Determine a Jordan Canonical basis for $T^2$ Question: Suppose that $T : V \to V$ is a linear operator on a complex vector space and that $\{v_1,v_2,\dots,v_n\}$ is a basis for $V$ that is a single Jordan chain (in other words, a cycle of generalized eigenvectors) for $T$. Determine a Jordan canonical basis for $T^2$.
My attempt: We can easily show, using the JCF of $T$, that the only eigenvalue of $T^2$ is $\lambda^2$, where $\lambda$ is the eigenvalue corresponding to the Jordan chain in the problem statement. Let $A$ be a matrix representation of $T$. Now, we also have that $(A^2 -\lambda^2 I) = (A-\lambda I)(A+\lambda I)$. Let $w_n = (A+\lambda I) v_n$ (note that we are using the convention that $v_{2} = (A-\lambda I)v_1,$ $v_{3} = (A-\lambda I)^2v_{1},$ $\dots, v_n = (A-\lambda I)^{n-1}v_1$). 
We also have that $(A^2 -\lambda^2)^2 = (A-\lambda I)^2(A+\lambda I)^2$, and, in general, $(A^2 -\lambda^2)^k = (A-\lambda I)^k(A+\lambda I)^k$.
This all feels promising, but I can't seem to really find the basis - this tells us that
$$(A^2 - \lambda^2 I)^nv_1 = (A-\lambda I)^n(A-\lambda I)^nv_1 = 0,$$
but I'm getting a little lost and don't know how to proceed from here.
 A: The classical Jordan canonical form is associated to a basis s.t. $(T-\lambda I_n)v_i=v_{i-1}$ (reverse the order). Anyway, the matrix associated to $T$ is a Jordan block $A=\lambda I_n+J_n$ where $J_n$ is the nilpotent Jordan block of dimension $n$. Then $A^2-\lambda^2I=(A-\lambda I)(A+\lambda I)$.
Case 1. $\lambda\not= 0$. Then $A+\lambda I$ is invertible and the CJF associated to $A^2$ is $\lambda^2I_n+J_n$.
Case 2. $\lambda=0$. Consider $dim(\ker(A^2)),dim(\ker(A^4)),\cdots$.  the CJF associated to $A^2$ is:
when $n=2p$: $diag(J_{p-1},J_{p-1})$.
when $n=2p+1$: $diag(J_p,J_{p-1})$. 
A: Let $v_1,\dotsc,v_n$ be a basis of $V$ such that $(T-\lambda I)v_i = v_{i+1}$ for $i=1,\dotsc,n-1$.
Assume first, that $\lambda\neq 0$. Put $S:= T+\lambda I$. Then $S$ is invertible and commutes with $T$. Now, put $w_i:= S^{i}v_i$. Then for each $i=1,\dotsc,n-1$, we have
$$\begin{align*}
(T^2-\lambda^2I)w_i &= (T+\lambda I)(T-\lambda I)S^{i}v_i\\
&= S^{i+1}(T-\lambda I)v_i\\
&= S^{i+1} v_{i+1}\\
&= w_{i+1}.
\end{align*}$$
Hence $w_1,\dotsc,w_n$ is your basis. Indeed, linear independence can be shown by descending induction on $n$: $w_n$ is not zero, hence linearly independent. Assuming $w_{i+1},\dotsc,w_n$ is linearly independent, let $\mu_i,\mu_{i+1},\dotsc,\mu_n$ be scalars such that $\sum_{j=i}^n\mu_jw_j = 0$. Applying $T^2-\lambda^2$ on this equation yields
$$
0 = (T^2-\lambda^2)\left(\sum_{j=i}^n\mu_jw_j\right) = \sum_{j=i}^n\mu_j (T^2-\lambda^2)w_j = \sum_{j=i}^n \mu_jw_{j+1} = \sum_{j=i+1}^n\mu_{j-1}w_j,
$$
where we put $w_{n+1}:= (T^2-\lambda^2)w_n=0$. By induction hypothesis, we have $\mu_i=\dotsb=\mu_{n-1} = 0$. Hence the above equation reads $\mu_nw_n = 0$, which implies $\mu_n = 0$ and thus, $w_i,\dotsc,w_n$ is linearly independent.
If $\lambda = 0$, then if $n$ is even, put $w_i = v_{2i}$ for $i=1,\dotsc, n/2$ and $w_{n/2+i} = v_{2i-1}$ for $i=1,\dotsc,n/2$.
If $n$ is odd, put $w_i = v_{2i}$ for $i=1,\dotsc,(n-1)/2$ and $w_{(n-1)/2 +i} = v_{2i-1}$ for $i=1,\dotsc, (n+1)/2$. Then $w_1,\dotsc,w_n$ is your basis (since it is just a permutation of $v_1,\dotsc,v_n$).
