# Show that if $T$ satisfies $T^*=-T$ then any eigenvalue of $T$ satisfies $\lambda^*=-\lambda.$

Let $V$ be a vector space over $\mathbb{C}$ with an inner product $\langle ,\rangle$ and $T\colon V \rightarrow V$ a linear map. Show that if $T$ satisfies $T^*=-T$ then any eigenvalue of $T$ satisfies $\lambda^*=-\lambda.$

So far I have said the following:

$$\langle \vec w,T\vec v \rangle=\langle T^* \vec w,\vec v \rangle=\langle-T\vec w, \vec v\rangle=\langle-\lambda\vec w, \vec v\rangle$$

Where do I go next?

EDIT: Let v=w. $$\langle v,Tv \rangle=\langle v,\lambda v\rangle=\langle \lambda ^*v,v \rangle = \lambda ^*||v|| ^2$$

$$\langle v,Tv \rangle=\langle T^*v,v \rangle =\langle-Tv,v \rangle=\langle -\lambda v,v \rangle=-\lambda||v|| ^2$$

Thus $\lambda ^*=-\lambda$

• Hint : Take $v=w$. – Captain Lama Apr 19 '16 at 16:40
• @CaptainLama Why are we able to do that? – MHW Apr 19 '16 at 17:02
• What do you mean ? If we want to take $v=w$ then we do it. You have not used any property of $v$ in your question so the expression is valid for any $v$ ; in particular for $v=w$. – Captain Lama Apr 19 '16 at 17:04
• @CaptainLama Is it correct now? – MHW Apr 19 '16 at 17:14
• Yes, if your inner product is linear in the first variable it's fine (I'm more used to inner products linear in the second variable but of coruse it's just a convention). – Captain Lama Apr 19 '16 at 17:16

Note that $\langle x, Tx \rangle = \langle T^*x, x \rangle = -\langle Tx, x \rangle =-\overline{\langle x, Tx \rangle}$, hence $\langle x, Tx \rangle$ is purely imaginary for any $x$.