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Is it true that product of two normal operators is normal if and only if they commute. If yes How can I show that.

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Being commutative is the sufficient, not necessary condition. –  chaohuang Nov 20 '12 at 20:35

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It is not true. For instance, unitaries are normal and the product of unitaries is unitary, so normal. But they don't have to commute. For example, $$ V=\frac1{\sqrt2}\begin{bmatrix} 1 &1\\-1&1\end{bmatrix}, \ W=\begin{bmatrix}1&0\\0&-1\end{bmatrix}; $$ then $$ VW=\frac1{\sqrt2}\begin{bmatrix} 1&-1\\-1&-1\end{bmatrix},\ WV=\frac1{\sqrt2}\begin{bmatrix} 1&1\\1&-1\end{bmatrix} $$

Edit: for the other implication, which is indeed true. If $A,B$ are normal and they commute, then $AB=BA$, so $A^*B^*=B^*A^*$. Now you can use Flugede-Putnam or the fact that commuting normal operators can be simultaneously unitarily diagonalized, to deduce that $AB^*=B^*A$. Then $$ (AB)^*AB=B^*A^*AB=ABB^*A^*=AB(AB)^*. $$

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Thank you for answer. But How can I prove that if they commute then the product has to be normal. –  shane Nov 21 '12 at 17:32
    
I edited the answer to include that. –  Martin Argerami Nov 21 '12 at 17:38

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