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1. 1.Computation · 7 min

   Two matrices:

   $$A=\left(\begin{array}{cc}5& -1\\ 1& 3\end{array}\right),C=\left(\begin{array}{ccc}1& 1& 0\\ -1& 3& 0\\ 1& 1& 2\end{array}\right).$$

   1. (a)Find the characteristic polynomial of $A$, both multiplicities of each eigenvalue, and a basis of each eigenspace. Is $A$ diagonalizable?
   2. (b)Do the same for $C$. (Expanding $\det (\lambda I-C)$ along its third column saves work.)
   3. (c)Give a one-line argument, using nothing but the list of eigenvalues, that $C$ cannot be diagonalizable. Name the fact you are using.
   Show the solution to question 1

   (a) Characteristic polynomial and multiplicities. $\mathrm{tr}A=8$ and $\det A=15-(-1)(1)=16$, so

   $$p_A(\lambda )={\lambda }^2-8\lambda +16=(\lambda -4{)}^2.$$

   The only eigenvalue is $4$, with $m_a(4)=2$. For the geometric multiplicity,

   $$A-4I=\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)$$

   has rank $1$, so $m_g(4)=2-1=1$. The eigenspace is $\{(x,y):x-y=0\}=\mathrm{span}(1,1)$; check: $A(1,1)=(4,4)$. Since $m_g(4)=1<2=m_a(4)$, the eigenvalue is defective and $A$ is not diagonalizable: there is no basis of ${\mathbb{C}}^2$ made of eigenvectors.

   (b) The $3\times 3$ matrix. The third column of $\lambda I-C$ is $(0,0,\lambda -2)$, so expanding along it,

   $$p_C(\lambda )=(\lambda -2)\det \left(\begin{array}{cc}\lambda -1& -1\\ 1& \lambda -3\end{array}\right)=(\lambda -2)((\lambda -1)(\lambda -3)+1)=(\lambda -2)({\lambda }^2-4\lambda +4)=(\lambda -2{)}^3.$$

   So $2$ is the only eigenvalue and $m_a(2)=3$. Now

   $$C-2I=\left(\begin{array}{ccc}-1& 1& 0\\ -1& 1& 0\\ 1& 1& 0\end{array}\right).$$

   Rows $1$ and $2$ are equal; rows $1$ and $3$ are not proportional (their sum is $(0,2,0)$), so the rank is $2$ and $m_g(2)=3-2=1$. The eigenspace is cut out by $-x+y=0$ and $x+y=0$, forcing $x=y=0$: it is $\mathrm{span}(0,0,1)$, and indeed $C{e}_3=(0,0,2)=2e_3$. The deficit is $m_a-m_g=2$: two chairs are empty. $C$ is not diagonalizable.

   (c) The scalar test. $C$ has the single eigenvalue $2$ and $C\ne 2I$. If $C$ were diagonalizable, $P^{-1}CP$ would be a diagonal matrix whose entries are eigenvalues of $C$, hence all equal to $2$; so $P^{-1}CP=2I$ and $C=P(2I)P^{-1}=2I$, which is false. This is the proposition that a matrix with only one eigenvalue ${\lambda }_0$ is diagonalizable only if it is ${\lambda }_{0}I$. The same one-liner disposes of $A$ in (a): its only eigenvalue is $4$ and $A\ne 4I$.

2. 2.Proof · 8 min

   Recall that a subspace $W\subseteq {\mathbb{C}}^n$ is $A$-invariant if $Aw\in W$ for every $w\in W$. Let $W=\mathrm{span}(e_1,e_2)\subseteq {\mathbb{C}}^3$ and

   $$A=\left(\begin{array}{ccc}1& 2& 4\\ 3& 4& 5\\ 0& 0& 6\end{array}\right).$$

   1. (a)Show that $W$ is $A$-invariant, write down the matrix $A_1$ of the restriction $A{|}_W$ in the basis $e_1,e_2$, and verify directly that $p_A(\lambda )=p_{A_1}(\lambda )p_{A_2}(\lambda )$, where $A_2$ is the remaining diagonal block.
   2. (b)Prove: if $W$ is $A$-invariant then $W$ is $q(A)$-invariant for every polynomial $q$; and if $W_1$ and $W_2$ are $A$-invariant then so are $W_1\cap W_2$ and $W_1+W_2$.
   3. (c)Prove: if $A$ is invertible and $W$ is $A$-invariant, then $W$ is $A^{-1}$-invariant. (Consider the restriction $A{|}_W:W\to W$.)
   Show the solution to question 2

   (a) A concrete invariant subspace. $A{e}_1=(1,3,0)$ and $A{e}_2=(2,4,0)$ both have third coordinate $0$, so they lie in $W$; a general $w=x{e}_1+y{e}_2$ has $Aw=xA{e}_1+yA{e}_2\in W$ by linearity. The columns of $A_1$ are the coordinates of $A{e}_1,A{e}_2$ in the basis $e_1,e_2$:

   $$A_1=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right),p_{A_1}(\lambda )={\lambda }^2-5\lambda +(4-6)={\lambda }^2-5\lambda -2.$$

   Here $P=I$ already, and $A$ is block upper triangular with $A_2=(6)$, so $p_{A_2}(\lambda )=\lambda -6$. Directly, expanding $\det (\lambda I-A)$ along the third row,

   $$p_A(\lambda )=(\lambda -6)\det \left(\begin{array}{cc}\lambda -1& -2\\ -3& \lambda -4\end{array}\right)=(\lambda -6)((\lambda -1)(\lambda -4)-6)=(\lambda -6)({\lambda }^2-5\lambda -2),$$

   which is $p_{A_1}{p}_{A_2}$, as the block triangular theorem predicts.

   (b) Polynomials, intersections, sums. First, $A^{j}w\in W$ for every $j\ge 0$ and $w\in W$, by induction: $A^{0}w=w\in W$, and if $A^{j}w\in W$ then $A^{j+1}w=A(A^{j}w)\in W$ by invariance. For $q(\lambda )={\sum }_j{c}_j{\lambda }^j$ we get $q(A)w={\sum }_j{c}_j{A}^{j}w$, a linear combination of vectors of $W$, hence in $W$.

   If $w\in W_1\cap W_2$ then $Aw\in W_1$ (as $w\in W_1$) and $Aw\in W_2$ (as $w\in W_2$), so $Aw\in W_1\cap W_2$. If $w=w_1+w_2$ with $w_i\in W_i$ then $Aw=A{w}_1+A{w}_2$ with $A{w}_i\in W_i$, so $Aw\in W_1+W_2$. $■$

   (c) Invariance under the inverse. The restriction $A{|}_W:W\to W$ is a linear map from $W$ to itself. It is injective, because $A$ itself is injective ($A$ is invertible, so $Aw=0$ forces $w=0$). An injective linear map from a finite-dimensional space to itself is surjective, by rank–nullity: $\dim \mathrm{im}=\dim W-\dim \ker =\dim W$.

   Now let $w\in W$. By surjectivity there is $w^{\prime }\in W$ with $A{w}^{\prime }=w$, and then $A^{-1}w=w^{\prime }\in W$. So $W$ is $A^{-1}$-invariant. $■$

   (Once Cayley–Hamilton is available there is a second proof: $A^{-1}$ is a polynomial in $A$ whenever $\det A\ne 0$, so (b) applies directly.)

3. 3.Computation · 7 min

   Let $A=\left(\begin{array}{cc}2& 1\\ 4& -1\end{array}\right)$, and let $B$ be a $3\times 3$ matrix with characteristic polynomial $p_B(\lambda )={\lambda }^3-2{\lambda }^2-5\lambda +6$.

   1. (a)Write down $p_A$ and the identity that Cayley–Hamilton gives for $A$. Use it — without computing $A^3$ or $A^4$ directly — to express $A^4$ in the form $\alpha A+\beta I$. Check your answer against each eigenvalue of $A$.
   2. (b)Use the same identity to write $A^{-1}$ in the form $\gamma A+\delta I$, and verify by multiplying.
   3. (c)Show that $B$ is invertible, find $\det B$, and express $B^{-1}$ as a polynomial in $B$ of degree $2$. Check the expression at each root of $p_B$.
   Show the solution to question 3

   (a) Reduce the power. $\mathrm{tr}A=1$ and $\det A=-2-4=-6$, so $p_A(\lambda )={\lambda }^2-\lambda -6=(\lambda -3)(\lambda +2)$. Cayley–Hamilton says $A^2-A-6I=0$, i.e.

   $$A^2=A+6I.$$

   Multiply by $A$: $A^3=A^2+6A=(A+6I)+6A=7A+6I$. Multiply again: $A^4=7A^2+6A=7(A+6I)+6A=13A+42I$. So $\alpha =13$, $\beta =42$.

   Check at the eigenvalues, which must satisfy every polynomial identity that $A$ does: at $\lambda =3$, $3^4=81$ and $13\cdot 3+42=81$; at $\lambda =-2$, $(-2{)}^4=16$ and $13(-2)+42=16$. Both agree. (Explicitly $A^4=\left(\begin{array}{cc}68& 13\\ 52& 29\end{array}\right)$, and squaring $A^2=\left(\begin{array}{cc}8& 1\\ 4& 5\end{array}\right)$ gives the same matrix.)

   (b) The inverse. From $A^2-A=6I$ we get $A(A-I)=6I$, so

   $$A^{-1}=\frac{1}{6}(A-I)=\frac{1}{6}\left(\begin{array}{cc}1& 1\\ 4& -2\end{array}\right),\gamma =\frac{1}{6},\delta =-\frac{1}{6}.$$

   Verify: $A(A-I)=\left(\begin{array}{cc}2& 1\\ 4& -1\end{array}\right)\left(\begin{array}{cc}1& 1\\ 4& -2\end{array}\right)=\left(\begin{array}{cc}2+4& 2-2\\ 4-4& 4+2\end{array}\right)=6I$.

   (c) The $3\times 3$ case. The constant term is $c_0=p_B(0)=\det (0\cdot I-B)=\det (-B)=(-1{)}^3\det B$, so $\det B=-c_0=-6\ne 0$ and $B$ is invertible. (Cross-check: $p_B(\lambda )=(\lambda -1)(\lambda +2)(\lambda -3)$, and the product of the eigenvalues is $1\cdot (-2)\cdot 3=-6$.)

   Cayley–Hamilton: $B^3-2B^2-5B+6I=0$, so $B(B^2-2B-5I)=-6I$ and

   $$B^{-1}=-\frac{1}{6}\left(B^2-2B-5I\right).$$

   Check at each eigenvalue, where the identity must read ${\lambda }^{-1}=-\frac{1}{6}({\lambda }^2-2\lambda -5)$: at $\lambda =1$, $-\frac{1}{6}(1-2-5)=1$; at $\lambda =-2$, $-\frac{1}{6}(4+4-5)=-\frac{1}{2}$; at $\lambda =3$, $-\frac{1}{6}(9-6-5)=\frac{1}{3}$. All three are the reciprocals of the eigenvalues, as they must be, since applying the identity to an eigenvector $v$ with $Bv=\lambda v$ gives ${\lambda }^{-1}v$ on the left and $-\frac{1}{6}({\lambda }^2-2\lambda -5)v$ on the right.

4. 4.Computation · 8 min

   Throughout, $m_X$ denotes the minimal polynomial of a matrix $X$ and $p_X$ its characteristic polynomial.

   1. (a)Let $E$ be the $3\times 3$ matrix every entry of which is $1$. Show that $E^2=3E$, find $m_E$, decide whether $E$ is diagonalizable, and then determine $p_E$ without computing any determinant.
   2. (b)A matrix $B$ satisfies $B^3=B$. Prove that $B$ is diagonalizable, and list every polynomial that could be $m_B$, giving a matrix that realises each one.
   3. (c)A $3\times 3$ matrix $C$ has $p_C(\lambda )=(\lambda -1{)}^2(\lambda -4)$. Determine $m_C$ if $\mathrm{rank}(C-I)=2$, and determine it if instead $\mathrm{rank}(C-I)=1$. Justify each step from the theorems on the minimal polynomial.
   Show the solution to question 4

   (a) The all-ones matrix. Each entry of $E^2$ is a row of ones dotted with a column of ones, namely $3$; so $E^2=3E$. Hence $q(\lambda )={\lambda }^2-3\lambda =\lambda (\lambda -3)$ annihilates $E$, and $m_E$ divides $q$. The monic divisors of degree $\ge 1$ are $\lambda$, $\lambda -3$ and $\lambda (\lambda -3)$. $m_E=\lambda$ would mean $E=0$ and $m_E=\lambda -3$ would mean $E=3I$; neither holds. So

   $$m_E(\lambda )=\lambda (\lambda -3),$$

   a product of distinct linear factors, and by the diagonalizability criterion $E$ is diagonalizable.

   For $p_E$: the roots of $p_E$ are the roots of $m_E$, namely $0$ and $3$. $E$ has rank $1$, so $m_g(0)=3-1=2$; since $E$ is diagonalizable, $m_a(0)=m_g(0)=2$, and the remaining eigenvalue is $3$ with multiplicity $1$. Thus $p_E(\lambda )={\lambda }^2(\lambda -3)$. Sanity check: $\mathrm{tr}E=3=0+0+3$, and $E(1,1,1)=(3,3,3)$.

   (b) $B^3=B$. The polynomial ${\lambda }^3-\lambda =\lambda (\lambda -1)(\lambda +1)$ annihilates $B$, so $m_B$ divides it. Every monic divisor of degree $\ge 1$ of a product of distinct linear factors is itself a product of distinct linear factors, so $m_B$ has no repeated root and $B$ is diagonalizable by the criterion. The candidates are the seven nontrivial monic divisors:

   $$\lambda ,\lambda -1,\lambda +1,\lambda (\lambda -1),\lambda (\lambda +1),(\lambda -1)(\lambda +1),\lambda (\lambda -1)(\lambda +1),$$

   realised respectively by $0$, $I$, $-I$, $\mathrm{diag}(0,1)$, $\mathrm{diag}(0,-1)$, $\mathrm{diag}(1,-1)$ and $\mathrm{diag}(0,1,-1)$. (A diagonal matrix has minimal polynomial $\prod (\lambda -d)$ over its distinct diagonal entries $d$, and each of these satisfies $B^3=B$ because $d^3=d$ for $d\in \{0,1,-1\}$.)

   (c) Reading $m_C$ off a rank. $m_C$ divides $p_C$ and has the same roots $1$ and $4$, so $m_C$ is either $(\lambda -1)(\lambda -4)$ or $(\lambda -1{)}^2(\lambda -4)$.

   If $\mathrm{rank}(C-I)=2$ then $m_g(1)=3-2=1<2=m_a(1)$, so $C$ is not diagonalizable. By the criterion $m_C$ must then have a repeated root, which rules out $(\lambda -1)(\lambda -4)$. Hence $m_C(\lambda )=(\lambda -1{)}^2(\lambda -4)$.

   If $\mathrm{rank}(C-I)=1$ then $m_g(1)=2=m_a(1)$, and $m_g(4)=1=m_a(4)$ automatically, so $C$ is diagonalizable and $m_C$ has distinct roots: $m_C(\lambda )=(\lambda -1)(\lambda -4)$. In both cases the exponent of $(\lambda -1)$ in $m_C$ is the size of the largest Jordan block for $1$: two blocks of size $1$ in the second case, one block of size $2$ in the first.

5. 5.Computation · 10 min

   Let

   $$A=\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 0\\ 1& 0& 2\end{array}\right),N=A-I.$$

   1. (a)Compute $p_A$ and both multiplicities of every eigenvalue. Which eigenvalue is defective, and by how much?
   2. (b)Compute $N^2$, describe the generalized eigenspace $K_1=\ker N^2$ as a set, and check that its dimension is what the theory predicts.
   3. (c)Find a generalized eigenvector $v$ of height exactly $2$ for the eigenvalue $1$, write down its Jordan chain $Nv,v$, verify that $Nv$ is an eigenvector, and compute $Av$ to confirm that $v$ itself is not one.
   4. (d)Find $K_2$ and exhibit a basis of ${\mathbb{C}}^3$ adapted to the primary decomposition ${\mathbb{C}}^3=K_1\oplus K_2$, verifying independence with a determinant.
   Show the solution to question 5

   (a) Eigenvalues and multiplicities. The third column of $\lambda I-A$ is $(0,0,\lambda -2)$; expanding along it,

   $$p_A(\lambda )=(\lambda -2)\det \left(\begin{array}{cc}\lambda -1& -1\\ 0& \lambda -1\end{array}\right)=(\lambda -2)(\lambda -1{)}^2.$$

   So $m_a(1)=2$ and $m_a(2)=1$. For $\lambda =2$, $1\le m_g\le m_a=1$ forces $m_g(2)=1$. For $\lambda =1$,

   $$N=A-I=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 1& 0& 1\end{array}\right)$$

   has the two independent rows $(0,1,0)$ and $(1,0,1)$, so $\mathrm{rank}N=2$ and $m_g(1)=3-2=1$. The eigenvalue $1$ is defective with deficit $2-1=1$; its eigenspace is $\{y=0,x+z=0\}=\mathrm{span}(1,0,-1)$.

   (b) The generalized eigenspace. Row $i$ of $N^2$ is row $i$ of $N$ applied to $N$: row $1$ is $(0,1,0)N=$ row $2$ of $N$ $=(0,0,0)$; row $2$ is $0$; row $3$ is $(1,0,1)N=$ row $1$ + row $3$ $=(1,1,1)$. Thus

   $$N^2=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 1& 1& 1\end{array}\right),K_1=\ker N^2=\{(x,y,z):x+y+z=0\}.$$

   Its dimension is $3-\mathrm{rank}{N}^2=3-1=2=m_a(1)$, exactly as the theorem $\dim K_{\lambda }=m_a(\lambda )$ predicts. (The chain of kernels has stabilised: $N^3=N\cdot N^2$ has rows $0$, $0$ and $(1,0,1)N^2=(1,1,1)$, so $N^3=N^2$ and $\ker N^3=\ker N^2$.)

   (c) A chain of length $2$. Height exactly $2$ means $v\in \ker N^2\setminus \ker N$. Take $v=(1,-1,0)$: its coordinates sum to $0$, so $v\in K_1$, while $Nv=(-1,0,1)\ne 0$. The chain is

   $$u_1=Nv=(-1,0,1),u_2=v=(1,-1,0).$$

   Check the bottom: $N{u}_1=(0,0,-1+1)=(0,0,0)$, so $u_1$ is an eigenvector (a multiple of $(1,0,-1)$ from (a)). And $Av=(1-1,-1,1)=(0,-1,1)$, which is not a multiple of $v$; instead $Av=v+Nv=u_2+u_1$ — the eigenvalue times $v$ plus something of lower height, which is the whole point of a generalized eigenvector.

   (d) The primary decomposition. $m_a(2)=1$, so $K_2=\ker (A-2I)$. With

   $$A-2I=\left(\begin{array}{ccc}-1& 1& 0\\ 0& -1& 0\\ 1& 0& 0\end{array}\right),$$

   row $2$ gives $y=0$ and row $3$ gives $x=0$, so $K_2=\mathrm{span}(0,0,1)$; indeed $A{e}_3=(0,0,2)$.

   Adapted basis: $u_1,u_2$ for $K_1$ and $e_3$ for $K_2$. Independence: expanding along the third column,

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

   So ${\mathbb{C}}^3=K_1\oplus K_2$ with $\dim K_1+\dim K_2=2+1=3$. In this basis $A$ is block diagonal: on $K_1$ it acts as $I+N{|}_{K_1}$, which in the chain basis is $J_2(1)$, and on $K_2$ as the $1\times 1$ block $(2)$.

6. 6.Proof · 9 min

   Let $N$, $M$ and $A$ be $n\times n$ complex matrices. Recall that $N$ is nilpotent if $N^k=0$ for some $k\ge 1$.

   1. (a)Prove that a nilpotent matrix has trace $0$ and determinant $0$.
   2. (b)Prove that if $N$ is nilpotent and $AN=NA$, then $AN$ is nilpotent.
   3. (c)Prove that if $N$ and $M$ are nilpotent and $NM=MN$, then $N+M$ is nilpotent. (If $N^k=0$ and $M^l=0$, consider $(N+M{)}^{k+l-1}$.)
   4. (d)Show by an explicit $2\times 2$ example that (c) fails without the hypothesis $NM=MN$, and that the product $NM$ in your example is not nilpotent either.
   Show the solution to question 6

   (a) Trace and determinant. If $Nv=\lambda v$ with $v\ne 0$ then $0=N^{k}v={\lambda }^{k}v$, so $\lambda =0$: the only eigenvalue of a nilpotent matrix is $0$. Over $\mathbb{C}$ the characteristic polynomial splits and all its roots are $0$, so $p_N(\lambda )={\lambda }^n$. But for any matrix the coefficient of ${\lambda }^{n-1}$ in $\det (\lambda I-N)$ is $-\mathrm{tr}N$ and the constant term is $(-1{)}^n\det N$. Comparing with ${\lambda }^n$, both vanish: $\mathrm{tr}N=0$ and $\det N=0$. $■$

   (For the determinant alone there is a one-liner: $(\det N{)}^k=\det N^k=\det 0=0$.)

   (b) A commuting product. Since $A$ and $N$ commute, $(AN{)}^j=A^j{N}^j$ for every $j\ge 1$, by induction: $(AN{)}^{j+1}=(AN{)}^j(AN)=A^j{N}^{j}AN=A^{j}A{N}^{j}N=A^{j+1}{N}^{j+1}$, using $N^{j}A=A{N}^j$ (itself an induction from $NA=AN$). With $N^k=0$ this gives $(AN{)}^k=A^k{N}^k=0$. $■$

   (c) A commuting sum. Let $N^k=0$ and $M^l=0$. Because $N$ and $M$ commute, the binomial theorem applies:

   $$(N+M{)}^{k+l-1}=\sum _{j=0}^{k+l-1}(\genfrac{}{}{0ex}{}{k+l-1}{j})N^j{M}^{k+l-1-j}.$$

   For each term, either $j\ge k$ or $k+l-1-j\ge l$: otherwise $j\le k-1$ and $k+l-1-j\le l-1$, whose sum is $k+l-1\le k+l-2$, absurd. In the first case $N^j=0$, in the second $M^{k+l-1-j}=0$; so every term vanishes and $(N+M{)}^{k+l-1}=0$. $■$

   (d) Without commutativity. Take

   $$N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),M=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),$$

   both with square $0$. They do not commute: $NM=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$ while $MN=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$. The sum $N+M=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ satisfies $(N+M{)}^2=I$, so $(N+M{)}^j$ is $I$ or $N+M$ for every $j$ and never $0$ (equivalently, its eigenvalues are $\pm 1$, not $0$). And $NM$ is a projection, $(NM{)}^2=NM\ne 0$, with eigenvalue $1$; so neither the sum nor the product is nilpotent. The commuting hypothesis in (b) and (c) cannot be dropped.

7. 7.Sketch · 9 min

   A nilpotent $9\times 9$ matrix $N$ has

   $$\mathrm{rank}N=5,\mathrm{rank}{N}^2=2,\mathrm{rank}{N}^3=1,N^4=0.$$

   You will draw its chain diagram: one column per Jordan chain, one dot per basis vector, the eigenvector at the bottom of each column, columns ordered tallest first. Write $r_k=\mathrm{rank}{N}^k$ with $r_0=9$.

   1. (a)Compute $b_{\ge k}=r_{k-1}-r_k$ for $k=1,\dots ,5$ and draw the diagram. Mark with an arrow what $N$ does to each dot.
   2. (b)Read off from the diagram: the block sizes, the Jordan form of $N$, its index of nilpotency, $m_N$, and $\dim \ker N$.
   3. (c)Explain, using only the picture, why the number of dots in row $k$ (counting from the bottom) is $r_{k-1}-r_k$. Begin by saying which dots $N^k$ sends to $0$.
   4. (d)Draw the chain diagram of $N^2$ by tracing where $N^2$ sends each dot, read off the Jordan form of $N^2$, and confirm it with the rank formula applied to the sequence $\mathrm{rank}(N^2{)}^k$.
   Show the solution to question 7

   (a) The rank differences and the picture. With $(r_0,r_1,r_2,r_3,r_4)=(9,5,2,1,0)$ and $r_5=0$,

   $$b_{\ge 1}=9-5=4,b_{\ge 2}=5-2=3,b_{\ge 3}=2-1=1,b_{\ge 4}=1-0=1,b_{\ge 5}=0.$$

   $b_{\ge k}$ is the number of columns of height at least $k$, i.e. the number of dots in row $k$. So the diagram has, from the bottom up, rows of $4$, $3$, $1$, $1$ dots: four columns of heights $4,2,2,1$, drawn tallest first. Every arrow points straight down one square — $N$ sends each dot to the dot beneath it — and each of the four bottom dots is sent to $0$ (draw those arrows leaving the diagram).

   (b) Reading the diagram. The column heights are the block sizes: $4,2,2,1$ (sum $9$, as it must be). So

   $$N\sim J_4(0)\oplus J_2(0)\oplus J_2(0)\oplus J_1(0).$$

   The index of nilpotency is the tallest column, $4$ (the dot at the top of that column survives $N^3$ and dies under $N^4$), so $m_N(\lambda )={\lambda }^4$. The number of columns is the number of eigenvectors in the basis, $\dim \ker N=4=b_{\ge 1}$. Consistency: $b_{=1}=4-3=1$, $b_{=2}=3-1=2$, $b_{=3}=1-1=0$, $b_{=4}=1-0=1$, matching the heights.

   (c) Why the rows count rank drops. Each dot is a basis vector, and $N^k$ moves every dot down $k$ squares. A dot at height $h$ therefore lands on the dot at height $h-k$ if $h>k$, and is sent to $0$ if $h\le k$. The dots that survive are exactly those in rows $k+1$ and above, and their images are distinct basis vectors — the bottom $h-k$ dots of each column — hence independent. So $\mathrm{rank}{N}^k=r_k$ is the number of dots in rows $k+1,k+2,\dots$; equivalently $r_{k-1}$ is the number of dots in rows $k,k+1,\dots$. Subtracting,

   $$r_{k-1}-r_k=\#\{\text{dots in row }k\}=\#\{\text{columns of height}\ge k\}=b_{\ge k}.$$

   Check against the data: rows $2$ and above hold $3+1+1=5=r_1$ dots; rows $3$ and above hold $2=r_2$; row $4$ holds $1=r_3$.

   (d) The diagram of $N^2$. $N^2$ moves each dot down two squares, so a column of $N$ splits into the dots of even height and the dots of odd height, each of which is a chain for $N^2$. The column of height $4$ splits into two chains of length $2$ (heights $4\to 2$ and $3\to 1$); each column of height $2$ splits into two chains of length $1$; the column of height $1$ stays a chain of length $1$. So the diagram of $N^2$ has seven columns of heights $2,2,1,1,1,1,1$, and

   $$N^2\sim J_2(0)\oplus J_2(0)\oplus J_1(0{)}^{\oplus 5}.$$

   Rank formula check: the powers of $N^2$ have ranks $\mathrm{rank}{N}^2=2$ and $\mathrm{rank}{N}^4=0$, so the sequence is $(9,2,0,\dots )$ and $b_{\ge 1}=7$, $b_{\ge 2}=2$, $b_{\ge 3}=0$: two blocks of size $2$ and five of size $1$, total $4+5=9$. The picture and the formula agree.

8. 8.Critique · 8 min

   A $4\times 4$ matrix $A$ has $p_A(\lambda )=(\lambda -1{)}^4$ and $\mathrm{rank}(A-I)=2$. A student writes:

   Step 1. The number of Jordan blocks for $\lambda =1$ is $\mathrm{rank}(A-I)=2$.

   Step 2. The block sizes add up to $m_a(1)=4$ and there are two blocks, so each block has size $4/2=2$.

   Step 3. Hence $J=J_2(1)\oplus J_2(1)$ and $m_A(\lambda )=(\lambda -1{)}^2$; since $m_A$ has a repeated root, $A$ is not diagonalizable.

   The number in Step 1 is correct but the reasoning is not; Step 2 is wrong; the last clause of Step 3 is correct.

   1. (a)State the correct formula for the number of blocks, explain why the student's version happens to give the right number here, and give a $4\times 4$ matrix for which it gives the wrong number.
   2. (b)Explain precisely why Step 2 does not follow, and list every Jordan form consistent with the given data, together with the minimal polynomial of each.
   3. (c)Name one further piece of data that decides between the possibilities in (b), and say what conclusion each of its possible values gives.
   4. (d)Give a correct proof that $A$ is not diagonalizable which uses only the data given, without knowing the Jordan form.
   Show the solution to question 8

   (a) The block count. The number of Jordan blocks for $\lambda$ is the geometric multiplicity,

   $$b_{\ge 1}=r_0-r_1=n-\mathrm{rank}(A-\lambda I),$$

   not the rank itself. Here $n-\mathrm{rank}(A-I)=4-2=2$, and the student's $\mathrm{rank}(A-I)=2$ agrees only because $4-2=2$: the two formulas coincide exactly when the rank is $n/2$. For $A=I_4$ the rank of $A-I$ is $0$, so the student would count $0$ blocks, whereas there are $4$ blocks of size $1$; for $A=J_4(1)$ the rank is $3$ and the student would count $3$ blocks, whereas there is $1$.

   (b) Equal sizes are not forced. The rank formula with $k=1$ gives only the number of blocks; it says nothing about their sizes, which are governed by the ranks of the higher powers $(A-I{)}^2,(A-I{)}^3,\dots$. Two blocks with sizes summing to $4$ can be $3+1$ or $2+2$ — the two partitions of $4$ into two parts — and nothing given so far excludes either. The consistent Jordan forms are

   $$J_3(1)\oplus J_1(1)\text{with}{m}_A(\lambda )=(\lambda -1{)}^3,\text{or}{J}_2(1)\oplus J_2(1)\text{with}{m}_A(\lambda )=(\lambda -1{)}^2,$$

   since the exponent in the minimal polynomial is the largest block size.

   (c) What decides it. The rank of $(A-I{)}^2$. By the rank-of-a-block lemma, $\mathrm{rank}(J_m(1)-I{)}^2=\max (m-2,0)$, so $r_2=1+0=1$ for $J_3(1)\oplus J_1(1)$ and $r_2=0+0=0$ for $J_2(1)\oplus J_2(1)$. Thus: if $\mathrm{rank}(A-I{)}^2=1$ then $b_{\ge 2}=r_1-r_2=1$ and $b_{\ge 3}=r_2-r_3=1$, giving $J_3(1)\oplus J_1(1)$; if $\mathrm{rank}(A-I{)}^2=0$ then $b_{\ge 2}=2$ and $b_{\ge 3}=0$, giving $J_2(1)\oplus J_2(1)$. Equivalently, one may ask whether $(A-I{)}^2=0$, or which of $(\lambda -1{)}^2$ and $(\lambda -1{)}^3$ is the minimal polynomial.

   (d) Non-diagonalizability from the data alone. $m_g(1)=4-\mathrm{rank}(A-I)=2$ while $m_a(1)=4$. The eigenvalue $1$ is therefore defective, the eigenvectors of $A$ span only a $2$-dimensional subspace of ${\mathbb{C}}^4$, and no basis of eigenvectors exists. (Alternatively: $A$ has the single eigenvalue $1$, and $A\ne I$ because $\mathrm{rank}(A-I)=2\ne 0$, so the scalar test applies.) The student's conclusion stands, but it never needed the Jordan form.

9. 9.Computation · 12 min

   Let

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

   whose characteristic polynomial is $(\lambda -2{)}^3$ (you may take this as given, but check it against $\mathrm{tr}A$ and $\det A$). Write $N=A-2I$.

   1. (a)Compute $\mathrm{rank}N$ and $N^2$. Deduce the number and sizes of the Jordan blocks, the Jordan form $J$, and $m_A$.
   2. (b)Build a Jordan basis, longest chain first: choose a suitable $v$, form its chain, then complete the basis. Write down $P$ with $P^{-1}AP=J$ (you need not compute $P^{-1}$).
   3. (c)Verify, for each column $u_j$ of $P$, the relation $A{u}_j=2u_j$ or $A{u}_j=2u_j+u_{j-1}$ that certifies $P^{-1}AP=J$.
   4. (d)Show that $e_2=(0,1,0)$ is an eigenvector of $A$ that does not lie in the image of $N$. Explain what goes wrong for someone who starts from $e_2$ and tries to solve $Nv=e_2$ for the top of a chain, and what goes wrong for someone who assigns the eigenvector $(1,2,1)$ to the $1\times 1$ block first.
   Show the solution to question 9

   (a) Block structure. Consistency of the given polynomial: $\mathrm{tr}A=6=3\cdot 2$, and expanding $\det A$ along the second column, $\det A=2\det \left(\begin{array}{cc}3& -1\\ 1& 1\end{array}\right)=2\cdot 4=8=2^3$. Now

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

   Every row is a multiple of $(1,0,-1)$, so $\mathrm{rank}N=1$. Better, $N=u{w}^{\mathsf{T}}$ with $u=(1,2,1{)}^{\mathsf{T}}$ and $w=(1,0,-1{)}^{\mathsf{T}}$, and $w^{\mathsf{T}}u=1+0-1=0$, so $N^2=u(w^{\mathsf{T}}u)w^{\mathsf{T}}=0$. (Row by row: row $1$ of $N^2$ is row $1$ minus row $3$ of $N$, which is $0$, and the other rows are multiples of that.)

   Rank sequence $(r_0,r_1,r_2)=(3,1,0)$: $b_{\ge 1}=2$, $b_{\ge 2}=1$, $b_{\ge 3}=0$. Two blocks, one of size $2$ and one of size $1$:

   $$J=J_2(2)\oplus J_1(2)=\left(\begin{array}{ccc}2& 1& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right),m_A(\lambda )=(\lambda -2{)}^2,$$

   the exponent being the largest block (equivalently, $N\ne 0=N^2$).

   (b) The basis, longest chain first. A chain of length $2$ needs a top $v\in \ker N^2\setminus \ker N={\mathbb{C}}^3\setminus \ker N$, where $\ker N=\{x-z=0\}$. Take $v=e_1=(1,0,0)$: $Nv=(1,2,1)\ne 0$. The chain, eigenvector first, is

   $$u_1=Nv=(1,2,1),u_2=v=(1,0,0).$$

   The $1\times 1$ block needs an eigenvector independent of $u_1$. Any vector with $x=z$ is in $\ker N$; $u_3=(0,1,0)$ qualifies and is not a multiple of $(1,2,1)$. Independence of all three: expanding along the third row,

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

   So

   $$P=(u_1\mid u_2\mid u_3)=\left(\begin{array}{ccc}1& 1& 0\\ 2& 0& 1\\ 1& 0& 0\end{array}\right),P^{-1}AP=J.$$

   (c) The certifying relations. $A{u}_1=(3-1,2+4-2,1+1)=(2,4,2)=2u_1$. $A{u}_2=A{e}_1=(3,2,1)=2(1,0,0)+(1,2,1)=2u_2+u_1$. $A{u}_3=A{e}_2=(0,2,0)=2u_3$. Column $j$ of $P^{-1}AP$ is the coordinate vector of $A{u}_j$ in the basis $u_1,u_2,u_3$, namely $2e_1$, $e_1+2e_2$ and $2e_3$ — precisely the columns of $J$, with the $1$ above the diagonal because the chain is listed eigenvector first.

   (d) The two traps. The second column of $N$ is zero, so $N{e}_2=0$ and $e_2$ is an eigenvector. Since $N=u{w}^{\mathsf{T}}$, every vector in the image of $N$ is a multiple of $u=(1,2,1)$; $e_2$ is not, so $e_2\notin \mathrm{im}N$ and the equation $Nv=e_2$ has no solution (concretely, $Nv=(v_1-v_3)(1,2,1)$ can never equal $(0,1,0)$). Someone working bottom-up from $e_2$ would be stuck: an eigenvector is the bottom of a longer chain only if it lies in $\mathrm{im}N\cap \ker N$, which here is the single line $\mathrm{span}(1,2,1)$.

   That is also why the short block must be assigned last. If one first reserved $(1,2,1)$ for the $1\times 1$ block, the length-$2$ chain would need a different bottom in $\mathrm{im}N\cap \ker N$, and there is none — the long chain could never be completed. Longest chains first, each started from its top, avoids both problems.

10. 10.Computation · 10 min

    Let $A$ be the matrix of Problem 5,

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

    and take the Jordan basis $u_1=(-1,0,1)$, $u_2=(1,-1,0)$, $u_3=(0,0,1)$, so that $P=(u_1\mid u_2\mid u_3)$ satisfies $P^{-1}AP=J=J_2(1)\oplus J_1(2)$. (If your chain in Problem 5 was different, either basis works; the solution uses this one.)

    1. (a)Write down $J^k$ for general $k\ge 0$, quoting the formula for powers of a Jordan block.
    2. (b)Compute $P^{-1}$, then $A^k=P{J}^k{P}^{-1}$ in closed form. Check your formula at $k=1$ and $k=2$ by direct multiplication.
    3. (c)Let $x_k=A^k{x}_0$. For $x_0=(1,-1,0)$ show that $\Vert x_k\Vert$ grows linearly in $k$ although the only eigenvalue involved is $1$; for $x_0=(0,1,0)$ show that it grows like $2^k$. Explain both behaviours in terms of the Jordan blocks.
    Show the solution to question 10

    (a) Powers of $J$. For a block $J_m(\lambda )=\lambda I+S$ the entry in position $(i,i+j)$ of $J_m(\lambda {)}^k$ is $\left(\genfrac{}{}{0ex}{}{k}{j}\right){\lambda }^{k-j}$. With $m=2$, $\lambda =1$ this gives $J_2(1{)}^k=\left(\begin{array}{cc}1& k\\ 0& 1\end{array}\right)$, and the $1\times 1$ block gives $2^k$. Powers of a block diagonal matrix are block diagonal, so

    $$J^k=\left(\begin{array}{ccc}1& k& 0\\ 0& 1& 0\\ 0& 0& 2^k\end{array}\right).$$

    (b) Conjugating back. $\det P=1$ (Problem 5), so $P^{-1}$ is the adjugate. The cofactors of $P=\left(\begin{array}{ccc}-1& 1& 0\\ 0& -1& 0\\ 1& 0& 1\end{array}\right)$ are $C_{11}=-1$, $C_{12}=0$, $C_{13}=1$, $C_{21}=-1$, $C_{22}=-1$, $C_{23}=1$, $C_{31}=0$, $C_{32}=0$, $C_{33}=1$, and transposing,

    $$P^{-1}=\left(\begin{array}{ccc}-1& -1& 0\\ 0& -1& 0\\ 1& 1& 1\end{array}\right).$$

    Check $P{P}^{-1}$: row $1$ of $P$ against the columns gives $(1+0+0,1-1+0,0)=(1,0,0)$; row $2$ gives $(0,1,0)$; row $3$ gives $(-1+1,-1+1,1)=(0,0,1)$. Correct.

    Now $P{J}^k$ has columns $P{e}_1=u_1$, $P(k{e}_1+e_2)=k{u}_1+u_2$ and $P(2^k{e}_3)=2^k{u}_3$:

    $$P{J}^k=\left(\begin{array}{ccc}-1& 1-k& 0\\ 0& -1& 0\\ 1& k& 2^k\end{array}\right).$$

    Multiplying by $P^{-1}$ row by row: row $1$ is $-1\cdot (-1,-1,0)+(1-k)(0,-1,0)=(1,1-(1-k),0)=(1,k,0)$; row $2$ is $-1\cdot (0,-1,0)=(0,1,0)$; row $3$ is $(-1,-1,0)+k(0,-1,0)+2^k(1,1,1)=(2^k-1,2^k-1-k,2^k)$. Hence

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

    At $k=1$ this is $\left(\begin{array}{ccc}1& 1& 0\\ 0& 1& 0\\ 1& 0& 2\end{array}\right)=A$. At $k=2$ the formula gives rows $(1,2,0)$, $(0,1,0)$, $(3,1,4)$; directly, row $1$ of $A^2$ is row $1$ + row $2$ of $A$ $=(1,2,0)$, row $2$ is $(0,1,0)$, and row $3$ is row $1$ + $2\cdot$row $3$ $=(1,1,0)+(2,0,4)=(3,1,4)$. Agreed. (Also $\mathrm{tr}{A}^k=2+2^k$, the sum of the $k$-th powers of the eigenvalues $1,1,2$.)

    (c) Two growth rates. For $x_0=(1,-1,0)=u_2$, the formula gives

    $$x_k=A^k{u}_2=(1-k,-1,(2^k-1)-(2^k-1-k))=(1-k,-1,k),$$

    so $\Vert x_k{\Vert }^2=(1-k{)}^2+1+k^2\sim 2k^2$ and $\Vert x_k\Vert \sim \sqrt{2}k$: linear growth. In Jordan coordinates $x_0=e_2$ and $J^k{e}_2=(k,1,0)$, i.e. $x_k=k{u}_1+u_2$ — the $k$ is the off-diagonal entry $\left(\genfrac{}{}{0ex}{}{k}{1}\right)1^{k-1}$ of $J_2(1{)}^k$. A diagonalizable matrix with eigenvalue $1$ would leave $x_0$ fixed; the size-$2$ block at eigenvalue $1$ is what produces the factor $k$.

    For $x_0=e_2=(0,1,0)$, $x_k$ is the second column of $A^k$, $(k,1,2^k-1-k)$, whose last coordinate dominates: $\Vert x_k\Vert \sim 2^k$. The reason is that $e_2$ has a component in $K_2$: $P^{-1}{e}_2$ is the second column of $P^{-1}$, $(-1,-1,1)$, so $e_2=-u_1-u_2+u_3$ (check: $(1,0,-1)+(-1,1,0)+(0,0,1)=(0,1,0)$), and the $u_3$ component is multiplied by $2^k$ at each step while the $K_1$ component only grows linearly. Growth is governed by the largest eigenvalue present in $x_0$'s decomposition, with a polynomial factor $k^{m-1}$ from a block of size $m$.

11. 11.Critique · 10 min

    Let $A=\left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)$. A student computes $e^A$ as follows.

    Step 1. Split $A=D+N$ with $D=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)$ and $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$.

    Step 2. $N^2=0$, so $e^N=I+N=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, and $e^D=\left(\begin{array}{cc}e& 0\\ 0& e^2\end{array}\right)$.

    Step 3. Therefore $e^A=e^D{e}^N=\left(\begin{array}{cc}e& e\\ 0& e^2\end{array}\right)$.

    Steps 1 and 2 are correct.

    1. (a)Identify the error in Step 3, name the property of the exponential being misused, and verify by a direct computation that its hypothesis fails here.
    2. (b)Compute $e^{At}$ correctly for all $t$ (diagonalize), state $e^A$, and compare its $(1,2)$ entry with the student's. Check your $e^{At}$ by confirming that its derivative at $t=0$ equals $A$.
    3. (c)Now let $B=\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)$ and split it as $B=2I+N$. Explain why the same three-step method is valid for $B$, and write down $e^{Bt}$.
    Show the solution to question 11

    (a) The flaw. Step 3 uses $e^{D+N}=e^D{e}^N$. That identity is the property “if $MN=NM$ then $e^{M+N}=e^M{e}^N$”, and it requires the two summands to commute. Here

    $$DN=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),ND=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)=\left(\begin{array}{cc}0& 2\\ 0& 0\end{array}\right),$$

    so $DN\ne ND$ and the property may not be invoked. A symptom the student could have noticed: the other order gives $e^N{e}^D=\left(\begin{array}{cc}e& e^2\\ 0& e^2\end{array}\right)$, a different matrix, whereas $e^A$ is one definite matrix.

    (b) The correct exponential. $A$ is upper triangular with distinct eigenvalues $1$ and $2$, so it is diagonalizable. Eigenvectors: for $\lambda =1$, $A-I=\left(\begin{array}{cc}0& 1\\ 0& 1\end{array}\right)$ gives $(1,0)$; for $\lambda =2$, $A-2I=\left(\begin{array}{cc}-1& 1\\ 0& 0\end{array}\right)$ gives $(1,1)$. So

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

    Then $e^{At}=P{e}^{Dt}{P}^{-1}$ (the exponential of a diagonal matrix is the diagonal of exponentials, and conjugation passes through the series):

    $$e^{At}=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}{e}^t& e^{2t}\\ 0& e^{2t}\end{array}\right)\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}{e}^t& e^{2t}-e^t\\ 0& e^{2t}\end{array}\right).$$

    At $t=1$, $e^A=\left(\begin{array}{cc}e& e^2-e\\ 0& e^2\end{array}\right)$. The $(1,2)$ entry is $e^2-e\approx 4.67$, not the student's $e\approx 2.72$. (Note that $\det$ does not catch the error: both matrices have determinant $e^3=e^{\mathrm{tr}A}$.)

    Check: $\frac{d}{dt}{e}^{At}{|}_{t=0}={\left(\begin{array}{cc}{e}^t& 2e^{2t}-e^t\\ 0& 2e^{2t}\end{array}\right)}_{t=0}=\left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)=A$, as $\frac{d}{dt}{e}^{At}=A{e}^{At}$ requires. The student's matrix, read as $e^{At}=\left(\begin{array}{cc}{e}^t& t{e}^t\\ 0& e^{2t}\end{array}\right)$ or in any other way, cannot pass this test and the eigenvector test simultaneously: $e^A$ must send $(1,1)$ to $e^2(1,1)$, and the student's matrix sends it to $(2e,e^2)$.

    (c) When the split is legal. $B=2I+N$, and $2I$ commutes with every matrix, in particular with $N$. So the commuting-sum property applies: $e^{Bt}=e^{2tI}{e}^{Nt}$. Now $e^{2tI}=e^{2t}I$ (the series for a scalar matrix is the scalar series), and $e^{Nt}=I+tN$ because $N^2=0$ truncates the series. Hence

    $$e^{Bt}=e^{2t}\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right),$$

    which is the general formula $e^{J_2(\lambda )t}=e^{\lambda t}\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)$ with $\lambda =2$. Check: its derivative at $t=0$ is $\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)=B$. The difference between $A$ and $B$ is exactly that $A$'s diagonal part is not a scalar, so it fails to commute with the nilpotent part; the diagonal–plus–nilpotent split only exponentiates termwise inside a single Jordan block.

12. 12.Proof · 10 min

    Let $A$ be an $n\times n$ complex matrix with eigenvalues ${\lambda }_1,\dots ,{\lambda }_n$ listed with algebraic multiplicity. You may use that $A$ is similar to an upper triangular matrix $T$ whose diagonal is ${\lambda }_1,\dots ,{\lambda }_n$, that $e^{P^{-1}MP}=P^{-1}{e}^{M}P$, and that the series defining $e^M$ converges entry by entry.

    1. (a)Prove that if $T$ is upper triangular with diagonal entries $t_1,\dots ,t_n$, then $T^k$ is upper triangular with diagonal entries $t_1^k,\dots ,t_n^k$, and deduce that $e^T$ is upper triangular with diagonal entries $e^{t_1},\dots ,e^{t_n}$.
    2. (b)Prove that the eigenvalues of $e^A$, with multiplicity, are $e^{{\lambda }_1},\dots ,e^{{\lambda }_n}$, and that $\det e^A=e^{\mathrm{tr}A}$.
    3. (c)Deduce that $e^A$ is invertible for every $A$, that $\det e^A>0$ whenever $A$ is real, and that $\det e^{At}=1$ for all $t$ whenever $\mathrm{tr}A=0$.
    Show the solution to question 12

    (a) Triangular powers and the exponential. Let $S$ and $T$ be upper triangular. For $i>j$ the entry $(ST{)}_{ij}={\sum }_m{S}_{im}{T}_{mj}$ vanishes: $S_{im}\ne 0$ needs $m\ge i$ and $T_{mj}\ne 0$ needs $m\le j$, and no $m$ satisfies $i\le m\le j$ when $i>j$. For $i=j$ the only surviving term is $m=i$, giving $(ST{)}_{ii}=S_{ii}{T}_{ii}$. So a product of upper triangular matrices is upper triangular with the products of the diagonals on its diagonal. By induction on $k$ (with $T^0=I$), $T^k$ is upper triangular with diagonal $t_1^k,\dots ,t_n^k$.

    The partial sums $E_K={\sum }_{k=0}^K{T}^k/k!$ are therefore upper triangular with diagonal entries ${\sum }_{k=0}^K{t}_i^k/k!$. Passing to the limit entry by entry: every below-diagonal entry is $0$ for every $K$, hence $0$ in the limit, and the $i$-th diagonal entry tends to $e^{t_i}$. So $e^T$ is upper triangular with diagonal $e^{t_1},\dots ,e^{t_n}$. $■$

    (b) Eigenvalues and determinant of $e^A$. Write $A=PT{P}^{-1}$ with $T$ upper triangular and diagonal ${\lambda }_1,\dots ,{\lambda }_n$. Then $e^A=P{e}^T{P}^{-1}$, so $e^A$ is similar to $e^T$ and has the same characteristic polynomial. By (a), $e^T$ is upper triangular with diagonal $e^{{\lambda }_1},\dots ,e^{{\lambda }_n}$, and the characteristic polynomial of a triangular matrix is ${\prod }_i(\lambda -e^{{\lambda }_i})$. So the eigenvalues of $e^A$, with multiplicity, are $e^{{\lambda }_1},\dots ,e^{{\lambda }_n}$.

    For the determinant, $\det e^A=\det P\cdot \det e^T\cdot \det P^{-1}=\det e^T$, and the determinant of a triangular matrix is the product of its diagonal:

    $$\det e^A=\prod _{i=1}^n{e}^{{\lambda }_i}=e^{{\lambda }_1+\cdots +{\lambda }_n}=e^{\mathrm{tr}T}=e^{\mathrm{tr}A},$$

    using that $\mathrm{tr}$ is a similarity invariant (it is minus the coefficient of ${\lambda }^{n-1}$ in the characteristic polynomial, which $A$ and $T$ share). $■$

    (c) Consequences. $e^{\mathrm{tr}A}\ne 0$ for every complex number $\mathrm{tr}A$, so $\det e^A\ne 0$ and $e^A$ is invertible — consistent with the direct argument that $e^{-A}$ is its inverse. If $A$ is real then $\mathrm{tr}A$ is real and $e^{\mathrm{tr}A}>0$; so, for instance, no real matrix has $e^A=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)$. Finally $\mathrm{tr}(At)=t\mathrm{tr}A$, so if $\mathrm{tr}A=0$ then $\det e^{At}=e^0=1$ for every $t$: the flow of ${\mathbf{x}}^{\prime }=A\mathbf{x}$ preserves volume exactly when the trace vanishes.