Spectrum of an invertible operator

Question

Let $H$ be a Hilbert space.

Let $A, T \in B(H)$ with $A$ invertible.

Show $\sigma(A^{-1}T A) = \sigma (T)$.

Attempt:

$\subseteq$:

We will show the contrapositive. Let $\lambda \in \sigma(T)$ such that $(\lambda I -T)$ is invertible.

So $(\lambda I -T)^{-1}$ exists...

$\supseteq:$

Let $\lambda \in \sigma(A^{-1}TA )$. We want to show that $\lambda\in \sigma(T)$.

We know that $(\lambda I - A^{-1}TA) $ is not invertible. And I want to show $(\lambda I -T)$ is not invertible either...

So basically for both directions I tried contrapositive and direct proof, but not sure how to show the other part is invertible or not invertible...

Thank you for your help!

Answer 1

I'll perhaps write a longer explanation than Igor Rivin.

If $A,B,C$ are invertible bounded operators, then $A\circ B \circ C$ is also invertible.

To show $\lambda \in \sigma(T)$ iff $\lambda\in\sigma(A^{-1}T A)$, you can show instead that $T-\lambda I$ is invertible iff $A^{-1}TA-\lambda I$ is invertible.

You can then use the identity Igor Rivin wrote:

$$A^{-1}(T-\lambda I)A= A^{-1}TA-\lambda I $$

for both directions. Where you should remember that

$$ \big(A^{-1} \big)^{-1} \Big( A^{-1}TA-\lambda I \Big)A^{-1}=T-\lambda I $$

for one direction.

Answer 2

Hint: $\lambda I - A^{-1} T A = A^{-1}(\lambda I - T) A.$