Singular matrix characterization I have never seen this result before:

If $A$ is singular, then there is $\alpha_0 > 0$ such that $B_{\alpha} = A + \alpha I\;$ is non-singular, for every $0 < |\alpha| < \alpha_0$.

Somebody knows how prove it? Thanks in advance.
 A: $B_\alpha$ is singular iff $-\alpha$ is an eigenvalue of $A$.
If $A=0$, then any $\alpha \neq 0$ will do, so you can choose $\alpha_0 = \infty$.
Otherwise, let $\alpha_0 = \min \{ |\lambda| \, | \,  \lambda \text{ is an eigenvalue of } A, \lambda \neq 0 \}$. Then if $0 < |\alpha| < \alpha_0$, $-\alpha$ is not an eigenvalue of $A$, and hence $B_\alpha$ is non-singular.
Note: This result is really just about the fact that the eigenvalues of $A$ form a finite set. It doesn't require that $A$ be singular. If $A$ is non-singular, the range of $\alpha$ can be extended to $|\alpha| < \alpha_0$ (ie, $0$ is included).
A: The matrix $A\in\mathbb{R}^{n\times n}$ is singular, if only if, your colums $A_1,\ldots A_n\in\mathbb{R}^n$ are L.D. i.e. there is $c\in\mathbb{R}^n$ such that $c\neq 0$ and
$$
c_1A_1+\ldots +c_nA_n=0. 
$$
Hint. But we know that $\alpha>0$ is large enough such that the column vectors $A_i+ \alpha e_i$ of $A+\alpha I$  vectors are linearly independent. If there is any $\alpha_0\in (0, \alpha)$ such that the vectors $A_i+ \alpha_0 e_i$  are LD then we can conclude that the vectors $e_1,\ldots e_n$ are linearly dependent. 
