# Can an invertible matrix have an eigenvalue equal to 0?

Nov 9, 2016

No.

A matrix is nonsingular (i.e. invertible) iff its determinant is nonzero.

To prove this, we note that to solve the eigenvalue equation

$A \vec{v} = \lambda \vec{v}$,

we have that

$\lambda \vec{v} - A \vec{v} = \vec{0}$

$\implies \left(\lambda I - A\right) \vec{v} = \vec{0}$

and hence, for a nontrivial solution,

$| \lambda I - A | = 0$.

Let $A$ be an $N \times N$ matrix. If we did have $\lambda = 0$, then:

$| 0 \cdot I - A | = 0$

$| - A | = 0$

$\implies {\left(- 1\right)}^{n} | A | = 0$

Note that a matrix inverse can be defined as:

${A}^{- 1} = \frac{1}{|} A | a \mathrm{dj} \left(A\right)$,

where $| A |$ is the determinant of $A$ and $a \mathrm{dj} \left(A\right)$ is the classical adjoint, or the adjugate, of $A$ (the transpose of the cofactor matrix).

Clearly, ${\left(- 1\right)}^{n} \ne 0$. Thus, the evaluation of the above yields $0$ iff $| A | = 0$, which would invalidate the expression for evaluating the inverse, since $\frac{1}{0}$ is undefined.

So, if the determinant of $A$ is $0$, which is the consequence of setting $\lambda = 0$ to solve an eigenvalue problem, then the matrix is not invertible.