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Let $E$ be a complex Hilbert space and $S\in \mathcal{L}(E)$.

Assume that $S$ is normal. Why $S$ has the following unique decomposition $$S=A_1+iA_2 ?$$ with $A_1$ and $A_2$ are self-adjoint operators.

Thank you.

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Even without normality, if we have an operator $S$, we can decompose it into a self-adjoint part and a skew-adjoint part, $S=(S+S^*)/2 + (S-S^*)/2=A_1+A_2$. Since $A_2^*=-A_2$, $(iA_2)^*=(-i)(-A_2)=iA_2$, and so $iA_2$ is self adjoint, and so we can write $S=A_1-i(iA_2)$.

Normality gives that the two terms actually commute with each other.

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  • $\begingroup$ Thank you. But why the decomposition is unique? $\endgroup$ – Student Dec 1 '17 at 10:57
  • $\begingroup$ If you assume you have such a decomposition, then you can recover $A_1$ and $A_2$ from $S$ and $S^*$ by simple linear algebra, and you will get that the formulas we used are the only correct ones. $\endgroup$ – Aaron Dec 1 '17 at 10:59
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Assume there exist self-adjoint operators $A_1$ and $A_2$ such that $$S = A_1 + iA_2$$

We have:

$$S^* = (A_1 + iA_2)^*= A_1^* + (iA_2)^* = A_1 - iA_2^*$$

Summing the two equalities gives:

$$S + S^* = 2A_1 \implies A_1 = \frac{S + S^*}{2}$$

And subtracting them gives:

$$S - S^* = 2iA_1 \implies A_2 = \frac{S - S^*}{2i} = -i\frac{S - S^*}{2}$$

This shows that $A_1$ and $A_2$ are unique. Now just verify that these $A_1$ and $A_2$ are indeed self-adjoint.

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If $S$ is normal, we put $$A_1=\frac{S+S^*}{2},\;\;A_2=-i\frac{S-S^*}{2}.$$ We see that $A_1$ and $A_2$ are bounded self-adjoint operators.

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