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Question:

Does a matrix $A \in M_{3 \times 3}(F)$ exist s.t. $A^4= \begin{bmatrix} 0&0&1\\0&0&0\\0&0&0\end{bmatrix}$

What I thought:

I think it doesn't. How do you start a proof of such a thing (prefer hints at first). Thx

jreing
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    Try computing the characteristic polynomial of $A$; you know how to do it, because you can compute its eigenvalues. Then, what does the Hamilton-Cayley theorem say? – egreg May 19 '13 at 19:48
  • @egreg I see how that means $A^12 = O, but what does that say about A? – rurouniwallace May 19 '13 at 19:58
  • Assuming there has been no answer, what can you say about the eigenvalues of $A$? Remember that if $\lambda$ is an eigenvalue of $A$, then $\lambda^n$ is an eigenvalue of $A^n$. – egreg May 19 '13 at 20:10

2 Answers2

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Hints:

Suppose such a matrix exists, then

(1) $\;A^8=0\;$ ;

(2) $\,A\,$ is nilpotent and of order $\,3\times 3\implies A^3=0\;$ , contradiction.

DonAntonio
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    Isn't that a bit more than just a hint ? – Dominic Michaelis May 19 '13 at 20:00
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    No. Point (1) is easy and just calculations, point(2) though requires to prove something: a nilpotent matrix always has a nilpotency degree of at most its order. In the case $,3\times 3,$ it maybe easier than in the general case, yet something's still must be proved. This, in my book, is a hint. – DonAntonio May 19 '13 at 20:03
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    Is this how this site is supposed to work? The OP has asked for a hint, I gave one; then you jump in giving the (almost) complete solution before the OP has acknowledged my hint. – egreg May 19 '13 at 20:08
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    Do you really think, @egreg, that I saw your comment before I posted my answer? Then you don't really know how this site works: many people writes down calmly, taking several minutes and up to dozens of minutes to complete an answer. The difference between your comment and my answer is hardly two minutes, and besides: can you say your comment and my answer are very alike? Mine is more computational, yours is more theoretical. – DonAntonio May 19 '13 at 20:13
  • @DonAntonio I don't want to be polemic. The time stamps say all. – egreg May 19 '13 at 20:17
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It's easy to compute $(A^4)^2=O$ (the null matrix).

I believe you already know the following result, but I'll prove it anyway, as it's useful in many situations.

If $\lambda$ is an eigenvalue of $A$ (in the complex numbers), then $\lambda^n$ is an eigenvalue of $A^n$: just do it by induction. The base step $n=1$ is obvious; then, if $v$ is an eigenvector for $\lambda$, you have, for $n>1$,

$$ A^nv=A(A^{n-1}v)=A(\lambda^{n-1}v)=\lambda^{n-1}Av=\lambda^nv. $$

In our case this means that the only eigenvalue of $A$ is $0$. So the characteristic polynomial of $A$ has a very simple form. Apply Hamilton-Cayley, to conclude that $A^3=\dots$ and get a contradiction.

egreg
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    Hi. I've gone to bed and woke to see the whole discussion. Actually we haven't got to Eigenvalues yet, so the solution with the nil potent matrix is more suitable for my class. But I'm always glad to see something new :-) Thanks! – jreing May 20 '13 at 05:03