mapsto (-1,-1).$$

   (b) The drawing. The square has corners $(0,0),(1,0),(1,1),(0,1)$. The image is the parallelogram with vertices $(0,0),(1,0),(0,-1),(-1,-1)$: it shares the bottom edge from $(0,0)$ to $(1,0)$ with the square, and its opposite edge is the segment from $(-1,-1)$ to $(0,-1)$ lying on the line $y=-1$. It leans to the left and sits entirely below the $x$-axis. Label $(1,0)$ as the image of ${\mathbf{e}}_1$, $(0,-1)$ as the image of ${\mathbf{e}}_1+{\mathbf{e}}_2$, and $(-1,-1)$ as the image of ${\mathbf{e}}_2$.

   Walking the square $(0,0)\to (1,0)\to (1,1)\to (0,1)$ goes counterclockwise. Walking the images $(0,0)\to (1,0)\to (0,-1)\to (-1,-1)$ goes clockwise. The orientation is reversed.

   (c) Determinant. $\det A=1\cdot (-1)-(-1)\cdot 0=-1$; equivalently $\det F_x\cdot \det H_{-1}=(-1)(1)$. Its size $|-1|=1$ is the area of the parallelogram: base $1$ along the $x$-axis and vertical height $1$ (the far edge lies on $y=-1$), so area $1\cdot 1=1$, unchanged. Its sign is the reversal of orientation found in (b): a shear preserves orientation and a reflection reverses it, so the composite reverses it.

   (d) $T\circ T$ is the identity.

   $$A^2=\left(\begin{array}{cc}1& -1\\ 0& -1\end{array}\right)\left(\begin{array}{cc}1& -1\\ 0& -1\end{array}\right)=\left(\begin{array}{cc}1& -1+1\\ 0& 1\end{array}\right)=I.$$

   The picture says the same thing. $T$ fixes $(0,0)$ and $(1,0)$. It sends ${\mathbf{e}}_2=(0,1)$ to $(-1,-1)$, and $A(-1,-1{)}^{\mathsf{T}}=(-1+1,1{)}^{\mathsf{T}}=(0,1{)}^{\mathsf{T}}$ sends it straight back; likewise $(1,1)\mapsto (0,-1)$ and $A(0,-1{)}^{\mathsf{T}}=(1,1{)}^{\mathsf{T}}$. So every vertex is either fixed or exchanged with its image, and applying $T$ twice restores the square. Geometrically $T$ is an oblique reflection in the $x$-axis: points on the axis stay put and everything else is flipped across it along the direction $(1,2)$, which $T$ reverses — $A(1,2{)}^{\mathsf{T}}=(1-2,-2{)}^{\mathsf{T}}=-(1,2{)}^{\mathsf{T}}$.

3. 3.Computation · 7 min

   A linear map $T:{\mathbb{R}}^3\to {\mathbb{R}}^2$ satisfies

   $$T(1,0,1)=(2,3),T(0,1,1)=(1,-1),T(1,1,0)=(0,4).$$

   1. (a)Show that the three input vectors form a basis of ${\mathbb{R}}^3$, and explain why this guarantees that $T$ is completely determined by the three values given.
   2. (b)Find $T({\mathbf{e}}_1)$, $T({\mathbf{e}}_2)$, $T({\mathbf{e}}_3)$ and hence the standard matrix $A$ of $T$. Check $A$ against all three pieces of data.
   3. (c)Compute $T(2,-1,3)$ twice: once from $A$ and once by expressing $(2,-1,3)$ in the given basis. Then decide whether $T$ is one-to-one and whether it is onto, naming the results you use.
   Show the solution to question 3

   (a) A basis, so $T$ is pinned down. Write ${\mathbf{u}}_1=(1,0,1)$, ${\mathbf{u}}_2=(0,1,1)$, ${\mathbf{u}}_3=(1,1,0)$. Three vectors in ${\mathbb{R}}^3$ form a basis exactly when the matrix with those columns is invertible:

   $$\det \left(\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 1& 1& 0\end{array}\right)=1(1\cdot 0-1\cdot 1)-0+1(0\cdot 1-1\cdot 1)=-1-1=-2\ne 0.$$

   So they are a basis. Every $\mathbf{x}\in {\mathbb{R}}^3$ is then a unique combination $a{\mathbf{u}}_1+b{\mathbf{u}}_2+c{\mathbf{u}}_3$, and because linear maps preserve linear combinations, $T(\mathbf{x})=aT({\mathbf{u}}_1)+bT({\mathbf{u}}_2)+cT({\mathbf{u}}_3)$ is forced. A linear map is determined by its values on a basis.

   (b) Express each ${\mathbf{e}}_j$ in the basis, then apply $T$. Solving $a{\mathbf{u}}_1+b{\mathbf{u}}_2+c{\mathbf{u}}_3=(p,q,r)$ means solving $a+c=p$, $b+c=q$, $a+b=r$. Adding the three equations gives $a+b+c=\frac{p+q+r}{2}$, and subtracting each original equation from that yields

   $$a=\frac{p-q+r}{2},b=\frac{-p+q+r}{2},c=\frac{p+q-r}{2}.$$

   For ${\mathbf{e}}_1=(1,0,0)$: $(a,b,c)=(\frac{1}{2},-\frac{1}{2},\frac{1}{2})$, so $T({\mathbf{e}}_1)=\frac{1}{2}(2,3)-\frac{1}{2}(1,-1)+\frac{1}{2}(0,4)=(\frac{1}{2},4)$.

   For ${\mathbf{e}}_2=(0,1,0)$: $(a,b,c)=(-\frac{1}{2},\frac{1}{2},\frac{1}{2})$, so $T({\mathbf{e}}_2)=-\frac{1}{2}(2,3)+\frac{1}{2}(1,-1)+\frac{1}{2}(0,4)=(-\frac{1}{2},0)$.

   For ${\mathbf{e}}_3=(0,0,1)$: $(a,b,c)=(\frac{1}{2},\frac{1}{2},-\frac{1}{2})$, so $T({\mathbf{e}}_3)=\frac{1}{2}(2,3)+\frac{1}{2}(1,-1)-\frac{1}{2}(0,4)=(\frac{3}{2},-1)$.

   Stack as columns:

   $$A=\left(\begin{array}{ccc}\frac{1}{2}& -\frac{1}{2}& \frac{3}{2}\\ 4& 0& -1\end{array}\right).$$

   Checks: $A{\mathbf{u}}_1=(\frac{1}{2}+\frac{3}{2},4-1)=(2,3)$ ✓; $A{\mathbf{u}}_2=(-\frac{1}{2}+\frac{3}{2},0-1)=(1,-1)$ ✓; $A{\mathbf{u}}_3=(\frac{1}{2}-\frac{1}{2},4+0)=(0,4)$ ✓.

   (c) Two routes to $T(2,-1,3)$. From the matrix: $A(2,-1,3{)}^{\mathsf{T}}=(1+\frac{1}{2}+\frac{9}{2},8-0-3{)}^{\mathsf{T}}=(6,5{)}^{\mathsf{T}}$.

   From the basis, with $(p,q,r)=(2,-1,3)$: $a=\frac{2+1+3}{2}=3$, $b=\frac{-2-1+3}{2}=0$, $c=\frac{2-1-3}{2}=-1$. Check: $3{\mathbf{u}}_1-{\mathbf{u}}_3=(3,0,3)-(1,1,0)=(2,-1,3)$ ✓. Then $T(2,-1,3)=3(2,3)+0(1,-1)-(0,4)=(6,5)$. ✓ The two routes agree.

   One-to-one: no. By the dimension obstruction, no linear map ${\mathbb{R}}^3\to {\mathbb{R}}^2$ is one-to-one, because three columns in ${\mathbb{R}}^2$ are always dependent. Concretely, rank-nullity gives nullity $3-2=1$; solving $A\mathbf{x}=0$ gives $z=4x$ and $y=13x$, so $\ker (T)=\mathrm{span}\{(1,13,4)\}$ — and indeed $A(1,13,4{)}^{\mathsf{T}}=(\frac{1}{2}-\frac{13}{2}+6,4-4{)}^{\mathsf{T}}=0$.

   Onto: yes. The columns $(\frac{1}{2},4)$ and $(-\frac{1}{2},0)$ are not multiples of each other, so they are independent and span ${\mathbb{R}}^2$; by the existence-and-uniqueness-via-columns theorem, $T$ is onto.

4. 4.Computation · 8 min

   Let

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

   and let $T:{\mathbb{R}}^4\to {\mathbb{R}}^3$ be $T(\mathbf{x})=A\mathbf{x}$.

   1. (a)Row reduce $A$. State $\mathrm{rank}(T)$ and $\mathrm{nullity}(T)$, give a basis for $\ker (T)$ and a basis for $\mathrm{im}(T)$, and check one vector of each basis directly against $A$.
   2. (b)Show that $\mathbf{b}=(3,5,0)$ lies in $\mathrm{im}(T)$, and describe the complete solution set of $T(\mathbf{x})=\mathbf{b}$ in the form ${\mathbf{v}}_0+\ker (T)$. Name the result that justifies this form, and verify one solution other than ${\mathbf{v}}_0$.
   3. (c)The pivot columns of your reduced matrix are $(1,0,0)$ and $(0,1,0)$. Prove that these two vectors do not span $\mathrm{im}(T)$, and say what this shows about reading an image basis off a reduced matrix.
   Show the solution to question 4

   (a) Row reduce. $R_2\to R_2-2R_1$ and $R_3\to R_3+R_1$ give rows $(1,-1,2,0)$, $(0,0,-1,1)$, $(0,0,3,-3)$. Then $R_3\to R_3+3R_2$ zeroes the last row. Scale $R_2\to -R_2$ and clear above the pivot with $R_1\to R_1-2R_2$:

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

   Pivots sit in columns $1$ and $3$, so $\mathrm{rank}(T)=2$; the free variables are $x_2,x_4$, so $\mathrm{nullity}(T)=2$. Check: $2+2=4=\dim {\mathbb{R}}^4$ ✓ (rank-nullity).

   Kernel: the reduced rows say $x_1=x_2-2x_4$ and $x_3=x_4$. Taking $(x_2,x_4)=(1,0)$ and $(0,1)$ gives the basis $(1,1,0,0)$ and $(-2,0,1,1)$. Check the second: $A(-2,0,1,1{)}^{\mathsf{T}}=(-2+2,-4+3+1,2+1-3{)}^{\mathsf{T}}=0$ ✓.

   Image: the pivot columns of the original $A$, namely $(1,2,-1)$ and $(2,3,1)$. The first is $A{\mathbf{e}}_1$, so it is certainly an output. ✓

   (b) The solution set is a translate of the kernel. Try $\mathbf{b}=\alpha (1,2,-1)+\beta (2,3,1)$: the third coordinate gives $-\alpha +\beta =0$, so $\beta =\alpha$; the first gives $3\alpha =3$, so $\alpha =\beta =1$; and the second checks, $2+3=5$ ✓. Hence $\mathbf{b}=A{\mathbf{e}}_1+A{\mathbf{e}}_3=A(1,0,1,0{)}^{\mathsf{T}}$, so $\mathbf{b}\in \mathrm{im}(T)$ with particular solution ${\mathbf{v}}_0=(1,0,1,0)$.

   By the proposition on the solution set of $T(\mathbf{x})=\mathbf{b}$ — every solution is the particular one plus something in the kernel — the complete solution set is

   $$(1,0,1,0)+s(1,1,0,0)+t(-2,0,1,1),s,t\in \mathbb{R}.$$

   Verify $s=t=1$: $\mathbf{x}=(0,1,2,1)$ and $A\mathbf{x}=(-1+4,-2+6+1,1+2-3)=(3,5,0)$ ✓.

   (c) Row operations move the column space. $\mathrm{im}(T)$ is the plane spanned by ${\mathbf{c}}_1=(1,2,-1)$ and ${\mathbf{c}}_3=(2,3,1)$. Suppose $(1,0,0)=\alpha {\mathbf{c}}_1+\beta {\mathbf{c}}_3$. The third coordinate gives $\beta =\alpha$, the first then gives $3\alpha =1$, so $\alpha =\beta =\frac{1}{3}$; but the second coordinate would be $2\alpha +3\beta =\frac{5}{3}\ne 0$. Contradiction, so $(1,0,0)\notin \mathrm{im}(T)$, and a set containing it cannot be a basis of the image.

   A cleaner way to see the whole plane: a normal vector is ${\mathbf{c}}_1\times {\mathbf{c}}_3=(2\cdot 1-(-1)\cdot 3,(-1)\cdot 2-1\cdot 1,1\cdot 3-2\cdot 2)=(5,-3,-1)$, so $\mathrm{im}(T)=\{5x-3y-z=0\}$. Then $\mathbf{b}=(3,5,0)$ satisfies $15-15-0=0$ ✓, and $(1,0,0)$ gives $5\ne 0$.

   The lesson: row operations preserve the dependence relations among the columns — which is why they correctly identify columns $1$ and $3$ as the ones to keep — but they change the columns themselves, and the reduced columns span a different subspace of ${\mathbb{R}}^3$ (here the $xy$-plane). Take the pivot columns from the original matrix.

5. 5.Proof · 10 min

   Throughout, $V$ is finite-dimensional with $\dim V=n$. Quote the rank-nullity theorem by name each time you use it, and say which map and which domain you are applying it to.

   1. (a)Let $T:V\to V$ be linear with $T\circ T=0$. Prove that $\mathrm{im}(T)\subseteq \ker (T)$, and deduce that $\mathrm{rank}(T)\le n/2$.
   2. (b)Write down an operator $T$ on ${\mathbb{R}}^4$ with $T\circ T=0$ and $\mathrm{rank}(T)=2$, verify both claims, and say what this shows about the bound in (a).
   3. (c)Let $T:U\to V$ and $S:V\to W$ be linear, with $U$ finite-dimensional. Prove that $\mathrm{rank}(S\circ T)\ge \mathrm{rank}(T)-\mathrm{nullity}(S)$. (Apply rank-nullity to the restriction of $S$ to the subspace $\mathrm{im}(T)$.)
   Show the solution to question 5

   (a) The image sits inside the kernel. Let $\mathbf{w}\in \mathrm{im}(T)$, say $\mathbf{w}=T(\mathbf{v})$. Then $T(\mathbf{w})=T(T(\mathbf{v}))=(T\circ T)(\mathbf{v})=0$, so $\mathbf{w}\in \ker (T)$. Hence $\mathrm{im}(T)\subseteq \ker (T)$.

   A subspace has dimension at most that of the space containing it, so $\mathrm{rank}(T)=\dim \mathrm{im}(T)\le \dim \ker (T)=\mathrm{nullity}(T)$. Now apply rank-nullity to $T$ on its domain $V$:

   $$n=\mathrm{rank}(T)+\mathrm{nullity}(T)\ge \mathrm{rank}(T)+\mathrm{rank}(T)=2\mathrm{rank}(T),$$

   so $\mathrm{rank}(T)\le n/2$. $■$

   (b) The bound is sharp. Take $T(x_1,x_2,x_3,x_4)=(x_3,x_4,0,0)$, whose standard matrix has $1$'s in positions $(1,3)$ and $(2,4)$ and zeros elsewhere. Then $T(T(\mathbf{x}))=T(x_3,x_4,0,0)=(0,0,0,0)$, so $T\circ T=0$. The image is $\{(a,b,0,0)\}=\mathrm{span}\{{\mathbf{e}}_1,{\mathbf{e}}_2\}$, of dimension $2$, so $\mathrm{rank}(T)=2=4/2$. The bound in (a) is attained, so it cannot be improved. Note that here $\ker (T)=\{x_3=x_4=0\}=\mathrm{span}\{{\mathbf{e}}_1,{\mathbf{e}}_2\}$ as well: equality in (a) happens exactly when $\mathrm{im}(T)=\ker (T)$.

   (c) Restrict $S$ to the image of $T$. Let $S^{\prime }$ be the restriction of $S$ to the subspace $\mathrm{im}(T)$, a linear map $S^{\prime }:\mathrm{im}(T)\to W$ (restricting a linear map to a subspace keeps both axioms). Its domain is finite-dimensional: by rank-nullity for $T$ on $U$, $\dim \mathrm{im}(T)=\dim U-\mathrm{nullity}(T)<\infty$.

   Its image is $S(\mathrm{im}T)=\{S(T(\mathbf{u})):\mathbf{u}\in U\}=\mathrm{im}(S\circ T)$, of dimension $\mathrm{rank}(S\circ T)$.

   Its kernel is $\{\mathbf{w}\in \mathrm{im}(T):S(\mathbf{w})=0\}=\mathrm{im}(T)\cap \ker (S)$, a subspace of $\ker (S)$, so its dimension is at most $\mathrm{nullity}(S)$.

   Rank-nullity for $S^{\prime }$ on the domain $\mathrm{im}(T)$ now reads

   $$\mathrm{rank}(T)=\dim \mathrm{im}(T)=\dim (\mathrm{im}(T)\cap \ker (S))+\mathrm{rank}(S\circ T)\le \mathrm{nullity}(S)+\mathrm{rank}(S\circ T),$$

   and rearranging gives $\mathrm{rank}(S\circ T)\ge \mathrm{rank}(T)-\mathrm{nullity}(S)$. $■$

   Sanity check on (b) with $S=T$: the inequality predicts $\mathrm{rank}(T^2)\ge 2-2=0$, and indeed $T^2=0$. The inequality is the lower half of the familiar $\mathrm{rank}(S\circ T)\le \min \{\mathrm{rank}S,\mathrm{rank}T\}$: composing can lose rank, but at most $\mathrm{nullity}(S)$ of it.

6. 6.Computation · 9 min

   Let $T:P_2\to P_2$ be the forward-difference operator $T(p)(t)=p(t+1)-p(t)$, and let $\mathcal{B}=\{1,t,t^2\}$.

   1. (a)Explain briefly why $T$ is linear and why its output really lies in $P_2$. Then compute $[T{]}_{\mathcal{B}}$.
   2. (b)Use the matrix to compute $T(3-2t+5t^2)$, and confirm the answer by direct substitution.
   3. (c)Read $\ker (T)$ and $\mathrm{im}(T)$ off the matrix, describe both as sets of polynomials, and check rank-nullity.
   4. (d)Now let $\mathcal{C}=\{1,1+t,1+t+t^2\}$. Compute $[T{]}_{\mathcal{C}}^{\mathcal{B}}$, and point to the single column that changed and why.
   Show the solution to question 6

   (a) Linearity and the matrix. For polynomials $p,q$ and a scalar $c$,

   $$T(cp+q)(t)=(cp+q)(t+1)-(cp+q)(t)=c(p(t+1)-p(t))+(q(t+1)-q(t))=(cT(p)+T(q))(t),$$

   which is the one-line test. If $\mathrm{deg}p\le 2$ then $p(t+1)$ is again a polynomial of degree at most $2$ (expand the binomials), and a difference of two such polynomials has degree at most $2$, so $T(p)\in P_2$.

   Apply $T$ to each basis vector and read off $\mathcal{B}$-coordinates:

   $$T(1)=1-1=0\to (0,0,0),T(t)=(t+1)-t=1\to (1,0,0),T(t^2)=(t+1{)}^2-t^2=2t+1\to (1,2,0).$$

   Stacking as columns,

   $$[T{]}_{\mathcal{B}}=\left(\begin{array}{ccc}0& 1& 1\\ 0& 0& 2\\ 0& 0& 0\end{array}\right).$$

   (b) Differencing a specific polynomial. $[3-2t+5t^2{]}_{\mathcal{B}}=(3,-2,5{)}^{\mathsf{T}}$, and

   $$\left(\begin{array}{ccc}0& 1& 1\\ 0& 0& 2\\ 0& 0& 0\end{array}\right)\left(\begin{array}{c}3\\ -2\\ 5\end{array}\right)=\left(\begin{array}{c}-2+5\\ 10\\ 0\end{array}\right)=\left(\begin{array}{c}3\\ 10\\ 0\end{array}\right),$$

   which decodes as $3+10t$. Directly: $p(t+1)=3-2(t+1)+5(t+1{)}^2=3-2t-2+5t^2+10t+5=5t^2+8t+6$, and subtracting $p(t)=5t^2-2t+3$ leaves $10t+3$. ✓

   (c) Kernel and image from the matrix. The system $[T{]}_{\mathcal{B}}\mathbf{c}=0$ says $c_2+c_3=0$ and $2c_3=0$, so $c_2=c_3=0$ with $c_1$ free: $\ker (T)=\mathrm{span}\{1\}$, the constant polynomials, and $\mathrm{nullity}(T)=1$. (These are the only polynomials with $p(t+1)=p(t)$ for all $t$.) The image is the column space, spanned by $(1,0,0)$ and $(1,2,0)$, which is $\{(a,b,0)\}$ — decoded, the polynomials $a+bt$, so $\mathrm{im}(T)=P_1$ and $\mathrm{rank}(T)=2$. Rank-nullity: $1+2=3=\dim P_2$ ✓. Like differentiation, differencing lowers the degree by one and kills exactly the constants.

   (d) Same map, new output basis. The outputs $T(1)=0$, $T(t)=1$, $T(t^2)=1+2t$ are unchanged; only step two of the recipe — express them in $\mathcal{C}$ — is redone. $0$ has coordinates $(0,0,0)$ in any basis. $1$ is the first vector of $\mathcal{C}$, so $[1{]}_{\mathcal{C}}=(1,0,0)$. For $1+2t=a\cdot 1+b(1+t)+c(1+t+t^2)$: the $t^2$ coefficient forces $c=0$, the $t$ coefficient gives $b=2$, and the constant term gives $a+b=1$, so $a=-1$. Thus $[1+2t{]}_{\mathcal{C}}=(-1,2,0)$, and

   $$[T{]}_{\mathcal{C}}^{\mathcal{B}}=\left(\begin{array}{ccc}0& 1& -1\\ 0& 0& 2\\ 0& 0& 0\end{array}\right).$$

   Check: $-1\cdot 1+2(1+t)=1+2t$ ✓. Only the third column changed, because $0$ and $1$ happen to have the same coordinates in $\mathcal{B}$ and $\mathcal{C}$ (both bases start with $1$) while $1+2t$ does not. Writing $T(t^2)$ itself as the column, $(1,2,0)$, would be the classic error: the column must be the $\mathcal{C}$-coordinates, not the vector.

7. 7.Computation · 10 min

   Let $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ be $T(x,y)=(4x-2y,x+y)$, with standard matrix $A$, and let $\mathcal{B}=\{(1,1),(2,1)\}$.

   1. (a)Compute $[T{]}_{\mathcal{B}}$ directly from the definition, by applying $T$ to each basis vector and expressing the result in $\mathcal{B}$.
   2. (b)Write down $P=P_{\mathcal{E}\leftarrow \mathcal{B}}$, compute $P^{-1}$, and verify $P^{-1}AP=[T{]}_{\mathcal{B}}$ by explicit multiplication. Confirm that trace and determinant agree between $A$ and $[T{]}_{\mathcal{B}}$, and name the theorem that says they must.
   3. (c)Find $[\mathbf{v}{]}_{\mathcal{B}}$ for $\mathbf{v}=(5,4)$, use $[T{]}_{\mathcal{B}}$ to obtain $[T(\mathbf{v}){]}_{\mathcal{B}}$, decode, and compare with $T(5,4)$ computed directly.
   4. (d)Compute $A^5$ exactly. Check your answer against the trace and determinant that $[T{]}_{\mathcal{B}}$ predicts for $A^5$.
   Show the solution to question 7

   (a) Directly from the definition. $T(1,1)=(4-2,1+1)=(2,2)=2(1,1)+0(2,1)$, so the first column is $(2,0{)}^{\mathsf{T}}$. $T(2,1)=(8-2,2+1)=(6,3)=0(1,1)+3(2,1)$, so the second column is $(0,3{)}^{\mathsf{T}}$. Hence

   $$[T{]}_{\mathcal{B}}=\left(\begin{array}{cc}2& 0\\ 0& 3\end{array}\right).$$

   The basis was chosen so that $T$ merely stretches each basis vector — by $2$ along $(1,1)$ and by $3$ along $(2,1)$.

   (b) The change-of-basis formula. The standard matrix is $A=\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right)$ (one row per output component). In ${\mathbb{R}}^2$ the matrix $P_{\mathcal{E}\leftarrow \mathcal{B}}$ is just the $\mathcal{B}$ vectors as columns:

   $$P=\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right),\det P=1-2=-1,P^{-1}=\frac{1}{-1}\left(\begin{array}{cc}1& -2\\ -1& 1\end{array}\right)=\left(\begin{array}{cc}-1& 2\\ 1& -1\end{array}\right).$$

   (Check: $P{P}^{-1}=\left(\begin{array}{cc}-1+2& 2-2\\ -1+1& 2-1\end{array}\right)=I$ ✓.) Multiply right to left, the direction a vector travels:

   $$AP=\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right)\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}2& 6\\ 2& 3\end{array}\right),P^{-1}(AP)=\left(\begin{array}{cc}-1& 2\\ 1& -1\end{array}\right)\left(\begin{array}{cc}2& 6\\ 2& 3\end{array}\right)=\left(\begin{array}{cc}-2+4& -6+6\\ 2-2& 6-3\end{array}\right)=\left(\begin{array}{cc}2& 0\\ 0& 3\end{array}\right).$$

   ✓ This matches (a), as the change-of-basis theorem for an operator, $[T{]}_{\mathcal{B}}=P^{-1}[T{]}_{\mathcal{E}}P$, requires. Invariants: $\mathrm{tr}A=4+1=5=2+3$ and $\det A=4+2=6=2\cdot 3$. They must agree by the similarity invariants theorem, since $A$ and $[T{]}_{\mathcal{B}}$ are similar.

   (c) Chasing one vector. Getting into $\mathcal{B}$-coordinates costs the inverse: $[\mathbf{v}{]}_{\mathcal{B}}=P^{-1}\mathbf{v}=(-5+8,5-4{)}^{\mathsf{T}}=(3,1{)}^{\mathsf{T}}$. Check: $3(1,1)+1(2,1)=(5,4)$ ✓. Then $[T(\mathbf{v}){]}_{\mathcal{B}}=[T{]}_{\mathcal{B}}[\mathbf{v}{]}_{\mathcal{B}}=(2\cdot 3,3\cdot 1{)}^{\mathsf{T}}=(6,3{)}^{\mathsf{T}}$, which decodes as $6(1,1)+3(2,1)=(12,9)$. Directly, $T(5,4)=(20-8,5+4)=(12,9)$. ✓

   (d) Powers through the friendly basis. From $D=P^{-1}AP$ with $D=\mathrm{diag}(2,3)$ we get $A=PD{P}^{-1}$, and by part (5) of the invariants theorem $A^5=P{D}^5{P}^{-1}$ with $D^5=\mathrm{diag}(32,243)$. First

   $$P{D}^5=\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)\left(\begin{array}{cc}32& 0\\ 0& 243\end{array}\right)=\left(\begin{array}{cc}32& 486\\ 32& 243\end{array}\right),$$

   then

   $$A^5=\left(\begin{array}{cc}32& 486\\ 32& 243\end{array}\right)\left(\begin{array}{cc}-1& 2\\ 1& -1\end{array}\right)=\left(\begin{array}{cc}-32+486& 64-486\\ -32+243& 64-243\end{array}\right)=\left(\begin{array}{cc}454& -422\\ 211& -179\end{array}\right).$$

   Checks. $A^5$ is similar to $D^5$, so its trace must be $32+243=275$: indeed $454-179=275$ ✓. Its determinant must be $32\cdot 243=6^5=7776$: $454\cdot (-179)-(-422)\cdot 211=-81266+89042=7776$ ✓. And $A^5(1,1{)}^{\mathsf{T}}$ should be $2^5(1,1{)}^{\mathsf{T}}$: $454-422=32$ and $211-179=32$ ✓.

8. 8.Critique · 9 min

   Let $T:{\mathbb{R}}^3\to {\mathbb{R}}^2$, $T(x,y,z)=(x+y,y+z)$, and $S:{\mathbb{R}}^2\to {\mathbb{R}}^3$, $S(a,b)=(a,b,a-b)$. A student writes the following.

   Step 1. $M_T=\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\end{array}\right)$ and $M_S=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 1& -1\end{array}\right)$.

   Step 2. The matrix of $S\circ T$ is $M_T{M}_S=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$, multiplying in the order the maps are written.

   Step 3. $\det \left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)=-1\ne 0$, so $S\circ T$ is invertible.

   Step 4. Therefore $T$ is one-to-one: if $T(\mathbf{u})=T(\mathbf{v})$ then $S(T(\mathbf{u}))=S(T(\mathbf{v}))$, and $S\circ T$ invertible forces $\mathbf{u}=\mathbf{v}$.

   Step 1 is correct.

   1. (a)Identify every error, naming the theorem or rule each one violates. One of the remaining steps is a valid inference from a false premise — say which, and explain why its conclusion is nevertheless false.
   2. (b)Compute the correct standard matrix of $S\circ T$, find its rank, and give a basis for $\ker (S\circ T)$.
   3. (c)Decide, with proof, which of the four statements hold: $S$ is one-to-one; $S$ is onto; $T$ is one-to-one; $T$ is onto. Use the matrix $M_T{M}_S$ for the two that are true.
   Show the solution to question 8

   (a) Where it goes wrong. Step 2 violates the composition theorem, $[S\circ T]=[S][T]$: the map applied first sits nearest the vector, on the right, so the matrix of $S\circ T$ is $M_S{M}_T$, not $M_T{M}_S$. The product the student formed, $M_T{M}_S$, is the matrix of $T\circ S:{\mathbb{R}}^2\to {\mathbb{R}}^2$. The shapes alone should have raised the alarm: $S\circ T$ is a map ${\mathbb{R}}^3\to {\mathbb{R}}^3$ and needs a $3\times 3$ matrix, and a $2\times 2$ determinant cannot be its determinant.

   Step 3 is arithmetically right but is about the wrong map: it proves that $T\circ S$ is invertible (a determinant test is legitimate for an operator on ${\mathbb{R}}^2$, by the invertibility criteria), and says nothing about $S\circ T$.

   Step 4 is the valid inference from a false premise. The argument "if $S\circ T$ is one-to-one then $T$ is one-to-one" is correct exactly as written. But its premise is false — see (b) — so the conclusion is unsupported, and in fact it is false: by the dimension obstruction, no linear map ${\mathbb{R}}^3\to {\mathbb{R}}^2$ is one-to-one, since three columns in ${\mathbb{R}}^2$ are always dependent.

   (b) The correct composite.

   $$M_{S\circ T}=M_S{M}_T=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 1& -1\end{array}\right)\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\end{array}\right)=\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 1\\ 1& 0& -1\end{array}\right).$$

   The third row is the first minus the second, so the rows are dependent and $\det =0$; the first two rows are independent, so the rank is $2$ and, by rank-nullity on ${\mathbb{R}}^3$, the nullity is $1$. Solving $M_{S\circ T}\mathbf{x}=0$: $x+y=0$ and $y+z=0$ (the third equation is redundant), so $x=-y$, $z=-y$, and

   $$\ker (S\circ T)=\mathrm{span}\{(1,-1,1)\}.$$

   Check: $(1-1,-1+1,1-1)=0$ ✓. So $S\circ T$ is not invertible, and Step 3's conclusion was false.

   (c) The four statements. The student's product is the matrix of $T\circ S$: $M_T{M}_S=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$, with determinant $-1\ne 0$, so $T\circ S$ is an invertible operator on ${\mathbb{R}}^2$.

   $S$ is one-to-one — true. If $S(\mathbf{a})=S(\mathbf{b})$ then $T(S(\mathbf{a}))=T(S(\mathbf{b}))$, and $T\circ S$ one-to-one gives $\mathbf{a}=\mathbf{b}$. (Also visible directly: the columns $(1,0,1)$ and $(0,1,-1)$ of $M_S$ are independent.)

   $T$ is onto — true. Given $\mathbf{w}\in {\mathbb{R}}^2$, put $\mathbf{a}=(T\circ S{)}^{-1}(\mathbf{w})$; then $T(S(\mathbf{a}))=\mathbf{w}$, so $\mathbf{w}\in \mathrm{im}(T)$.

   $T$ is one-to-one — false, by the dimension obstruction ($3>2$); explicitly $T(1,-1,1)=(0,0)$. Because $S$ is one-to-one, $\ker (S\circ T)=\ker (T)$, which is why the kernel in (b) is exactly this line.

   $S$ is onto — false, by the dimension obstruction ($2<3$); explicitly $\mathrm{im}(S)=\{(a,b,a-b)\}$ is the plane $z=x-y$, and $(0,0,1)$ is not on it.

   The moral: a composite that is invertible tells you the inner map is one-to-one and the outer map is onto — and nothing more.

9. 9.Proof · 10 min

   Let $V$ and $W$ be vector spaces with $\dim V=n\ge 1$ and $\dim W=m$, and write $\mathcal{L}(V,W)$ for the space of linear maps $V\to W$ with the pointwise operations. You may use that $\dim \mathcal{L}(V,W)=mn$, and that a linear map may be prescribed arbitrarily on a basis and extended by linearity.

   1. (a)Fix a nonzero ${\mathbf{v}}_0\in V$ and let $U=\{T\in \mathcal{L}(V,W):T({\mathbf{v}}_0)=0\}$. Prove that $U$ is a subspace of $\mathcal{L}(V,W)$ and that $\dim U=m(n-1)$.
   2. (b)Fix $T\in \mathcal{L}(V)$ and define ${\Phi }_T:\mathcal{L}(V)\to \mathcal{L}(V)$ by ${\Phi }_T(S)=T\circ S$. Prove that ${\Phi }_T$ is linear, and that ${\Phi }_T$ is one-to-one if and only if $T$ is one-to-one.
   Show the solution to question 9

   (a) $U$ is a kernel, and evaluation is onto. Define the evaluation map $E:\mathcal{L}(V,W)\to W$ by $E(T)=T({\mathbf{v}}_0)$. It is linear because the operations on $\mathcal{L}(V,W)$ are pointwise:

   $$E(cS+T)=(cS+T)({\mathbf{v}}_0)=cS({\mathbf{v}}_0)+T({\mathbf{v}}_0)=cE(S)+E(T).$$

   By definition $U=\ker (E)$, and the kernel of a linear map is a subspace. That proves the first claim.

   For the dimension, compute the rank of $E$. Since ${\mathbf{v}}_0\ne 0$, the set $\{{\mathbf{v}}_0\}$ is linearly independent and extends to a basis $\{{\mathbf{v}}_0,{\mathbf{b}}_2,\dots ,{\mathbf{b}}_n\}$ of $V$. Given any $\mathbf{w}\in W$, prescribe $T({\mathbf{v}}_0)=\mathbf{w}$ and $T({\mathbf{b}}_j)=0$ for $j\ge 2$, and extend by linearity; then $E(T)=\mathbf{w}$. So $E$ is onto and $\mathrm{rank}(E)=m$. Rank-nullity for $E$ on the domain $\mathcal{L}(V,W)$ gives

   $$mn=\dim \mathcal{L}(V,W)=\mathrm{nullity}(E)+\mathrm{rank}(E)=\dim U+m,$$

   so $\dim U=mn-m=m(n-1)$. $■$

   The count agrees with the knob-counting picture: a map in $U$ is free to send each of ${\mathbf{b}}_2,\dots ,{\mathbf{b}}_n$ anywhere in $W$ — that is $n-1$ choices of $m$ numbers — while its value on ${\mathbf{v}}_0$ is fixed.

   (b) Left composition with $T$. Linearity. For $S_1,S_2\in \mathcal{L}(V)$, a scalar $c$, and any $\mathbf{v}\in V$,

   $${\Phi }_T(c{S}_1+S_2)(\mathbf{v})=T(c{S}_1(\mathbf{v})+S_2(\mathbf{v}))=cT(S_1(\mathbf{v}))+T(S_2(\mathbf{v}))=(c{\Phi }_T(S_1)+{\Phi }_T(S_2))(\mathbf{v}),$$

   using linearity of $T$. Two maps agreeing at every $\mathbf{v}$ are equal, so ${\Phi }_T(c{S}_1+S_2)=c{\Phi }_T(S_1)+{\Phi }_T(S_2)$.

   ($\Leftarrow$) If $T$ is one-to-one, so is ${\Phi }_T$. Since ${\Phi }_T$ is linear it suffices to show its kernel is trivial. Suppose ${\Phi }_T(S)=T\circ S=0$. Then for every $\mathbf{v}$, $T(S(\mathbf{v}))=0$, so $S(\mathbf{v})\in \ker (T)=\{0\}$ by the trivial-kernel proposition. Hence $S(\mathbf{v})=0$ for all $\mathbf{v}$, i.e. $S=0$.

   ($\Rightarrow$) If $T$ is not one-to-one, neither is ${\Phi }_T$. Pick ${\mathbf{v}}_0\ne 0$ with $T({\mathbf{v}}_0)=0$, and any basis $\{{\mathbf{b}}_1,\dots ,{\mathbf{b}}_n\}$ of $V$. Prescribe $S({\mathbf{b}}_1)={\mathbf{v}}_0$ and $S({\mathbf{b}}_j)=0$ for $j\ge 2$, extended by linearity. Then $S\ne 0$, because $S({\mathbf{b}}_1)\ne 0$. But $(T\circ S)({\mathbf{b}}_j)$ equals $T({\mathbf{v}}_0)=0$ for $j=1$ and $T(0)=0$ otherwise, and a linear map that vanishes on a basis is the zero map, so $T\circ S=0$. Thus $S$ is a nonzero element of $\ker ({\Phi }_T)$, and ${\Phi }_T$ is not one-to-one. $■$

   Remark: ${\Phi }_T$ is an operator on the finite-dimensional space $\mathcal{L}(V)$, so by the square dichotomy it is one-to-one iff onto iff invertible; the proof shows all three happen exactly when $T$ is invertible, and then ${\Phi }_T^{-1}(S)=T^{-1}\circ S$.

10. 10.Proof · 10 min

    Let $T:V\to V$ be a linear operator on a finite-dimensional space, and write $T^2=T\circ T$. You may use that for subspaces $X,Y$ of $V$, $\dim (X+Y)=\dim X+\dim Y-\dim (X\cap Y)$, and that a subspace of $V$ with the same dimension as $V$ is all of $V$.

    1. (a)Prove that $\ker (T)\subseteq \ker (T^2)$ and $\mathrm{im}(T^2)\subseteq \mathrm{im}(T)$.
    2. (b)Prove that $\ker (T)=\ker (T^2)$ if and only if $\mathrm{im}(T)=\mathrm{im}(T^2)$.
    3. (c)Prove that if $\ker (T)=\ker (T^2)$ then $\ker (T)\cap \mathrm{im}(T)=\{0\}$, and deduce that $\ker (T)+\mathrm{im}(T)=V$.
    4. (d)Give an operator on ${\mathbb{R}}^2$ for which $\ker (T)\ne \ker (T^2)$, and show that both conclusions of (c) fail for it.
    Show the solution to question 10

    (a) Two inclusions. If $\mathbf{v}\in \ker (T)$ then $T^2(\mathbf{v})=T(T(\mathbf{v}))=T(0)=0$, because a linear map fixes the origin; so $\mathbf{v}\in \ker (T^2)$. If $\mathbf{w}\in \mathrm{im}(T^2)$, say $\mathbf{w}=T(T(\mathbf{v}))$, then $\mathbf{w}=T(\mathbf{u})$ with $\mathbf{u}=T(\mathbf{v})$, so $\mathbf{w}\in \mathrm{im}(T)$. $■$

    (b) Equal kernels iff equal images. Let $n=\dim V$. Both $T$ and $T^2$ have domain $V$, so rank-nullity gives

    $$n=\mathrm{nullity}(T)+\mathrm{rank}(T)=\mathrm{nullity}(T^2)+\mathrm{rank}(T^2),$$

    hence $\mathrm{nullity}(T)=\mathrm{nullity}(T^2)$ if and only if $\mathrm{rank}(T)=\mathrm{rank}(T^2)$.

    Now use (a). A subspace $X\subseteq Y$ with $\dim X=\dim Y$ equals $Y$, because a basis of $X$ is an independent set in $Y$ of size $\dim Y$, hence a basis of $Y$. Applied to $\ker (T)\subseteq \ker (T^2)$: equality holds iff the nullities agree. Applied to $\mathrm{im}(T^2)\subseteq \mathrm{im}(T)$: equality holds iff the ranks agree. Chaining,

    $$\ker (T)=\ker (T^2)⟺\mathrm{nullity}(T)=\mathrm{nullity}(T^2)⟺\mathrm{rank}(T)=\mathrm{rank}(T^2)⟺\mathrm{im}(T)=\mathrm{im}(T^2).■$$

    (c) Kernel and image become complementary. Assume $\ker (T)=\ker (T^2)$ and let $\mathbf{v}\in \ker (T)\cap \mathrm{im}(T)$. Being in the image, $\mathbf{v}=T(\mathbf{u})$ for some $\mathbf{u}$; being in the kernel, $T(\mathbf{v})=0$. So $T^2(\mathbf{u})=T(\mathbf{v})=0$, which puts $\mathbf{u}\in \ker (T^2)=\ker (T)$, and therefore $\mathbf{v}=T(\mathbf{u})=0$. Hence $\ker (T)\cap \mathrm{im}(T)=\{0\}$.

    For the sum, use the dimension formula with the intersection now zero-dimensional, then rank-nullity for $T$ on $V$:

    $$\dim (\ker (T)+\mathrm{im}(T))=\mathrm{nullity}(T)+\mathrm{rank}(T)-0=n=\dim V.$$

    A subspace of $V$ of dimension $\dim V$ is $V$ itself, so $\ker (T)+\mathrm{im}(T)=V$. $■$ (Together, the two conclusions say $V=\ker (T)\oplus \mathrm{im}(T)$: every vector splits uniquely into a part that $T$ destroys and a part that $T$ produced.)

    (d) A counterexample when the kernels differ. Take $T(x,y)=(y,0)$, with standard matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. Then $\ker (T)=\{y=0\}=\mathrm{span}\{{\mathbf{e}}_1\}$, while $T^2(x,y)=T(y,0)=(0,0)$, so $T^2=0$ and $\ker (T^2)={\mathbb{R}}^2\ne \ker (T)$.

    The image is $\mathrm{im}(T)=\{(y,0)\}=\mathrm{span}\{{\mathbf{e}}_1\}$, the same line as the kernel. So $\ker (T)\cap \mathrm{im}(T)=\mathrm{span}\{{\mathbf{e}}_1\}\ne \{0\}$ and $\ker (T)+\mathrm{im}(T)=\mathrm{span}\{{\mathbf{e}}_1\}\ne {\mathbb{R}}^2$: both conclusions of (c) fail. Consistent with (b), the images differ too: $\mathrm{im}(T^2)=\{0\}\ne \mathrm{im}(T)$. Rank-nullity still holds for each map separately ($1+1=2$ and $2+0=2$); what fails is only the splitting.

11. 11.Critique · 8 min

    Two arguments about similarity follow. Each reaches a false conclusion.

    Argument 1. “$A=\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)$ and $B=\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)$ have the same trace $4$, the same determinant $4$, the same rank $2$ and the same characteristic polynomial $(\lambda -2{)}^2$. By the similarity invariants theorem, $A\sim B$.”

    Argument 2. “$A=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ and $B=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$ are not similar. Similar matrices represent the same operator, but $A$ is the horizontal shear $(x,y)\mapsto (x+y,y)$ and $B$ is the vertical shear $(x,y)\mapsto (x,x+y)$, and these are different operators on ${\mathbb{R}}^2$.”

    1. (a)For Argument 1: state exactly what the similarity invariants theorem says and which direction of implication the student used. Then decide whether $A\sim B$, with a complete proof.
    2. (b)For Argument 2: identify the sentence that is wrong and explain the misreading of ‘represent the same operator’. Then decide whether $A\sim B$; if they are similar, exhibit an invertible $P$ with $B=P^{-1}AP$, verify it by multiplication, and describe the basis $\mathcal{B}$ of ${\mathbb{R}}^2$ in which the horizontal shear has matrix $B$.
    Show the solution to question 11

    (a) The converse of a theorem is not the theorem. The similarity invariants theorem says: if $B=P^{-1}AP$ for some invertible $P$, then $A$ and $B$ have the same determinant, trace, rank and characteristic polynomial. It is a one-way implication, similar $\Rightarrow$ invariants agree, and its practical use is the contrapositive: invariants differ $\Rightarrow$ not similar. The student used the converse — invariants agree $\Rightarrow$ similar — which the theorem does not assert and which is false, as this very pair shows.

    $A\nsim B$. Here $B=2I$, and for every invertible $P$,

    $$P^{-1}(2I)P=2P^{-1}IP=2P^{-1}P=2I.$$

    So the only matrix similar to $2I$ is $2I$ itself. Since $A\ne 2I$, $A$ is not similar to $B$. $■$ In operator language: $B$ represents "double every vector", which looks identical in every basis, whereas $A$ sends ${\mathbf{e}}_2$ to $(1,2)$, not to $2{\mathbf{e}}_2$, so $A$ is not doubling and no change of basis can make it so. Agreement on all the listed invariants can only ever fail to rule out similarity; it never establishes it.

    (b) Same operator, different rulers. The wrong sentence is the last one: "these are different operators on ${\mathbb{R}}^2$, so the matrices are not similar." Similar matrices represent one operator in two different bases. The operator that $A$ describes in the standard basis and the operator that $B$ describes in the standard basis are indeed different maps — but that is not the question. The question is whether the horizontal shear, read in some other basis, has matrix $B$.

    It does. Take $P=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, which is invertible ($\det P=-1$) with $P^{-1}=P$. Then

    $$AP=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right),P^{-1}(AP)=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)=B.✓$$

    So $A\sim B$. The columns of $P=P_{\mathcal{E}\leftarrow \mathcal{B}}$ are the vectors of $\mathcal{B}$, so $\mathcal{B}=\{{\mathbf{e}}_2,{\mathbf{e}}_1\}$: the standard basis in the opposite order.

    Check this directly from the definition of $[T{]}_{\mathcal{B}}$ for the horizontal shear $T(x,y)=(x+y,y)$. First basis vector ${\mathbf{e}}_2$: $T({\mathbf{e}}_2)=(1,1)=1\cdot {\mathbf{e}}_2+1\cdot {\mathbf{e}}_1$, coordinates $(1,1)$. Second basis vector ${\mathbf{e}}_1$: $T({\mathbf{e}}_1)={\mathbf{e}}_1=0\cdot {\mathbf{e}}_2+1\cdot {\mathbf{e}}_1$, coordinates $(0,1)$. Hence $[T{]}_{\mathcal{B}}=\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)=B$ ✓. Swapping the names of the two axes turns a horizontal shear into a vertical one — the geometry is unchanged, only the labelling moved. By symmetry of $\sim$ (here even with the same $P$, since $P^{-1}=P$), $B\sim A$ as well.

12. 12.Computation · 11 min

    Let $D:P_3\to P_2$ be differentiation, $D(p)=p^{\prime }$, and let $S:P_2\to P_3$ be integration from $0$, $S(p)(t)={\int }_0^{t}p(u)du$. Use the bases $\mathcal{B}=\{1,t,t^2,t^3\}$ of $P_3$ and $\mathcal{C}=\{1,t,t^2\}$ of $P_2$.

    1. (a)Compute $[D{]}_{\mathcal{C}}^{\mathcal{B}}$ and $[S{]}_{\mathcal{B}}^{\mathcal{C}}$, stating the shape of each before you fill it in.
    2. (b)Compute both products $[D{]}_{\mathcal{C}}^{\mathcal{B}}[S{]}_{\mathcal{B}}^{\mathcal{C}}$ and $[S{]}_{\mathcal{B}}^{\mathcal{C}}[D{]}_{\mathcal{C}}^{\mathcal{B}}$. Name the theorem that makes them the matrices of $D\circ S$ and $S\circ D$, identify both composite maps in words, and explain from the calculus point of view why one of them is the identity and the other is not.
    3. (c)Prove that no linear map $R:P_2\to P_3$ whatsoever satisfies $R\circ D={\mathrm{id}}_{P_3}$.
    4. (d)So $D$ has a right inverse but no left inverse. Explain precisely why this does not contradict the theorem listing equivalent criteria for the invertibility of an operator.
    Show the solution to question 12

    (a) Two rectangular matrices. $D$ goes from a $4$-dimensional space to a $3$-dimensional one, so $[D{]}_{\mathcal{C}}^{\mathcal{B}}$ is $3\times 4$. Apply $D$ to $\mathcal{B}$ and read $\mathcal{C}$-coordinates: $D(1)=0$, $D(t)=1$, $D(t^2)=2t$, $D(t^3)=3t^2$, giving

    $$[D{]}_{\mathcal{C}}^{\mathcal{B}}=\left(\begin{array}{cccc}0& 1& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 3\end{array}\right).$$

    $S$ goes from $3$ dimensions to $4$, so $[S{]}_{\mathcal{B}}^{\mathcal{C}}$ is $4\times 3$. Apply $S$ to $\mathcal{C}$: $S(1)=t$, $S(t)=\frac{1}{2}{t}^2$, $S(t^2)=\frac{1}{3}{t}^3$, with $\mathcal{B}$-coordinates $(0,1,0,0)$, $(0,0,\frac{1}{2},0)$, $(0,0,0,\frac{1}{3})$:

    $$[S{]}_{\mathcal{B}}^{\mathcal{C}}=\left(\begin{array}{ccc}0& 0& 0\\ 1& 0& 0\\ 0& \frac{1}{2}& 0\\ 0& 0& \frac{1}{3}\end{array}\right).$$

    (b) The two composites. By the theorem that composition is matrix multiplication, $[D\circ S{]}_{\mathcal{C}}^{\mathcal{C}}=[D{]}_{\mathcal{C}}^{\mathcal{B}}[S{]}_{\mathcal{B}}^{\mathcal{C}}$ and $[S\circ D{]}_{\mathcal{B}}^{\mathcal{B}}=[S{]}_{\mathcal{B}}^{\mathcal{C}}[D{]}_{\mathcal{C}}^{\mathcal{B}}$; the inner basis $\mathcal{B}$ (respectively $\mathcal{C}$) matches in each product, as it must. Multiplying,

    $$[D{]}_{\mathcal{C}}^{\mathcal{B}}[S{]}_{\mathcal{B}}^{\mathcal{C}}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)=I_3,[S{]}_{\mathcal{B}}^{\mathcal{C}}[D{]}_{\mathcal{C}}^{\mathcal{B}}=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right).$$

    (For the first: row $i$ of $[D]$ has a single entry $i$ in column $i+1$, and column $i$ of $[S]$ has a single entry $\frac{1}{i}$ in row $i+1$, so they meet exactly on the diagonal with product $1$. For the second: row $1$ of $[S]$ is zero, and rows $2,3,4$ of $[S]$ pick out $1,\frac{1}{2},\frac{1}{3}$ times rows $1,2,3$ of $[D]$.)

    In words, $D\circ S={\mathrm{id}}_{P_2}$: differentiating ${\int }_0^{t}p$ returns $p$, which is the fundamental theorem of calculus. And $(S\circ D)(p)={\int }_0^t{p}^{\prime }(u)du=p(t)-p(0)$: integrating the derivative recovers $p$ minus its constant term. The matrix shows exactly that — it kills $1$ and fixes $t,t^2,t^3$. The asymmetry is the kernel of $D$: differentiation destroys the constants, and no subsequent operation can rebuild information that has already been lost.

    (c) No left inverse exists. Suppose some linear $R:P_2\to P_3$ had $R\circ D={\mathrm{id}}_{P_3}$. Then $D$ would be one-to-one: if $D(p)=D(q)$, apply $R$ to get $p=R(D(p))=R(D(q))=q$. But $D$ is not one-to-one — $D(1)=0=D(0)$ with $1\ne 0$, i.e. $\ker (D)=\mathrm{span}\{1\}$ is nontrivial. Contradiction. $■$

    A second proof by counting: $\mathrm{rank}(R\circ D)\le \mathrm{rank}(D)\le \dim P_2=3$, since every output of $R\circ D$ is an output of $R$ applied to a $3$-dimensional space, whereas $\mathrm{rank}({\mathrm{id}}_{P_3})=4$. Rank-nullity for $D$ on $P_3$ already forces $\mathrm{nullity}(D)\ge 4-3=1$ before any calculus is done.

    (d) Why there is no contradiction. The theorem of equivalent criteria — invertible $⟺$ one-to-one $⟺$ onto $⟺$ trivial kernel $⟺$ full rank $⟺$ nonzero determinant — is stated for an operator $T:V\to V$ on a single finite-dimensional space. Its engine is the square dichotomy from rank-nullity, which needs $\dim V$ on both sides, and it is there that a one-sided inverse becomes two-sided. $D:P_3\to P_2$ is not an operator: the domain has dimension $4$ and the codomain $3$. Rank-nullity makes $D$ onto (rank $3$ fills $P_2$) but never one-to-one (nullity at least $1$), so a right inverse can exist — $S$ is one — while a left inverse cannot. On the matrix side, $[D]$ is $3\times 4$, not square; it has no determinant, and a non-square matrix can satisfy $MN=I$ without $NM=I$, exactly as $[D][S]=I_3$ while $[S][D]\ne I_4$. Finally, by classification by dimension, $P_3\ncong P_2$ because $4\ne 3$, so no isomorphism between them exists at all — the most one can hope for is a one-sided inverse, and that is what $S$ provides.