Show that a matrix $A$ is singular if and only if $0$ is an eigenvalue.

$A$ singular $\iff\det(A)=0\iff\det(A-0\cdot I)=0\iff 0$ is eigenvalue of $A$.

Michael

Note that, the determinant of $n\times n$ matrix $A$ can be computed using the eigenvalues as

$$ |A|=\lambda_1\lambda_2\dots\lambda_n ,$$

which is the product of the eigenvalues.

We know that $0 \in \lambda(A)$ iff there exists some nonzero solution to the eigenvector equation $Ax = \lambda x = 0\cdot x = 0$. Thus $0$ is an eigenvalue iff $\exists b \in \mathrm{Ker}(A)$ with $b \neq 0$. But since $\mathrm{Ker}(A) \neq \{ 0 \}$ we conclude that $A$ must be singular.