Invertibility and Eigenvalues

Theorem

$A$ is invertible $⟺$ $\lambda =0$ is not an Eigenvalue.

Proof

Proving $⟹$

1. Suppose $A$ is invertible
2. Then, $\det (A)={\lambda }_1{\lambda }_2\dots {\lambda }_n\ne 0$
3. Thus, ${\lambda }_1,\dots ,{\lambda }_n$ is not zero
4. Therefore, $\lambda =0$ is not an eigenvalue of $A$

Proving $⟸$

1. Suppose $\lambda =0$ is not an eigenvalue
2. This gives ${\lambda }_1,\dots ,{\lambda }_n\ne 0$
3. We know $\det (A)={\lambda }_1\dots {\lambda }_n\ne 0$
4. And so, $A$ is invertible