What is the Jordan canonical form of $A^{2}$ if we know that of $A$? 
Let $A\in M_{n}$ have Jordan canonical form $J_{n_1}(\lambda_{1})\oplus\cdots\oplus J_{n_k}(\lambda_{k})$. If $A$ is non-singular ($\lambda_i\neq 0$), what is the Jordan canonical form of $A^{2}$?

I can prove that if the eigenvalues of $A$ are $\sigma(A)=\{\lambda_{1},\dots, \lambda_{n} \}$ then $\sigma(A^{2})=\{\lambda_{1}^{2},\dots, \lambda_{n}^{2} \}$, for this reason I have been trying to attack this problem using this fact, but I am getting nowhere. How should I proceed?
 A: Everything works blockwise, so you can simply assume that $A$ is one Jordan block...
So let $A=J_n(\lambda)$, which we can write as $\lambda I+N$ with $N=J_n(0)$. Then $A^2=\lambda^2I+2\lambda N+N^2$. The matrix $N'=2\lambda N+N^2$ is nilpotent and (because $\lambda\neq0$) has rank $n-1$, so it is conjugate to $N$. It follows that $A^2$ is conjugate to $\lambda^2I+N=J_n(\lambda^2)$.
A: Thank you @Mariano. Intuitively I believe this makes sense, but I just want to go through some details.
Given the Jordan canonical form of a matrix $A$, I want to show that an arbitrary Jordan block of $A$ corresponding to the eigenvalue $\lambda$, $J_{k}(\lambda)$, gives rise to precisely one Jordan block $J_{k}(\lambda^{2})$ for $J_{k}(\lambda)^{2}$.
Let $J_{k}(\lambda)=\lambda I + N$, where $N=J_{k}(0)$. Then $J_{k}(\lambda)^{2}=\lambda^{2}I+2\lambda N +N^{2}=\lambda^{2}I+N'$.
Now consider 
$$\mathrm{rank}(J_{k}(\lambda)-\lambda I)=\mathrm{rank}(N).$$ 
By construction, $N$ is the matrix with all zero entries except for $1$'s on the super diagonal, so $\mathrm{rank}(N^{i})=k-i$ for $i=1,2,...,k$. Initially the rank of $N$ is $k-1^{(*)}$ because the first column consists of all zeros and the rest of the columns contain nonzero entries. Each successive power of $N$ reduces the rank by $1$. Similarly, 
$$\mathrm{rank}(J_{k}(\lambda)^{2}-\lambda^{2} I)=\mathrm{rank}(N')$$ 
and for similar reasons $\mathrm{rank}(N'^{i})=k-i$ for $i=1,2,...,k$.
Therefore $$\mathrm{rank}(J_{k}(\lambda)-\lambda I)^{i}=\mathrm{rank}(J_{k}^{2}(\lambda)-\lambda^{2} I)^{i}$$ for $i=1,2,...,k$.
In particular,  $$\mathrm{rank}(J_{k}(\lambda)^{2}-\lambda^{2} I)^{0}=k$$ and $$\mathrm{rank}(J_{k}(\lambda)^{2}-\lambda^{2} I)^{1}=k-1.$$
This tells us that the Jordan canonical form of the single Jordan block $J_{k}(\lambda)^{2}$ is $J_{k}(\lambda^{2})$. (If $\mathrm{rank}(J_{k}(\lambda)^{2}-\lambda^{2} I)^{1}>k-1$ then the Jordan canonical form for $J_{k}(\lambda)^{2}$ would contain more than one block.)
$^{(*)}$ Note: I glossed over the proof that $N=J_{k}(0)$ has rank $k-1$. I think to prove this I can argue that since $I_{k-1}$ has rank $k-1$, by appending first a $1\times k-1$ zero row to $I_{k}$ and then a $k\times 1$ zero column, the rank remains unchanged. And by the property of nilpotent matrices, successive power reduce the rank by 1.
