nilpotent and linear transformation 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!!
 A: 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 $\dim V \ge 4$
A: 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
$\lambda ^4 = 0, \tag{3}$
forcing
$\lambda = 0. \tag{4}$
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!!!
