Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 7 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 $\lambda^2I+N=J_n(\lambda^2)$.

share|improve this answer
    
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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.