# show that every linear operator $A:E \rightarrow E$ is the sum of a self-adjoint with an anti-self-adjoint operator.

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 !!

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

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.

• oh i was going in that direction but ive lost a negative on the way
– Uzop
Jul 9, 2017 at 21:13