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I am working on a paper and I have a complex-valued $N \times K$ matrix $A$ and its complex conjugate transpose $A^*$. What happens when I take a product $A A^*$? What does the property state what happens when I multiply them other than having a real-valued matrix?

Here N>>K

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  • $\begingroup$ is the matrix square? Thin? Fat? Full column rank? Full row rank? $\endgroup$ – Rodrigo de Azevedo Nov 19 '16 at 12:18
  • $\begingroup$ N>>Kso it makes A a thin matrix $\endgroup$ – user3153034 Nov 19 '16 at 12:32
  • $\begingroup$ @RodrigodeAzevedo N>>K so it makes A a thin matrix $\endgroup$ – user3153034 Nov 19 '16 at 15:41
  • $\begingroup$ The product is Hermitian and positive semidefinite. $\endgroup$ – Rodrigo de Azevedo Nov 19 '16 at 15:46
  • $\begingroup$ If $\mathrm A^* \mathrm A = \mathrm I$, then $\mathrm A \mathrm A^*$ is a projection matrix. In that case, the trace of $\mathrm A \mathrm A^*$ equals its rank. $\endgroup$ – Rodrigo de Azevedo Nov 19 '16 at 15:52
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It may be instructive to look at the SVD of $A$. Let

$$ A = U \begin{bmatrix} \Sigma\\ 0 \end{bmatrix} V^*, $$

where $\Sigma$ is the $K\times K$ diagonal matrix of singular values. Then

$$ AA^* = U \begin{bmatrix} \Sigma^2 & 0\\ 0 & 0 \end{bmatrix} U^* = \sum_{i=1}^r \sigma_i^2 u_iu_i^*, $$ where $r \leq K$ is the rank of $A$. Thus, the rank of $AA^*$ is equal to the rank of $A$, its column space is the span of $U_{1:r}$ (the same as $A$), and its nullspace is the span of $U_{r+1:N}$.

I am not sure if this is what you were looking for but I hope it helps.

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  • $\begingroup$ no i was not looking for this answer all i want to know is what does this product represents for example if we take a derivative of something we find its slope/rate of change $\endgroup$ – user3153034 Nov 19 '16 at 15:45

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