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Question: Let $V$ be an inner product space and let $\alpha$ be an endomorphism of $V$ satisfying the condition that $\alpha^*\alpha=\sigma_0$. Show that $\alpha=\sigma_0$.

I know that $\sigma_0$ is the 0-functional where $V\mapsto 0_v$. In this instance, the question does not specify selfadjoint (which i believe the condition contradicts) or that it is normal. Thus im not really sure if there are propositions that could help me.

I am clearly missing something in the properties of $\alpha^*$ as I am also stuck on the following question: Show $\alpha$ is selfadjoint.

Any help would be greatly appreciated!

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Aksel'sRose
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1 Answers1

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Hint If $x \in V$ show that $\alpha^*\alpha=\sigma_0$ implies $$< \alpha(x), \alpha(x) >=0$$

Full solution

Let $v \in V$. Then, since $\alpha^*\alpha=\sigma_0$ you get $$< \alpha(v), \alpha(v) >=<\alpha^* \alpha(v), v>=<0,v>=0 $$

Now, since $< , >$ is an inner product space $$<u,u> =0 \Rightarrow u=0$$

This shows that $\alpha(v)=0= \sigma_0(v)$.

Thus $$\alpha(v) =\sigma_0(v) \forall v\in V$$

N. S.
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  • I began working in this strain, I will try again. Thank you. – Aksel'sRose Nov 13 '16 at 20:27
  • Okay, Im not getting it... Heres what I have: $\langle {\alpha^\alpha (v), v}\rangle= \langle {\alpha (v), \alpha(v)}\rangle$ which is =0 iff $\alpha(v)=0_v$. Or $\langle{\alpha^\alpha (v), v}\rangle=\langle{0_v, v}\rangle$. – Aksel'sRose Nov 13 '16 at 22:21
  • @Aksel'sRose If you write that properly, you get the solution. Check the edit to my post;) – N. S. Nov 13 '16 at 22:23