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

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- April 18th 2007, 07:23 AMruproteinEigenvalue 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

?? - April 18th 2007, 07:38 AMJhevon
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 - April 18th 2007, 10:06 AMThePerfectHacker
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.