A 2x2 matrix $M$ exists. Suppose $M^3=0$ show that (I want proof) $M^2=0$

Solution 1:

Here's a naive approach to the question in the title:

If $M$ has rank 2, then it's invertible and $M^3$ cannot be $0$. If $M$ has rank 0, then $M^2=0$ trivially. So the only interesting case is when $M$ has rank $1$.

In this case there's a vector $v$ such that $Mv$ is nonzero. Since $M^3v=0$, it must be that either $Mv$ or $M^2v$ is a nonzero vector that $M$ maps to zero. So the kernel and the image of $M$ has a nontrivial intersection -- but both of these subspaces have dimension $1$, so they must coincide.

In particular, because the image is contained in the kernel, $M^2v$ must be $0$ for all $v$, so $M^2=0$.

Solution 2:

I had a hard time following all the details of the arguments in the OP's question, but based upon the Hamilton-Cayley theorem there is a pretty simple solution to the query expressed in the title.

Hamilton-Cayley tells us that we have

$M^2 + bM + c = 0 \tag{1}$

for some scalars $b, c$. If $b= c = 0$, (1) reduces to

$M^2 = 0, \tag{2}$

and we are done. If $c \ne 0$, multiply (1) by $M^2$ yielding

$M^4 + bM^3 + cM^2 = 0, \tag{3}$

and using $M^4 = M^3 = 0$, we have

$cM^2 = 0,\tag{4}$

and we are done. If $c = 0$, $b \ne 0$, multiply (1) by $M$ to obtain

$M^3 + bM^2 = 0, \tag{5}$

$bM^2 = 0, \tag{6}$

and once again we are done. We have thus covered all cases, and we are done, and I mean done! QED.