Let $A \in M_n(C)$, $A^m = 0 \implies A^n = 0$ Consider a matrix $A$ of size $n$ over $C$. I tried to show that if $A^m =0$ for some $m$ then $A^n =0$
We may assume that WLOG there exist a vector $x \ne 0$ such that $A^{m-1} x \ne 0$.
Suppose $a_0x + a_1 Ax \cdots + a_{m-1}A^{m-1}x =0$ where $a_i$ are scalars.
Applying $A^{m-1}$ to the above equation we see that $a_0 A^{m-1}x = 0$ so $a_0 = 0$. Next by applying $A^{m-2}$ on first equation we can see that $a_1 =0$. By continuing the same process we see $a_i = 0$ for all $i$. So $x, Ax, \cdots A^{m-1}x$ are $m$ linearly independent vectors so $m \le n$. So $A^n = A^m A^{n-m} =0$.
Is this correct? I'm curious if any other methods also there?
 A: This argument looks correct, and yes, there are some other ways to demonstrate that your statement holds. Both of the following rely on the fact that all of the eigenvalues of $A$ must be equal to $0$. To see this, let $\vec v \neq \vec 0$ be an eigenvector of $A$ with eigenvalue $\lambda$. Then we have
$$\lambda^m \vec v = A^m \vec v = 0 \vec v = \vec 0$$
so $\lambda^m = 0$ and hence $\lambda = 0$.
Now for the other methods of proof.
Method 1: The Cayley-Hamilton theorem gives that $A$ must satisfy its own characteristic polynomial. All of the eigenvalues of $A$ are $0$, so the characteristic polynomial of $A$ is simply $x^n$. Thus, $A^n = 0$.
Method 2: Because the eigenvalues of $A$ are all $0$, the Jordan canonical form of $A$ - call this $J$ - has zero entries on and below the main diagonal. It follows that $J^n = 0$, so $A^n = (PJP^{-1})^n = PJ^nP^{-1} = 0$.
These methods both rely on some extra machinery to get the job done, which is what makes your solution nice.
A: If $q(\lambda)=\lambda^r$ and if $q(A)=0$, then the minimal polynomial $m$ for $A$ must divide $q$, which means that $m(\lambda)=\lambda^s$ for some $s \le n$. Consequently $A^{n}=A^{s}A^{n-s}=0$.
