## Monthly Archives: December 2011

### Compute the exponential of a matrix

Let $D = \begin{bmatrix} 1 & 2 & -4 & 4 \\ 2 & -1 & 4 & -8 \\ 1 & 0 & 1 & -2 \\ 0 & 1 & -2 & 3 \end{bmatrix}$. Compute $\mathsf{exp}(D)$.

Let $P = \begin{bmatrix} 0 & 1 & 2 & 0 \\ 2 & 0 & -2 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix}$. Evidently, $P^{-1}DP = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix}$ is in Jordan canonical form.

Now $\mathsf{exp}(P^{-1}DP) = \mathsf{exp} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \oplus \mathsf{exp} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$, using this previous exercise. By this previous exercise, we have $\mathsf{exp} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} e & e \\ 0 & e \end{bmatrix}$. So $P^{-1}\mathsf{exp}(D)P = e P^{-1}DP$, and we have $\mathsf{exp}(D) = eD$.

(WolframAlpha agrees.)

### Compute the exponential of a Jordan block

Let $N$ be an $n \times n$ matrix over $\mathbb{C}$ with 1 on the first superdiagonal and 0 elsewhere. Compute $\mathsf{exp}(N\alpha)$. Next, suppose $J$ is an $n \times n$ Jordan block with eigenvalue $\lambda$, and compute $\mathsf{exp}(J\alpha)$.

First, given a parameter $0 \leq t \leq n$, we define a matrix $N_t$ as follows: $N_t = [a_{i,j}]$, where $a_{i,j} = 1$ if $j = i+t$ and $0$ otherwise. (That is, $N_t$ is the $n \times n$ matrix with 1 on the $t$th superdiagonal and 0 elsewhere, and $N = N_1$.)

Lemma: $N_1^k = N_k$ for $k \in \mathbb{N}$. Proof: We proceed by induction. Certainly the result holds for $k = 0$ and $k = 1$. Now suppose it holds for some $k$. We have $N_1(K+1) = N_1N_1^k$ $= N_1N_k$ $= [a_{i,j}][b_{i,j}]$ $= [\sum_\ell a_{i,\ell}b_{\ell,j}]$. Consider entry $(i,j)$: $\sum_\ell a_{i,\ell}b_{\ell,j}$. If $\ell \neq i+1$, then $a_{i,\ell} = 0$. So this sum is $a_{i,i+1}b_{i+1,j}$. Now if $j \neq i+1+t$, then the whole sum is 0; otherwise, it is 1. So in fact $N_1^{k+1} = N_{k+1}$. $\square$

Now $\mathsf{exp}(N\alpha) = \sum_{k \in \mathbb{N}} \dfrac{1}{k!}(\alpha N)^k$ $= \sum_{k \in \mathbb{N}} \dfrac{1}{k!} \alpha^k N_k$. As we saw in this previous exercise, powers of $N$ past the $n-1$th are all zero; so in fact we have $\mathsf{exp}(N\alpha) = \sum_{k=0}^{n-1} \dfrac{1}{k!} \alpha^k N_k$. Note that the indices of nonzero entries of the $N_k$ are mutually exclusive. So $\mathsf{exp}(N\alpha)$ is the $n \times n$ matrix whose $(i,j)$ entry is $\dfrac{\alpha^{j-i}}{(j-i)!}$ if $j \geq i$ and 0 otherwise.

Now let $J$ be the $n \times n$ Jordan block with eigenvalue $\lambda$. Note that $J\alpha = I\alpha + N\alpha$; using this previous exercise, we have $\mathsf{exp}(J\alpha) = e^\alpha \mathsf{exp}(N\alpha)$. (Where $N$ is as above.)

### A fact about matrix exponentials

Let $\lambda \in \mathbb{C}$ and let $M$ be an $n \times n$ matrix over $\mathbb{C}$. Prove that $\mathsf{exp}(\lambda I + M) = e^{\lambda}\mathsf{exp}(M)$.

Note that $\lambda I$ and $M$ commute. Using two previous exercises (here and here), we have $\mathsf{exp}(\lambda I + M) = \mathsf{exp}(\lambda I)\mathsf{exp}(M)$ $= \mathsf{exp}(\lambda)I \mathsf{exp}(M)$ $= e^\lambda \mathsf{exp}(M)$.

### For commuting matrices, the exponential of a sum is the product of exponentials

Let $A$ and $B$ be commuting matrices. Prove that $\mathsf{exp}(A+B) = \mathsf{exp}(A)\mathsf{exp}(B)$, where $\mathsf{exp}(x) = \sum_{k \in \mathbb{N}} \dfrac{1}{k!}x^k$ is a formal power series over a field with characteristic zero.

We will work in the ring $F[[x]][[y]]$ of formal power series over $y$ with coefficients which are formal power series over $x$. (Note that $x$ and $y$ commute.) Note that if $G(x)$ is a formal power series and $p(x)$ a polynomial, then $G(p(x))$ is the formal power series obtained by ‘collecting like terms’. Remember the binomial theorem.

 $\mathsf{exp}(x+y)$ = $\sum_{t \in \mathbb{N}} \dfrac{1}{t!}(x+y)^t)$ = $\sum_{t \in \mathbb{N}} \dfrac{1}{t!} \sum_{k=0}^t {t \choose k} x^ky^{t-k}$ = $\sum_{t \in \mathbb{N}} \sum_{k = 0}^t \dfrac{1}{k!}x^k \cdot \dfrac{1}{(t-k)!}y^{t-k}$ = $\sum_{t \in \mathbb{N}} \sum_{h+k=t} \dfrac{1}{h!}x^h \cdot \dfrac{1}{k!}y^{k}$ = $\sum_{h \in \mathbb{N}} \sum_{k \in \mathbb{N}} \dfrac{1}{h!}x^h \cdot \dfrac{1}{k!}y^k$ = $\left( \sum_{h \in \mathbb{N}} \dfrac{1}{h!}x^h \right) \left( \sum_{k \in \mathbb{N}} \dfrac{1}{k!}y^{k} \right)$ = $\mathsf{exp}(x) \mathsf{exp}(y)$.

In particular, if $A$ and $B$ commute, then $\mathsf{exp}(A+B) = \mathsf{exp}(A)\mathsf{exp}(B)$.

### Facts about power series of matrices

Let $G(x) = \sum_{k \in \mathbb{N}} \alpha_k x^k$ be a power series over $\mathbb{C}$ with radius of convergence $R$. Let $A$ be an $n \times n$ matrix over $\mathbb{C}$, and let $P$ be a nonsingular matrix. Prove the following.

1. If $G(A)$ converges, then $G(P^{-1}AP)$ converges, and $G(P^{-1}AP) = P^{-1}G(A)P$.
2. If $A = B \oplus C$ and $G(A)$ converges, then $G(B)$ and $G(C)$ converge and $G(B \oplus C) = G(B) \oplus G(C)$.
3. If $D$ is a diagonal matrix with diagonal entries $d_i$, then $G(D)$ converges, $G(d_i)$ converges for each $d_i$, and $G(D)$ is diagonal with diagonal entries $G(d_i)$.

Suppose $G(A)$ converges. Then (by definition) the sequence of matrices $G_N(A)$ converges entrywise. Let $G_N(A) = [a_{i,j}^N]$, $P = [p_{i,j}]$, and $P^{-1} = [q_{i,j}]$. Now $G_N(P^{-1}AP) = P^{-1}G_N(A)P$ $= [\sum_\ell \sum_k q_{i,k} a_{k,\ell}^Np_{\ell,j}]$. That is, the $(i,j)$ entry of $G_N(P^{-1}AP)$ is $\sum_\ell \sum_k q_{i,k} a_{k,\ell}^Np_{\ell,j}$. Since each sequence $a_{k,\ell}^N$ converges, this sum converges as well. In particular, $G(P^{-1}AP)$ converges (again by definition). Now since $G_N(P^{-1}AP) = P^{-1}G_N(A)P$ for each $N$, the corresponding sequences for each $(i,j)$ are equal for each term, and so have the same limit. Thus $G(P^{-1}AP) = P^{-1}G(A)P$.

Now suppose $A = B \oplus C$. We have $G_N(B \oplus C) = \sum_{k=0}^N \alpha_k (B \oplus C)^k$ $= \sum_{k=0}^N \alpha_k B^k \oplus C^k$ $= (\sum_{k = 0}^N \alpha_k B^k) \oplus (\sum_{k=0}^N \alpha_k C^k)$ $= G_N(B) \oplus G_N(C)$. Since $G_N(A)$ converges in each entry, each of $G_N(B)$ and $G_N(C)$ converge in each entry. So $G(B)$ and $G(C)$ converge. Again, because for each $(i,j)$ the corresponding sequences $G_N(A)_{i,j}$ and $(G_N(B) \oplus G_N(C))_{i,j}$ are the same, they converge to the same limit, and thus $G(B \oplus C) = G(B) \oplus G(C)$.

Finally, suppose $D$ is diagonal. Then in fact we have $D = \bigoplus_{t=1}^n [d_t]$, and so by the previous part, $G(D) = \bigoplus_{t=1}^n G(d_t)$. In particular, $G(d_t)$ converges, and $G(D)$ is diagonal with diagonal entries $G(d_t)$ as desired.

### On the convergence of formal power series of matrices

Let $G(x) = \sum_{k \in \mathbb{N}} \alpha_kx^k$ be a formal power series over $\mathbb{C}$ with radius of convergence $R$. Let $||A|| = \sum_{i,j} |a_{i,j}|$ be the matrix norm introduced in this previous exercise.

Given a matrix $A$ over $\mathbb{C}$, we can construct a sequence of matrices by taking the $N$th partial sum of $G(A)$. That is, $G_N(A) = \sum_{k = 0}^N a_kA^k$. This gives us $n^2$ sequences $\{c_{i,j}^N\}$ where $c_{i,j}^N$ is the $(i,j)$ entry of $G_N(A)$. Suppose $c_{i,j}^N$ converges to $c_{i,j}$ for each $(i,j)$, and let $C = [c_{i,j}]$. In this situation, we say that $G_N(A)$ converges to $C$, and that $G(A) = C$. (In other words, $G_N(A)$ converges precisely when each entrywise sequence $G_N(A)_{i,j}$ converges.)

1. Prove that if $||A|| \leq R$, then $G_N(A)$ converges.
2. Deduce that for all matrices $A$, the following power series converge: $\mathsf{sin}(A) = \sum_{k \in \mathbb{N}} \dfrac{(-1)^k}{(2k+1)!}A^{2k+1}$, $\mathsf{cos}(A) = \sum_{k \in \mathbb{N}} \dfrac{(-1)^k}{(2k)!}A^{2k}$, and $\mathsf{exp}(A) = \sum_{k \in \mathbb{N}} \dfrac{1}{k!} A^k$.

[Disclaimer: My facility with even simple analytical concepts is laughably bad, but I’m going to give this a shot. Please let me know what’s wrong with this solution.]

We begin with a lemma.

Lemma: For all $(i,j)$, $(A^k)_{i,j} \leq ||A||^k$, where the subscripts denote taking the $(i,j)$ entry. Proof: By the definition of matrix multiplication, we have $A^k = [\sum_{t \in n^{k-1}} \prod_{i=0}^{k-1} a_{t(i), t(i+1)}]$, where $t(0) = i$ and $t(k+1) = j$. (I’m abusing the notation a bit here; $t$ is an element of $\{1,\ldots,n\}^{k-1}$, which we think of as a function on $\{1,\ldots,k-1\}$.) Now $|A^k_{i,j}| \leq \sum_{t \in n^{k-1}} \prod_{i=0}^{k-1} |a_{t(i), t(i+1)}|$ by the triangle inequality. Note that $||A||^k$ is the sum of all possible $k$-fold products of (absolute values of) entries of $A$, and that we have $|A^k_{i,j}|$ bounded above by a sum of some distinct $k$-fold products of (absolute values of) entries of $A$. In particular, $|A^k_{i,j}| \leq ||A||^k$ since the missing terms are all positive. Thus $|\alpha_k A^k_{i,j}| = |(\alpha_k A^k)_{i,j}| \leq |\alpha_k| ||A||^k$.

Let us define the formal power series $|G|(x)$ by $|G|(x) = \sum |\alpha_k| x^k$. What we have shown (I think) is that $\sum_{k=0}^N |\alpha_k A^k_{i,j}| \leq |G|_N(||A||)$, using the triangle inequality.

Now recall, by the Cauchy-Hadamard theorem, that the radius of convergence of $G(x)$ satisfies $1/R = \mathsf{lim\ sup}_{k \rightarrow \infty} |\alpha_k|^{1/k}$. In particular, $|G|$ has the same radius of convergence as $G$. So in fact the sequence $|G|_N(||A||)$ converges. Now the sequence $\sum_{k=0}^N |\alpha_k A^k_{i,j}|$ is bounded and monotone increasing, and so must also converge. That is, $\sum_{k=0}^N (\alpha_k A^k)_{i,j} = G_N(A)_{i,j}$ is absolutely convergent, and so is convergent. Thus $G(A)$ converges.

Now we will address the convergence of the series $\mathsf{sin}$. Recall that the radius of convergence $R$ of $\mathsf{sin}$ satisfies $R^{-1} = \mathsf{lim\ sup}_{k \rightarrow \infty} |\alpha_k|^{1/k}$ $\mathsf{lim}_{k \rightarrow \infty} \mathsf{sup}_{n \geq k} |\alpha_k|^{1/k}$. Now $\alpha_k = 0$ if $k$ is even and $(-1)^t/k!$ if $k = 2t+1$ is odd. So $|\alpha_k|^{1/k} = 0$ if $k$ is even and $|1/k!|^{1/k} = 1/(k!)^{1/k} > 0$ if $k$ is odd.

Brief aside: Suppose $k \geq 1$. Certainly $1 \leq k! < (k+1)^k$, so that $(k!)^{1/k} < k+1$. Now $1 \leq (k!)^{1 + 1/k} < (k+1)!$, so that $(k!)^{1/k} < ((k+1)!)^{1/(k+1)}$. In particular, if $n > k$, then $(k!)^{1/k} < (n!)^{1/n}$.

By our brief aside, we have that $\mathsf{sup}_{n \geq k} |\alpha_k|^{1/k} = 1/(k!)^{1/k}$ if $k$ is odd, and $1/((k+1)!)^{1/(k+1)}$ if $k$ is even. So (skipping redundant terms) we have $\mathsf{sin}(x)$ satisfies $R^{-1} = \mathsf{lim}_{k \rightarrow \infty} 1/(k!)^{1/k}$. We know from the Brief Aside that this sequence is monotone decreasing. Now let $\varepsilon > 0$. Consider now the sequence $\varepsilon$, $2 \varepsilon$, $3 \varepsilon$, et cetera. only finitely many terms of this sequence are less than or equal to 1. So there must exist some $k$ such that the product $\prod_{i=1}^k \varepsilon i$ is greater than 1. Now $\varepsilon^k k! > 1$, so that $\varepsilon > 1/(k!)^{1/k}$. Thus the sequence $1/(k!)^{1/k}$ converges to 0, and so the radius of convergence of $\mathsf{sin}(x)$ is $\infty$.

By a similar argument, the radii of convergence for $\mathsf{cos}(X)$ and $\mathsf{exp}(X)$ are also $\infty$.

### Properties of a matrix norm

Given an $n \times n$ matrix $A$ over $\mathbb{C}$, define $||A|| = \sum_{i,j} |a_{i,j}|$, where bars denote the complex modulus. Prove the following for all $A,B \in \mathsf{Mat}_n(\mathbb{C})$ and all $\alpha \in \mathbb{C}$.

1. $||A+B|| \leq ||A|| + ||B||$
2. $||AB|| \leq ||A|| \cdot ||B||$
3. $||\alpha A|| = |\alpha| \cdot ||A||$

Say $A = [a_{i,j}]$ and $B = [b_{i,j}]$.

We have $||A+B|| = \sum_{i,j} |a_{i,j} + b_{i,j}|$ $\leq \sum_{i,j} |a_{i,j}| + |b_{i,j}|$ by the triangle inequality. Rearranging, we have $||A+B|| \leq (\sum_{i,j} |a_{i,j}| + \sum_{i,j} |b_{i,j}|$ $= ||A|| + ||B||$ as desired.

Now $||AB|| = \sum_{i,j} |\sum_k a_{i,k}b_{k,j}|$ $\leq$latex \sum_{i,j} \sum_k |a_{i,k}|b_{k,j}|\$ by the triangle inequality. Now $||AB|| \leq \sum_{i,j} \sum_{k,t} |a_{i,k}| |b_{t,j}|$ since all the new terms are positive, and rearranging, we have $||AB|| \leq \sum_{i,k} \sum_{j,t} |a_{i,k}||b_{t,j}|$ $= (\sum_{i,k} |a_{i,k}|)(\sum_{j,t} |b_{t,j}|)$ $= ||A|| \cdot ||B||$.

Finally, we have $||\alpha A|| = \sum_{i,j} |\alpha a_{i,j}|$ $= |\alpha| \sum_{i,j} |a_{i,j}|$ $= |\alpha| \cdot ||A||$.

### Matrices with square roots over fields of characteristic 2

Let $F$ be a field of characteristic 2. Compute the Jordan canonical form of a Jordan block $J$ of size $n$ and eigenvalue $\lambda$ over $F$. Characterize those matrices $A$ over $F$ which are squares; that is, characterize $A$ such that $A = B^2$ for some matrix $B$.

Let $J = [b_{i,j}]$ be the Jordan block with eigenvalue $\lambda$ and size $n$. That is, $b_{i,j} = \lambda$ if $j = i$, $1$ if $j = i+1$, and 0 otherwise. Now $J^2 = [\sum_k b_{i,k}b_{k,j}]$; if $k \neq i$ or $k \neq i+1$, then $b_{i,k} = 0$. Evidently then we have $(J^2)_{i,j} = \lambda^2$ if $j = i$, $1$ if $j = i+2$, and 0 otherwise, Noting that $2 = 0$. So $J^2 - \lambda^2 I = \begin{bmatrix} 0 & I \\ 0_2 & 0 \end{bmatrix}$, where $0_2$ is the $2 \times 2$ zero matrix and $I$ is the $(n-2) \times (n-2)$ identity matrix. Now let $v = \begin{bmatrix} V_1 \\ V_2 \end{bmatrix}$, where $V_1$ has dimension $2 \times 1$. Now $(J^2 - \lambda^2 I)v = \begin{bmatrix} V_2 \\ 0 \end{bmatrix}$. That is, $J^2 - \lambda^2 I$ ‘shifts’ the entries of $v$– so $e_{i+2} \mapsto e_i$ and $e_1, e_2 \mapsto 0$. In particular, the kernel of $J^2 - \lambda^2 I$ has dimension 2, so that by this previous exercise, the Jordan canonical form of $J^2$ has two blocks (both with eigenvalue $\lambda^2$.

Now $J^2 - \lambda^2 I = (J + \lambda I)(J - \lambda I)$ $= (J - \lambda I)^2$, since $F$ has characteristic 2. Note that $J-\lambda I$ has order $n$, since (evidently) we have $(J - \lambda I)e_{i+1} = e_i$ and $(J - \lambda I)e_1 = 0$. So $J-\lambda I$ has order $n$. If $n$ is even, then $(J^2 - \lambda^2 I)^{n/2} = 0$ while $(J^2 - \lambda^2 I)^{n/2-1} \neq 0$, and if $n$ is odd, then $(J^2-\lambda^2 I)^{(n+1)/2} = 0$ while $(J^2 - \lambda^2 I)^{(n+1)/2-1} \neq 0$. So the minimal polynomial of $J^2$ is $(x-\lambda^2)^{n/2}$ if $n$ is even and $(x-\lambda^2)^{(n+1)/2}$ if $n$ is odd.

So the Jordan canonical form of $J^2$ has two Jordan blocks with eigenvalue $\lambda^2$. If $n$ is even, these have size $n/2,n/2$, and if $n$ is odd, they have size $(n+1)/2, (n-1)/2$.

Now let $A$ be an arbitrary $n \times n$ matrix over $F$ (with eigenvalues in $F$). We claim that $A$ is a square if and only if the following hold.

1. The eigenvalues of $A$ are square in $F$
2. For each eigenvalue $\lambda$ of $A$, the Jordan blocks with eigenvalue $\lambda$ can be paired up so that the sizes of the blocks in each pair differ by 0 or 1.

To see the ‘if’ part, suppose $P^{-1}AP = \bigoplus H_i \oplus K_i$ is in Jordan canonical form, where $H_i$ and $K_i$ are Jordan blocks having the same eigenvalue $\lambda_i$ and whose sizes differ by 0 or 1. By the first half of this exercise, $H_i \oplus K_i$ is the Jordan canonical form of $J_i^2$, where $J_i$ is a Jordan block with eigenvalue $\sqrt{\lambda_i}$. Now $A$ is similar to the direct sum of these $J_i^2$, and so $Q^{-1}AQ = J^2$. Then $A = (Q^{-1}JQ)^2 = B^2$ is square.

Conversely, suppose $A = B^2$ is square, and say $P^{-1}BP = J$ is in Jordan canonical form. So $P^{-1}AP = J^2$. Letting $J_i$ denote the Jordan blocks of $J$, we have $P^{-1}AP = \bigoplus J_i^2$. The Jordan canonical form of $J_i^2$ has two blocks with eigenvalue $\lambda_i^2$ and whose sizes differ by 0 or 1, by the first half of this exercise. So the Jordan blocks of $A$ all have eigenvalues which are square in $F$ and can be paired so that the sizes in each pair differ by 0 or 1.

### Matrices with square roots

Let $F$ be a field whose characteristic is not 2. Characterize those matrices $A$ over $F$ which have square roots. That is, characterize $A$ such that $A = B^2$ for some matrix $B$.

We claim that $A$ has a square root if and only if the following hold.

1. Every eigenvalue of $A$ is square in $F$.
2. The Jordan blocks of $A$ having eigenvalue 0 can be paired up in such a way that the sizes of the blocks in each pair differ by 0 or 1.

First we tackle the ‘if’ part. Suppose $P^{-1}AP = \bigoplus J_i \oplus \bigoplus H_i \oplus \bigoplus K_i$ is in Jordan canonical form, where $J_i$ are the Jordan blocks having nonzero eigenvalue $\lambda_i$ and $H_i$ and $K_i$ are the blocks with eigenvalue 0 such that the sizes of $H_i$ and $K_i$ differ by 0 or 1. As we showed in this previous exercise, $J_i = Q_iM_i^2Q_i^{-1}$ is the Jordan canonical form of the square of the Jordan block $M_i$ with eigenvalue $\sqrt{\lambda_i}$, and $H_i \oplus K_i = T_iN_i^2T_i^{-1}$ is the Jordan canonical form of the square of the Jordan block whose size is $\mathsf{dim}\ H_i + \mathsf{dim}\ K_i$ with eigenvalue 0. So we have $P^{-1}AP = C^2$, and so $A = (P^CP^{-1})^2$ $= B^2$ is a square.

Conversely, suppose $A = B^2$ is a square, and say $P^{-1}BP = J$ is in Jordan canonical form. Now $PAP^{-1} = J^2$. Now say $Q^{-1}J^2Q = K$ is in Jordan canonical form; then $Q^{-1}PAP^{-1}Q = K$. By the previous exercise, the nonzero eigenvalues of $K$ are square in $F$, and the Jordan blocks having eigenvalue 0 can be paired so that the sizes in each pair differ by 0 or 1.

### On the square of a Jordan block

Let $J$ be a Jordan block of size $n$ and eigenvalue $\lambda$ over a field $K$ whose characteristic is not 2.

1. Suppose $\lambda \neq 0$. Prove that the Jordan canonical form of $J^2$ is the Jordan block of size $n$ with eigenvalue $\lambda^2$.
2. Suppose $\lambda = 0$. Prove that the Jordan canonical form of $J^2$ has two blocks (with eigenvalue 0) of size $n/2, n/2$ if $n$ is even and of size $(n+1)/2, (n-1)/2$ if $n$ is odd.

First suppose $\lambda \neq 0$.

Lemma: Let $e_i$ denote the $i$th standard basis element (i.e. $e_i = [\delta_{i,1}\ \delta_{i,2}\ \ldots\ \delta_{i,n}]^\mathsf{T}$. If $1 \leq i < n-2$, then $(J^2-\lambda^2I)e_{i+2} = 2\lambda e_{i+1} + e_i$, $(J^2-\lambda^2I)e_{2} = 2\lambda e_1$, and $(J^2-\lambda^2I)e_1 = 0$. Proof: We have $J^2-\lambda^2I = (J+\lambda I)(J-\lambda I)$. Evidently, $(J-\lambda I)e_{i+2} = e_{i+1}$. Now $J+\lambda I = [b_{j,k}]$, where $b_{j,k} = 2\lambda$ if $j = k$ and $1$ if $k = j+1$ and 0 otherwise. So $(J+\lambda I)e_{i+1} = [\sum_k b_{j,k} \delta_{k,i+1}]$ $= [b_{j,i+1}]$ $= 2\lambda e_{i+1} + e_i$ if $i > 1$, and similarly $(J^2-\lambda^2 I)(e_2) = 2\lambda e_1$. $\square$

In particular, note that $(J^2 - \lambda^2 I)^{n-1} e_n = 2^{n-1}\lambda^{n-1} e_1$ is nonzero (here we use the noncharacteristictwoness of $K$), while $(J^2 - \lambda^2 I)^n = 0$. So the minimal polynomial of $J^2$ is $(x-\lambda^2)^n$. Thus the Jordan canonical form of $J^2$ has a single Jordan block, of size $n$, with eigenvalue $\lambda^2$.

Now suppose $\lambda = 0$. Evidently, we have $Je_{i+1} = e_i$ and $Je_1 = 0$. Now $J^n = 0$ and $J^{n-1} \neq 0$. If $n$ is even, we have $(J^2)^{n/2} = 0$ and $(J^2)^{n/2-1} = J^{n-2} \neq 0$, so that the minimal polynomial of $J^2$ is $x^{n/2}$. If $n$ is odd, we have $(J^2)^{(n+1)/2} = 0$ while $(J^2)^{(n+1)/2-1} = J^{n-1} \neq 0$, so the minimal polynomial of $J^2$ is $x^{(n+1)/2}$.

Next, we claim that $\mathsf{ker}\ J^2$ has dimension 2. To this end, note that $J^2e_{i+2} = e_i$ is nonzero, while $J^2e_2 = J^2e_1 = 0$. By this previous exercise, $J^2$ has 2 Jordan blocks; thus the blocks of $J^2$ both have eigenvalue 0 and are of size $n/2,n/2$ if $n$ is even, and $(n+1)/2, (n-1)/2$ if $n$ is odd.