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

##### 1 Answer

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

#|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

Clearly,

So, if the determinant of **not** invertible.