TITLE: Proving $A^2 = 0$ given $A^5 = 0$
QUESTION [5 upvotes]: 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

REPLY [3 votes]: 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 $A^3=A^2=0$.