$T,U$ self-adjoint, $U$ positive definite, then $TU$ has only real eigenvalues

Let $V$ be a finite dimensional inner product space over $\Bbb R$.
Let $T$ and $U$ be linear self-adjoint operator on $V$.
Assume $U$ is positive definite.
Show all the eigenvalues of $TU$ are real.

Let $\lambda$ be an eigenvalue of $TU$, with corresponding eigenvector $x$. Then \begin{align} \left<\lambda x,x \right> &= \left< \left(TU\right)x,x\right> \\ &= \left< x,\left(TU\right)^* x\right> \\ &= \left<x,\left(U^* T^*\right)x \right> \\ &= \left<x, \left(U T \right) x \right> \\ &= \left< x,\left(TU\right)x\right> \tag{1} \\ &= \left< x,\lambda x\right>, \end{align} So, $\lambda$ is real.
I don't see why step (1) holds.
That is, how to show $TU$ is self-adjoint.

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You cannot prove that $TU$ is self-adjoint because this is not true in general. You should try some other way to prove that all eigenvalues of $TU$ are real. For example, let us abuse the notations so that $T,U$ are also their respective matrix representation under the canonical basis.
1. As $U$ is positive definite, it can be orthogonally diagonalized as $U=QDQ^T$ for some positive diagonal matrix $D$ and some real orthogonal matrix $Q$.
2. Hence you may take a self-adjoint square root of $U$. That is, set $U^{1/2}=QD^{1/2}Q^T$, where $D^{1/2}$ is the entrywise square root of $D$.
3. Show that $TU$ is similar to $U^{1/2}TU^{1/2}$ and argue that the latter has real eigenvalues.
I don't see the last part of (3), that is how to show $U^{1/2} T U^{1/2}$ has real eigeinvalues. With $T=A^{-1}XA$ and $U^{1/2}=B^{-1}YB$, then $U^{1/2} T U^{1/2}=\left( B^{-1}Y B \right) \left( A^{-1}XA\right) \left( B^{-1}YB\right)$. From this, how can I infer the eigeinvalues are real? – Nicolas Essis-Breton Dec 29 '12 at 16:25
I'm not sure what your $A,B,X,Y$ refer to, but you just need to check that $M=U^{1/2}TU^{1/2}$ is a self-adjoint matrix. The eigenvalues of every self adjoint matrix are real. – user1551 Dec 29 '12 at 16:52