^^
Thank you very much!
In the basis that diagonalizes the matrix, the matrix can be written :
$\displaystyle \begin{pmatrix}
\lambda_1 & 0 & 0 &\ldots & 0 \\
0 & \lambda_2 & 0 & \ldots & 0 \\
0 & 0 & \lambda_3 & \ldots & 0 \\
\vdots & \vdots & \vdots & \ddots & 0 \\
0 & 0 & 0 & \ldots & \lambda_n \end{pmatrix}$
Where $\displaystyle \lambda_1, \dots, \lambda_n$ are the eigenvalues of the matrix.
The determinant in this base is obviously the product of the $\displaystyle \lambda_i$
So if one of the eigenvalues is 0, the determinant is 0.
And since the determinant doesn't change with the basis, we have proved that det A = 0, and hence it is not invertible.
Edit : waaaah 39th post
If $\displaystyle A$ has $\displaystyle 0$ as an eigenvalue then it means $\displaystyle A\bold{x} = 0\bold{x}$ for $\displaystyle \bold{x} \not = \bold{0}$.
Therefore, $\displaystyle A\bold{x} = \bold{0}$ for $\displaystyle \bold{x}\not = \bold{0}$.
This means $\displaystyle A$ is not invertible.