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?

  • $\begingroup$ 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 $A^*A=AA^*$. $\endgroup$ Apr 22, 2013 at 0:53
  • 2
    $\begingroup$ 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 $\Bbb{C}^2$. $\endgroup$ Apr 22, 2013 at 2:47

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .