Show that if a matrix A has zero as its eigenvalue it is not

Show that if a matrix A has zero as its eigenvalue, it is not invertible. Show that if V is an eigenvector for a matrix A associated to an eigenvalue A, then we Av is also an eigenvector for a matrix A associated to lambda.

Solution

proof-:Suppose A is square matrix and invertible and, for the sake of contradiction, let 0 be an eigenvalue. Consider, (AI)v=0 with =0

(A0I)v=0

(A0)v=0

Av=0

We know A is an invertible and in order for Av=0, v=0, but v must be non-trivial such that det(AI)=0. Here lies our contradiction. Hence, 0 cannot be an eigenvalue.

2) Suppose A is a real matrix with a real eigenvalue . Then there exists a real
column vector v is not = 0 such that Av =lambda v.

Start with Aw =lambda w where w is a non zero column vector with complex entries

Write w = v + v where both v, v are real vectors. We then have

Av + Av = (v + v)

Since w is not = 0, at least one of the two v, v must be a
non zero vector. Hence it is true .