Can an invertible matrix have an eigenvalue equal to 0?

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Can an invertible matrix have an eigenvalue equal to 0?

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Answer 1 · 2016-11-09T07:54:58

No.

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

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

$Avecv = lambdavecv$ ,

we have that

$lambdavecv - Avecv = vec0$

$=> (lambdaI - A)vecv = vec0$

and hence, for a nontrivial solution,

$|lambdaI - A| = 0$ .

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

$|0*I - A| = 0$

$|-A| = 0$

$=> (-1)^n|A| = 0$

Note that a matrix inverse can be defined as:

$A^(-1) = 1/|A| adj(A)$ ,

where $|A|$ is the determinant of $A$ and $adj(A)$ is the classical adjoint, or the adjugate, of $A$ (the transpose of the cofactor matrix).

Clearly, $(-1)^(n) ne 0$ . Thus, the evaluation of the above yields $0$ iff $|A| = 0$ , which would invalidate the expression for evaluating the inverse, since $1/0$ 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.

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Can an invertible matrix have an eigenvalue equal to 0?

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