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Prove that if $\sigma$ and $\psi$ are self-conjugate transformations over an Euclidean vector space $E$ , with nonnegative eigenvalues . One of them is nonsingular , then the eigenvalues of the transformation $\sigma\psi$ are real and nonnegative .

My attempt :

It's no matter if I consider the inner product over fields $C$ . Then the inner product is viewed as a Hermitian form .

Because $\phi$ and $\psi$ are self-conjugate so for each of them we can choose a standard basis and the matrices of them in these basis is digagonal with elements $diagol(\lambda_{1},...\lambda_{n})$ . So I can write these matrices as square of $digol(\sqrt{\lambda_{1}},...\sqrt{\lambda_{n}})$ which means exist two self-conjugate transformations $h,k$ satisfy : $h^{2} = \sigma , k^{2} = \psi$ .

Then I transform as follows , assume $\lambda$ is an eigenvalue of $\phi \psi$ and the corresponding eigenvector is $\alpha$ :

$$\left \langle \sigma \psi ( \alpha) ,\psi(\alpha) \right \rangle = \lambda \left \langle \alpha , \psi(\alpha) \right \rangle$$

Because of self-conjugate :

$$\left \langle h(h(\psi(\alpha)),\psi(\alpha) \right \rangle = \lambda \left \langle \alpha , \psi(\alpha) \right \rangle$$

$$\left \langle h(\psi(\alpha)),h(\psi(\alpha)) \right \rangle=\lambda \left \langle \alpha,\psi(\alpha) \right \rangle$$

Now $t= \left \langle \alpha ,\psi(\alpha) \right \rangle = \left \langle \psi(\alpha) , \alpha \right \rangle \neq 0$ then it must be a positive real number , hence $\lambda$ is a positive real number .

If $t = 0$ then $t = \left \langle k(\alpha),k(\alpha) \right \rangle =0 \Rightarrow k(\alpha)=0\Rightarrow \psi(\alpha)=0 \Rightarrow \sigma(\psi) = 0 = \lambda \alpha \Rightarrow \lambda = 0$

Did I miss something when I didn't use condition one of them is nonsingular ?

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  • $\begingroup$ What's a standard basis? Besides, how do you know that there is a basis (standard or not) with respect to which both endomorphisms are diagonalizable? $\endgroup$ – José Carlos Santos Jul 17 '17 at 14:44
  • $\begingroup$ It's a basis $(e_{i})$ satisfy $\left \langle e_{i},e_{i} \right \rangle = 1$ and $\left \langle e_{i},e_{j} \right \rangle = 0 \forall i \neq j$ . Now the spectral theorem asserts that a linear transformations is conjugate if and only if there is a standard basis $(e_{i})$ and its matrix in this basis is diagonal . About existence of basis , every vector euclidean space always has one by using Schmidt orthogonalization . $\endgroup$ – Gankedbymom Jul 17 '17 at 14:47
  • $\begingroup$ That's an orthonormal basis, not a standard basis. And, yes, the spectral theorem says that what you wrote is one for $\sigma$ and for $\psi$ individually, but not necessarily for both of them simultaneously. $\endgroup$ – José Carlos Santos Jul 17 '17 at 14:50
  • $\begingroup$ Then did I have any mistakes here ? $\endgroup$ – Gankedbymom Jul 17 '17 at 14:53
  • $\begingroup$ Yes. Like I saind in my previous comment, you don't know for sure that there is an orthonormal basis with respect to which both $\sigma$ and $\psi$ are diagonalizable. $\endgroup$ – José Carlos Santos Jul 17 '17 at 14:55

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