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$S$ is a non-zero $3$ by $3$ matrix. Is the statement "$S^4 = 0$ but $S^3 \neq 0$" necessarily false?

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Hint: What could be the characteristic polynomial of $S$? – Did Oct 30 '12 at 13:27
up vote 1 down vote accepted

Since $S^4=0$, $S$ is a nilpotent. The degree of an n × n nilpotent matrix is always less than or equal to n, so $S^3=0$.

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Thanks! Though I have taken introductory linear algebra, we never did nilpotent matrices. So, I didn't know this property! :/ – Legendre Oct 30 '12 at 13:58
Oh nevermind :) I missed that you also specified it was 3x3! – rschwieb Oct 30 '12 at 14:04

Since you know that $x^4$ is an annihilating polynomial of $S$, that means the minimal polynomial divides $x^4$. Since the minimal polynomial is always of smaller (or equal) degree to the characteristic polynomial, we know that it must be one of $$\left\{x,\ x^2,\ x^3\right\}$$ In particular this shows that the characteristic polynomial must be $p(x)=x^3$ where Cayley-Hamilton shows that $S^3 = 0$. You cannot correctly conclude that $S^2 = 0$ however, for example $$S=\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix}$$ Satisfies $S^3 = 0$ but $S^2 \neq 0$.

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My mistake was answering the title question correctly, but failing to read the additional hypotheses mentioned in the OP :/ – rschwieb Oct 30 '12 at 14:07
@rschwieb We're all guilty of that sometimes! :) – EuYu Oct 30 '12 at 14:10

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