Let $A^*$ denote the complex conjugate transpose of a matrix $A$. In the Euclidean norm (operator norm), if $$\|A^*A+AA^*\|=2\,\|A^*A\|$$ prove/disprove that $A$ is normal.
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What matrix norm are you using? The operator norm is not the Euclidean norm, but a norm induced by the Euclidean norm. If we view $M_n(\mathbb C)$ as $\mathbb C^{n^2}$, the Euclidean norm is the Frobenius norm. $A$ is indeed normal if the Frobenius norm is used in the given condition, but not necessarily normal if the operator norm is used. – user1551 Feb 02 '15 at 02:26
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Of course you are correct. I found some articles saying the Euclidean norm (meaning the operator norm), hence thought it was well-established. But, since you are the second person today, I must reconsider my view and be more careful. Thank you. – PeterA Feb 02 '15 at 08:39
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Consider the matrix $$ A= \begin{pmatrix} 0&0&0\\ 1 & 0 &0\\ 0 &0& 10\end{pmatrix}$$ then $$AA^* = \begin{pmatrix} 0&0&0\\ 0 & 1 &0\\ 0 &0& 100\end{pmatrix},A^*A \begin{pmatrix} 1&0&0\\ 0 & 0 &0\\ 0 &0& 100\end{pmatrix}\quad\text{ and }\quad A^*A +AA^*=\begin{pmatrix} 1&0&0\\ 0 & 1 &0\\ 0 &0& 200\end{pmatrix}. $$ So, $A^*A$ has $100$ as maximum eigenvalue and $A^*A+AA^*$ has $200$ as maximum eigenvalue. Since these are nonnegative symmetric matrices (namely $A^*A$ and $A^*A+AA^*$), we have $\|A^*A+AA^*\| = 200$ and $\|A^*A\|=100$ however $A$ is not normal.
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