Is it true that product of two normal operators is normal if and only if they commute. If yes How can I show that.
Product of Two Commuting Normal Operators
2
$\begingroup$
linear-algebra
-
0Being commutative is the sufficient, not necessary condition. – 2012-11-20
1 Answers
4
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)^*. $
-
0I edited the answer to include that. – 2012-11-21