Show that the square matrix A is invertible 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.
 A: There's a simple way to do this without manipulating an expansion of the polynomial, or knowing anything about determinants or characteristic polynomials.
First, recall that $A$ is non-invertible if and only if there exists some non-zero vector $\bf{x}$  such that $A{\bf x} = {\bf 0}$. Suppose that there exists such a vector.
Since $p(A)$ is a sum of matrices ($a_n A^n + \dots + a_1 A + a_0 I$) we may compute ${\bf x}^T p(A) \bf{x}$. The fact that $A {\bf x} = 0$ implies that ${\bf x}^T A^n {\bf x} = {\bf 0}$ for any $n>0$, so we have that ${\bf x}^T p(A) {\bf x} = {\bf x}^T a_0 {\bf x} = p(0) \left| {\bf x} \right|^2$. Since we are told that $p(A) = 0$, we have that $p(0) \left| {\bf x} \right|^2 = 0$. Since ${\bf x}$ is non-zero, we must have $p(0) = 0$, contradicting the assumption that $p(0) \ne 0$.
To summarise: we are given that $p(A) = 0$ and that $p(0) \ne 0$. If $A$ is not an invertible matrix, then the argument of the previous two paragraphs shows that the conditions $p(A) = 0$ and $p(0) \ne 0$ cannot both hold. So $A$ must be invertible.
A: $p(0)\neq 0$ implies that $0$ is not a root of characteristic polynomial $p(x)$ which is turn says that $0$ is not an eigenvalue of $A$. As $det(A)=$ product of eigenvalues gives  $det(A)\neq 0$ which suggests that $A$ is invertible.
A: If $p(0)$ is nonzero then $a_0$ is nonzero. Therefore, one has: $$I=-\sum_{i=1}^n\frac{a_i}{a_0}A^i=-A\sum_{i=0}^{n-1}\frac{a_{i+1}}{a_0}A^i.$$
A: Well,
if $p(x)=a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$ is the characteristic polynomial then it is known from theory that $a_0$ represents the determinant of the matrix. Since $p(0)=a_0 \neq 0$ this means that $\det A \neq 0$ . Thus $A$ is invertible. 
A: All eigenvalues of A must be roots of P. 
(See the eigenvector X and P(A)X) 
=> all eigennumbers of A is not zero => |A| is not zero => A is invertible
