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Can someone help me to show that every linear operator $A:E \rightarrow E $ is the sum of a self-adjoint with an anti-self-adjoint operator.

Even start the question would be awesome, i dont know what property to use so i can start the demonstration

Thanks !!

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    $\begingroup$ Here's a hint - think of how any function can be written as a sum of an even function and an odd function. $\endgroup$
    – Matt
    Jul 9, 2017 at 20:29
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    $\begingroup$ it has something to do with this property T = ((T + T^ ∗) /2) +( (T − T ^∗)/ 2 ) ? $\endgroup$
    – Uzop
    Jul 9, 2017 at 20:33
  • $\begingroup$ yes, what can you show about those operators? $\endgroup$
    – Matt
    Jul 9, 2017 at 20:35
  • $\begingroup$ that he is equal to T adjoint ? $\endgroup$
    – Uzop
    Jul 9, 2017 at 20:43
  • $\begingroup$ $T+T^*$ is self-adjoint. Can you show it? $\endgroup$
    – mfl
    Jul 9, 2017 at 20:45

1 Answer 1

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We will show that $A+A^*$ is self-adjoint. (By definition $\langle A^* x,y\rangle =\langle x,Ay\rangle$.) Now

$$\begin{align} \langle (A+A^*) x,y\rangle & = \langle A x,y\rangle+\langle A^* x,y\rangle \\ & =\langle x,A^*y\rangle+\langle x,Ay\rangle \\ & =\langle x, (A+A^*)y\rangle.\end{align}$$

We will show that $A-A^*$ is anti-self-adjoint. We have

$$\begin{align} \langle (A-A^*) x,y\rangle & = \langle A x,y\rangle-\langle A^* x,y\rangle \\ & =\langle x,A^*y\rangle-\langle x,Ay\rangle \\ & =-\langle x, (A-A^*)y\rangle.\end{align}$$

Finally, we have $A=\frac12 (A+A^*)+\frac12 (A-A^*)$ and we are done.

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  • $\begingroup$ oh i was going in that direction but ive lost a negative on the way $\endgroup$
    – Uzop
    Jul 9, 2017 at 21:13
  • $\begingroup$ I have editted the answer to give a complete answer. $\endgroup$
    – mfl
    Jul 9, 2017 at 21:17
  • $\begingroup$ thanks a lot it really helped me $\endgroup$
    – Uzop
    Jul 9, 2017 at 21:17

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