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Let $A\in M_{n}$ have Jordan canonical form $J_{n1}(\lambda_{1})\oplus\cdots\oplus J_{nk}(\lambda_{k})$. If $A$ is non-singular($\lambda\neq 0$), what is the Jordan canonical form of $A^{2}$?

I can prove that if the eigenvalues of A are $\sigma(A)=\{\lambda_{1}\cdots \lambda_{n} \}$ then $\sigma(A^{2})=\{\lambda_{1}^{2}\cdots \lambda_{n}^{2} \}$, for this reason I have been trying to attack this problem using this fact, but I am getting nowhere. How should I proceed?

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up vote 8 down vote accepted

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 $\lambda^2I+N=J_n(\lambda^2)$.

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I understand up until you said " so it is conjugate to $N$" I am not familiar with that term, and I cannot find a definition that makes this statement clear to me. Could you please elaborate? Thanks. – Edison Oct 14 '11 at 4:59
You can replace that phrase with «so its Jordan form is $N$» – Mariano Suárez-Alvarez Oct 14 '11 at 5:27
thank you for this explanation. there are some details that have been bugging me. I have written them out as a new answer, only because this space is too small. Are there any glaring mistakes? Thanks again. – Edison Oct 16 '11 at 7:20

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 $J_{k}^{2}(\lambda)$.

Let $J_{k}(\lambda)=\lambda I + N$ , where $N=J_{k}(0)$, then $J_{k}^{2}(\lambda)=\lambda^{2}I+2\lambda N +N^{2}=\lambda^{2}I+N^{'}$.

Now consider $$rank(J_{k}(\lambda)-\lambda I)$$ $$=rank(N)$$

By construction, $N$ is the matrix with all zero entries except for 1's on the super diagonal, so $rank(N)^{i}=k-1$ 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,

$$rank(J_{k}^{2}(\lambda)-\lambda^{2} I)$$ $$=rank(N')$$ and for similar reasons, $rank(N^{'})^{i}=k-1$ for $i=1,2,...k$.

Therefore $$rank(J_{k}(\lambda)-\lambda I)^{i}=rank(J_{k}^{2}(\lambda)-\lambda^{2} I)^{i}$$ for $i=1,2,...k$.

In particular, $$rank(J_{k}^{2}(\lambda)-\lambda^{2} I)^{0}=k$$ and $$rank(J_{k}^{2}(\lambda)-\lambda^{2} I)^{1}=k-1$$ This tells us that the Jordan canonical form of the single Jordan block $J_{k}^{2}(\lambda)$ is $J_{k}(\lambda^{2})$. (If $rank(J_{k}^{2}(\lambda)-\lambda^{2} I)^{1}>k-1$ then the Jordan canonical form for $J_{k}^{2}(\lambda)$ 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.

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