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 $<x, y>' = <T(x), y>$ To show that $TU$ is self-adjoint, repeat the argument with the inverse of $T$ in place of $T$.

I have tried

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

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 $<x,y>'=<T(x),y>$" is a bit contrived.


1 Answer 1


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.

  • $\begingroup$ So the condition " $T$ is positive-definite" is critical right? In general, if $T$ and $U$ are self-adjoint, $TU$ is not self-adjoint? $\endgroup$
    – SGKw
    Jul 18 at 4:01
  • 1
    $\begingroup$ $T$ and $U$ is self-adjoint, then $TU$ is self-adjoint iff $TU=UT$. $\endgroup$
    – fusheng
    Jul 18 at 4:21
  • $\begingroup$ I can see some connections. Thanks $\endgroup$
    – SGKw
    Jul 18 at 4:23

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