The question is: The square matrix $A$ satisfies $p(A) = 0$, where $p(x)$ is a polynomial such that $p(0) \ne 0$. Show that $A$ is invertible.
I'm lost, I don't know if there's something more I have to learn to do this. I've gotten this far (I'm most likely not on the right track):
$$ p(A) = a_0I+a_1A+a_2A^2+ …+a_nA^n $$
$$ p(0) = a_0I+(a_1\cdot 0)+(a_2\cdot 0^2)+\ldots +(a_n\cdot 0^n) $$
$$ p(0) = a_oI$$
$$ p(A) = p(0)+a_1A+a_2A^2 +\ldots +a_nA^n $$
I don't quite know what to do further. I know that if $AX=B$, where $A$ is the square matrix, $B$ is a matrix vector, if there's only one solution $X$ for all $B$, then $A$ is invertible.
Best Answer
If $p(0)$ is nonzero then $a_0$ is nonzero. Therefore, one has: $$I=-\sum_{i=1}^n\frac{a_i}{a_0}A^i=-A\sum_{i=0}^{n-1}\frac{a_{i+1}}{a_0}A^i.$$