# Eigenvalue Question

• Apr 18th 2007, 07:23 AM
ruprotein
Eigenvalue Question
Prove that a linear operator T on a finite dimensional vector space is invertible if and only if zero is not an eigenvalue of T

??
• Apr 18th 2007, 07:38 AM
Jhevon
Quote:

Originally Posted by ruprotein
Prove that a linear operator T on a finite dimensional vector space is invertible if and only if zero is not an eigenvalue of T

??

i'm a bit rusty in linear algebra (even though i took it just last semester) so here's your answer in a nutshell, you have to transform this into a rigorous proof.

Recall that eigen values and eigen vectors are "unchanged" under linear transformations. so the matrix resulting from applying a linear transformation on a finite demensional matrix will have the same eigenvalues and eigenvectors as the first matrix.

now, if we have an eigenvalue = 0, then det(I*lambda - A) = 0, this means that detA = 0, which means that the matrix is not invertible, since for a matrix A to be invertible, we must have detA not= 0

so that should get you started. sorry i can't help more, i was lost for most of that class
• Apr 18th 2007, 10:06 AM
ThePerfectHacker
Quote:

Originally Posted by ruprotein
Prove that a linear operator T on a finite dimensional vector space is invertible if and only if zero is not an eigenvalue of T

??

I prove it one way.

Let the standard matrix for T be A, that is [T]=A.

We say that A is invertible by hypothesis, meaning
det(A) not = 0.

The eigenvalues k are solutions to the polynomial,

det(k*I-A)=0

Now, if k is zero, then,

det(k*I-A)=det(0*I-A)=det(-A)=(-1)^n * det(A) !=0.
Because det(A)!=0.

Hence k=0 is not and eigenvalue.