Linear transformations and eigenvalues

Question

Let $T: \mathbb C^n \rightarrow \mathbb C^n$ be linear. Let $\beta$ and $\gamma$ be any two ordered bases. Prove that the eigenvalues of $[T]_\beta$ and $[T]_\gamma$ are the same.

Can anyone provide tips/hints in the right direction? I'm struggling as I try to understand this intuitively....thank you

what do you know about similar matrices? – Dec 05 '13 at 05:13 — , Dec 05 '13 at 05:13

score 1 · Accepted Answer · answered Dec 05 '13 at 05:10

1

Hint:

$[T]_\beta$ and $[T]_\gamma$ are similar matrices, i.e. there is a matrix $Q$ (change of coordinate matrix) so that $[T]_\beta = Q^{-1}[T]_\gamma Q$.

answered Dec 05 '13 at 05:10

Pratyush Sarkar

3,734

Oh okay, so they have the same characteristic polynomials and therefore same eigenvalues. Thanks! – A A Dec 05 '13 at 05:21

score 1 · Answer 2 · answered Dec 05 '13 at 05:31

The eigenvalues are a property of the operator $T$. Recall that $\lambda$ is an eigenvalue of $[T] $ if $0=\det([T]-\lambda\,I)=\det([T-\lambda\,I]) $. This happens precisely when the operator $T-\lambda\, I$ is not invertible, i.e. there exists $v\in\mathbb C$ with $Tv=\lambda v$. This shows that the eigenvalues of $[T]$ (in any basis) are precisely the eigenvalues of $T$.

score 1 · Answer 3 · answered Dec 05 '13 at 05:36

1

HINT $$det(B - \lambda I_n) = det(A - \lambda I_n)$$ for two similar matrices $A$ and $B$. (Why?)

answered Dec 05 '13 at 05:36

Don Larynx

4,703

Linear transformations and eigenvalues

3 Answers3