Non-zero matrix $S$. $S^4 = 0$ but $S^3 \neq 0$ is false?

$S$ 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

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 $S^3=0$.
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 $S^2 \neq 0$.