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A normal operator is defined as $AA^{*}\ =A^{*}A$

Where A is an operator

how do i show the sum of two normal operators is normal? Or find a counter example that shows this is false?

olliepower
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  • That is not a definition. For one thing, $x$ needs to be quantified. Will you please provide your assumptions? Operators on what? A complex Hilbert space? Typically a normal operator on a Hilbert space is defined to be an operator that commutes with its adjoint, and it can be proved that an operator $A$ on a Hilbert space $H$ is normal if and only if $|Ax|=|A^x|$ for all $x\in H$. It might be easier to answer this question using the usual definition, that $A^A=AA^*$. – Jonas Meyer Apr 22 '13 at 00:53
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    Notice that if $A,B$ are normal, you're interested in whether $(A+B)(A+B)^=(A+B)(A^+B^)=AA^+AB^+BA^+BB^=A^A+A^B+B^A+B^B$. But you're not given that $AB^=B^A$ or $BA^=A^*B$, so you should figure it's false. I'd look for counterexamples where $A$ and $B$ are unitary acting on $\Bbb{C}^2$. – Kevin Carlson Apr 22 '13 at 02:47

1 Answers1

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Try $U_1 = \begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}$, $U_2 = \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}$. Both are unitary, hence normal. However $U_1+U_2 = \begin{bmatrix} 0 & 2 \\ 0 & 0\end{bmatrix}$, and $(U_1+U_2) e_1 = 0$, but $(U_1+U_2)^* e_1 = 2 e_2$, hence $U_1+U_2$ is not normal.

copper.hat
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