prove that if gcd(f, minimal polyonimial of A) is not 1 then f(A) is not invertible A is square matrix and f is polynomial. prove that if gcd(f, minimal polyonimial of A) is not 1 then f(A) is not invertible. 
any hints please..
 A: (Assuming the underlying field is the reals or the complexes). (I think that @Pedro went a little far in saying that I was assuming the field was closed -- see final remark.)
Suppose that $f$ and the minimal polynomial have a nonconstant common factor $g$. And suppose that $\lambda$ is a root of $g$ (there's at least one by the fundamental theorem of algebra). Then $\lambda$ is an eigenvalue of $A$.
Now $f(A)$ contains the factor $g(A)$. Show that $g(A)$, applied to an eigenvector for $\lambda$,(**) gives zero. If $f(A)$ sends some nonzero vector to zero, can it be invertible? 
(**) Such an eigenvector may have to be taken in $C^n$ instead of $R^n$. 

Implicit here is the idea that a real matrix is invertible over the complexes if and only if it is invertible over the reals. Thus, for instance, the matrix
$$
A = \begin{bmatrix}
0 & 1 \\
-1 & 0 
\end{bmatrix}
$$
has $x^2 + 1$ as its minimal polynomial. If we take $f(x) =x^2 + 1$, then the gcd is evidently not one. And if we compute $f(A)$, we'll see that (in this extreme case) it's actually the zero matrix, hence not invertible. If you don't like that example, consider this one:
$$
A = \begin{bmatrix}
0 & 1 & & \\
-1 & 0 & & \\
& & 0 & 2 \\
& & -2 & 0 
\end{bmatrix}
$$
Now the minimal polynomial is $(x^2 + 1) (x^2 + 4)$; again using $f(x) =x^2 + 1$, we see that $f(A)$ is not invertible. And why not? Because (for instance) it sends the eigenvector for $\lambda = -i$, namely the vector $\begin{bmatrix}i \\ 1 \end{bmatrix}$, to zero. 
