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.
