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I was doing matrix multiplication and noticed the following,

$$ X^* X X^\top = X^\top \overline{X} X^\top $$ where $X,X^\top,X^*$, is a complex N x N matrix, its real transpose and its hermitian transpose respectively. Why is this true? I was able to reason that it can only be true if

$$ X^* X=\overline{(X^* X)}=X^\top \overline{X} $$ since $X^*=\overline{X^\top}$. So then we indeed have $$ X^* X X^\top = X^\top \overline{X} X^\top $$ My question is, why then is $$ X^* X = \overline{X^* X}? $$ Now If $z\equiv X^* X$, and $z$ was a scalar, $z=\overline{z}$ implies $z\in \mathbb{R}$. Does this hold for matrices also? Is it then true that $X^* X \in M_{N}(\mathbb{R})$? If so , can we prove that in general $$ X^* X \in M_{N}(\mathbb{R})? $$ or was this just an accidental case based on the matrices I was working with? Thanks a lot!

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Why then is $X^\ast X=\overline{X^\ast X}$?

It is not. Counterexample: $$ X=\pmatrix{1&i\\ 0&0},\ X^\ast X=\pmatrix{1&i\\ -i&1}\notin M_2(\mathbb R). $$

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    Okay so I realize there is a counter example thanks! So my case was just accidental based on what $X$ was ? – 2017-01-27
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    @Integrals Whether it's accidental or there's a deeper reason depend on what exact examples of $X$ you have tried. – 2017-01-27
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    Thanks a lot, thats exactly what I was looking for. – 2017-01-27