4
$\begingroup$

Let $V$ be a complex vector space, with Hermitian inner product $\langle z,w\rangle$. Let $T : V → V$ be a linear transformation. Show that $T$ is self adjoint if and only if $\langle Tz,z\rangle$ is real for every $z ∈ V$.

My solution is:

In the left side: $T$ is self-adjoint $\Leftrightarrow$ $T=T^*$ $\Leftrightarrow$ $T=UAU^*$ where $UU^*=I$ and A is a diagonal matrix.

In the right side:$\langle Tz,z\rangle$ is real for every $z ∈ V$ $\Leftrightarrow$ $z'T'\overline{z}$ is real $\Leftrightarrow$ $T=UU^*$. So there is some discrepency between the two sides.

Can you tell me which step is wrong?

$\endgroup$

1 Answer 1

5
$\begingroup$

$\Rightarrow)\quad$ We have for all $z$

$$\langle Tz,z\rangle=\langle z,T^*z\rangle=\langle z,Tz\rangle=\overline{\langle Tz,z\rangle}$$ so $\langle Tz,z\rangle$ is real.

$\Leftarrow)\quad$ Since $\langle Tz,z\rangle$ is real for all $z$ then we get $\langle Sz,z\rangle=0$ where $S=T-T^*$. Moreover, since $S$ is skew-hermitian then it's diagonalizable and from the equality $\langle Sz,z\rangle=0$ we see that their eigenvalues are $0$ so $S=0$. Finally we get $T=T^*$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .