# $T$ and $U$ self-adjoint on $V$, $T$ is positive definite. Prove $TU$ & $UT$ are diagonalizable linear operators that have only real eigenvalues.

This is a problem from Freidberg linear algebra (4th edition) chapter 6.4.21

The whole problem and the hints are

Let $$V$$ be a finite-dimensional inner product space, and let $$T$$ and $$U$$ be self-adjoint operators on $$V$$ such that $$T$$ is positive definite. Prove that both $$TU$$ and $$UT$$ are diagonalizable linear operators that have only real eigenvalues. Hint: Show that $$UT$$ is self-adjoint with respect to the inner product $$' = $$ To show that $$TU$$ is self-adjoint, repeat the argument with the inverse of $$T$$ in place of $$T$$.

I have tried

$$ = = = < x, U^*T(y) > = $$

But this didn't work.

I don't know why the hint is given, and I think there is another way of not using that hint.

Because I think "Show that $$UT$$ is self-adjoint with respect to the inner product $$'=$$" is a bit contrived.

We think of $$T$$ and $$U$$ as two matrices.

Because $$T$$ is positive-definite, $$T$$=$$P^{*}P$$. Thus $$TU=P^{*}PU$$=$$P^{*}PUP^{*}\left( P^* \right) ^{-1}$$ is similar to $$PUP^{*}$$. Because $$U$$ is self-adjiont, $$TU$$ has real eigenvalues and can be diagonalizable.

$$UT=UP^{*}P=P^{-1}PUP^{*}P$$ is the same case.

• So the condition " $T$ is positive-definite" is critical right? In general, if $T$ and $U$ are self-adjoint, $TU$ is not self-adjoint?
– SGKw
Jul 18 at 4:01
• $T$ and $U$ is self-adjoint, then $TU$ is self-adjoint iff $TU=UT$. Jul 18 at 4:21
• I can see some connections. Thanks
– SGKw
Jul 18 at 4:23