Proving $A^2 = 0$ given $A^5 = 0$ I have a class question where I must prove $A^2 = 0$ given $A^5 = 0$ with A being a 2x2 matrix.   I though that I could simply say that as $A^5 = 0$ then $A^2 \cdot A^3 = 0 \implies A^2 = 0$ as $A^2 = A\cdot A$ and $A^3 = A\cdot A\cdot A$ $\implies A = 0 \implies A^2 = 0$.
Could someone verify that my proof is valid and not just a circular argument (which I feel it may be)  Thanks 
 A: You can use the characteristic equation
$$ A^2-Tr(A)A+I_2\det{A}=O_2,$$ where $Tr(A)$ is the sum of the elements on the first diagonal while $\det{A}$ is the determinant of $A$ and it is zero because $A^5=0$. Hence $ A^2-Tr(A)A=O_2$. If $Tr(A)=0$ you are done. If not multiply the previous equation by$A^3$ and you will get $A^4=O_2$  and so on till $A^2=O_2$. 
A: Your reasoning is wrong because $AB=0$ does not imply that $A=0$ or $B=0$. Take for example
$$\left(\begin{matrix}1&0\\0&0\end{matrix}\right)\left(\begin{matrix}0&0\\0&1\end{matrix}\right)$$
But if $A^5=0$ then the rank of $A$ is not two, so it must be $1$ or $0$.
If it is $0$ then clearly $A=0$ and hence $A^2=0$.
If it is $1$ then the image $E$ of the linear function $f(x)=Ax$, $x\in\Bbb R^2$ has dimension $1$. Now there are two possibilities: $f(E)$ has dimension $0$ or $f(E)$ has dimension $1$. If $f(E)$ has dimension $0$ then $A^2=0$, as we wish to show, and if $f(E)=E$ then $f$ is clearly a bijection over $E$, and $A^n$ has rank $1$ for all $n$, a contradiction with the fact $A^5=0$.
A: If $A^4\ne0$  then there's $x_0\in\Bbb R^2$ such that $A^4x_0\ne0$ but in this case the family of vectors $(x_0,Ax_0,A^2x_0,A^3x_0)$ are linearly independent. In fact, if
$$\alpha_0 x_0+\alpha_1 Ax_0+\alpha_2 A^2x_0+\alpha_3 A^3x_0=0$$
then applying respectively $A^4,A^3,A^2$ and $A$ to the last equality we see that $\alpha_i=0$. This (i.e. the vectors are linearly independent) is  an obvious contradiction since $\dim\Bbb R^2=2$. Hence $A^4=0$.
Now repeat this proof and you get $A^3=A^2=0$.
A: Your argument breaks down where you say that $(A^2)(A^3) =0$  implies that either $A^2 = 0$ or $A^3 = 0$.   In general, there is no zero-product rule in linear algebra; it's easy for $AB$ to equal $0$ even if $A$ and $B$ are both non-zero.  In bigger spaces, your theorem would be false.  For instance, suppose you make a $5\times 5$ matrix which is all zeros except on the diagonal just above the main diagonal; for instance, let $A_{12}=A_{23}=A_{34}=A_{45}=1$ and everything else zeros.  Then  $A^5 =0$  but $A^2$ is not zero.
That being said, you are working only in the set of $2\times 2$ matrices.  That's a pretty special case and the statement is surely true for that case.  I don't know how far along in your textbook you are...one approach would be to observe that the minimal polynomial of a $2\times 2$ matrix can't be any larger than degree $2$; since the minimal polynomial must be a factor of $x^5$, it can only be $x^1$ or $x^2$.
If you haven't covered characteristic polynomials and minimal polynomials yet, you could say that, whatever the matrix, there is a basis in which the matrix is upper triangular.  In that basis, the diagonal entries of $A^5$ are just the various values of $a^5$, where $a$ is one of the diagonal entries.  Since $A^5$ is $0$, the diagonal entries in this basis must be $0$.  This leaves only one possible non-zero entry, in the northeast corner.  So $A^2$ must equal $0$ .   There are some easy generalizations to bigger spaces--e.g., if $A$ is $m  \times m$  and $A^n = 0$, where $n$ is bigger than $m$.  Something like that.  
A: This is not valid because for matrices $AB = 0$ doesn't imply $A=0\vee B=0$. Instead use the dimension of $A$ (It's crucial that $A\in\mathbb R^{2\times 2}$ here).

Hint
If $A^k = 0$ for some $k\ge n$ then $0$ must necessarily be the only eigenvalue of $A\in\mathbb R^{n\times n}$, since for any eigenpair $(\lambda, x)$
$$0 = A^k x = \lambda^k x = 0 \Leftrightarrow \lambda^k = 0 \Leftrightarrow \lambda = 0$$
Now if $0$ is the only eigenvalue of $A\in\mathbb R^{2\times 2}$ then...
