Question

Can an invertible matrix have an eigenvalue equal to 0?

Answer 1

No.

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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.