Let $A$ be $n \times n$ matrix and $A^2=A^*A$. Why is $A$ a Hermitian matrix?
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2What do we know from the definition of a Hermitian matrix? – Rammus May 28 '15 at 12:39
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1If the matrix is invertible, the proof is trivial : $A^2=A^* A\rightarrow A=A^*$ – Peter May 28 '15 at 12:41
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Since we are working with complex matrices, Jordan canonical form is a very powerful tool. – Quang Hoang May 28 '15 at 12:47
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Let us first do a calculation (where we use the assumption) $$ (A-A^*)^*(A-A^*)=(A^*-A)(A-A^*)=A^*A-(A^*)^2-A^2+AA^*=AA^*-(A^*)^2. $$ Now $$ ((A^*)^2)^*=A^2=A^*A, $$ so $$ (A^*)^2=(A^*A)^*=A^*A. $$ Inserting this into the calculation above, $$ (A-A^*)^*(A-A^*)=AA^*-A^*A $$
But then the trace of that matrix on the left-hand side is zero, since (here we use that the trace is linear and that $\text{tr}\,(AB)=\text{tr}\,(BA)$) $$ \text{tr}\,(AA^*-A^*A)=\text{tr}\,AA^*-\text{tr}\,A^*A=0. $$ Hence $$\text{tr}\,(A-A^*)^*(A-A^*)=0. $$ Thus, the square of the Frobenius norm of $A-A^*$ is zero. It follows that $A-A^*=0$, and thus that $A$ is Hermitian.
mickep
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I hope it is correct as it is written now (and thanks to you who pointed the mistake out). – mickep May 28 '15 at 13:34
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This question is the complex counterpart of
With the inner product $\langle X,Y\rangle=\operatorname{Re}\operatorname{tr}(XY^\ast)$ defined on the real linear space $M_n(\mathbb C)$, Hermitian matrices are orthogonal to skew-Hermitian matrices. Now, if we denote the Hermitian and skew-Hermitian parts of $A$ by respectively $H$ and $K$, the condition $AA^\ast=A^2$ implies that $\langle K,K\rangle=\langle K,H\rangle=0$. Therefore $K=0$ and $A$ is Hermitian.
