if $S$ is an $3$ by $3$ matrix with entries from real numbers

why it's not possible that $S^4=0$ but $S^3$ is not zero?

I'm thinking it's something related to nilpotent. And in nilpotent, there's a thm saying that, If $S$: $n$ by n matrix, $S$ is nilpotent Then $det(S)=0$ and $S^n=0$

But i don't think this thm can be of any help in this question. Anyone can point out the right direction? Thanks so much!!

  • $\begingroup$ Look at the grade of minimal polynomial of S... $\endgroup$ Oct 2, 2013 at 14:41

2 Answers 2


Note that if $S$ acts on the vector space $V$ (of dimension $3$) then $V\supset SV$. Applying $S$ to this relation gives $SV\supset S^2V$ so we have $V\supset SV \supset S^2V \supset S^3V \supset S^4V=0$

If any of the inclusions is an equality, the chain becomes stationary from that point. If none is an equality, each inclusion implies a strict reduction in dimension, so $\dim V \ge 4$

  • $\begingroup$ I still not follow.... :( $\endgroup$
    – Lily
    Oct 2, 2013 at 15:57

If $S$ is a $3 \times 3$ real matrix with $S^4 = 0$, we see that all the eigenvalues of $S$ vanish. For if $\lambda$ is an eigenvalue of $S$, there is a nonzero vector $v$ such that

$Sv = \lambda v, \tag{1}$

which is readily seen to imply that

$0 = S^4v = \lambda^4v, \tag{2}$

whence, since $v \ne 0$, we must have

$\lambda ^4 = 0, \tag{3}$


$\lambda = 0. \tag{4}$

The fact that all the eigenvalues of $S$ vanish implies that $\det S = 0$; thus the characteristic polynomial $p_S(\lambda)$ of $S$ must take the form

$p_S(\lambda) = \lambda^3 + a\lambda^2 + b\lambda, \tag{5}$

with $a,b \in \Bbb R$. By the Hamilton-Cayley theorem, $S$ satisfies its own characteristic polynomial, so that

$S^3 + aS^2 + bS = p_S(S) = 0. \tag{6}$

If now $a = b= 0$, we have $S^3 = 0$; we are done. If at least one of $a, b$ is nonzero, we multiply (6) by $S$ to obtain, using $S^4 = 0$,

$aS^3 + bS^2 = Sp_S(S) = 0; \tag{7}$

If $b = 0$, $a \ne 0$, (7) implies $S^3 = 0$; we are done. If $b \ne 0$, multiply (7) by $S$ to yield

$bS^3 = S^2p_S(S) = 0; \tag{8}$

this forces $S^3 = 0$ and we are done once and for all. QED.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

  • $\begingroup$ You don't need to talk about eigenvalues to conclude that $\det S=0$: just use that $0=\det S^4=(\det S)^4$. $\endgroup$
    – lhf
    Oct 2, 2013 at 17:59
  • $\begingroup$ @lhf: Curiously enough, you don't need $(\det S)^4 = \det S^4 = 0$ to conclude that $\det S = 0$, if you use the fact that $\det S$ is the product of the eigenvalues! I guess the real issue is which is the more complex argument: one based on $\det(AB) = \det A \det B$ or one based on $Sv = \lambda v \Rightarrow S^n v = \lambda^n v$. I paid my money, so I made my choice! ;) $\endgroup$ Oct 2, 2013 at 19:12
  • $\begingroup$ Thanks Robert's cannot-be-more-detailed work, also the comment that provide another way to get det(S)=0. And I think that in (5), it could be pS(λ)=λ^3+aλ^2+bλ+c, only that one more step are added to conclude S^3=0 $\endgroup$
    – Lily
    Oct 3, 2013 at 0:08

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