$\mathbf P=[\mathbf p_1, \mathbf p_2 ,... ,\mathbf p_N]$,and each $\mathbf p_i$ is a $t \times 1 $ complex matrix,that is ,the elements in $\mathbf p_i$ are all complex number.Now,let $trace\{\mathbf P\mathbf P^H\}=P$
How do we calculate $\mathbf P\mathbf P^H$ to this answer $\frac{P}{t} \mathbf I$
I mean ,$\mathbf P\mathbf P^H = \frac{P}{t} \mathbf I$
Does anyone know how to calculate it?
$\mathbf P\mathbf P^H = \frac{P}{t} \mathbf I$ is in $(9)$
Let $B=PP^H$, then the paper poses the following constrained optimization problem $$\eqalign{ \min_B &\phi(B) &= {\rm Tr}\big(B^{-1}\big) \cr {\rm st\;} &{\rm Tr}(B) &= {\cal P} }$$ Introduce an unconstrained variable $X$, and set $$\eqalign{ B &= \bigg(\frac{{\cal P}}{I:X}\bigg)X \quad\implies\quad B^{-1} &= \bigg(\frac{I:X}{{\cal P}}\bigg)X^{-1} \cr }$$ where a colon denotes the trace/Frobenius product, i.e. $\;A:B={\rm Tr}(A^TB)$
Obviously setting $A=I$ yields $$\eqalign{ {\rm Tr}(B) &= I:B \cr &= \bigg(\frac{{\cal P}}{I:X}\bigg)\big(I:X\big) = {\cal P} }$$ meaning that $B$ defined in terms of $X$ satisfies the constraint.
Now solve the unconstrained problem $$\eqalign{ {\cal P}\,\phi &= {\cal P}\;\,{\rm Tr}\big(B^{-1}\big) \cr &= (I:X)\;(I:X^{-1}) \cr &= (\alpha)\,(\beta) \cr {\cal P}\;d\phi &= \beta\,d\alpha + \alpha\,d\beta \cr &= \beta\,I:dX + \alpha\,I:dX^{-1} \cr &= \beta\,I:dX - \alpha\,I:(X^{-1}\,dX\,X^{-1}) \cr &= \Big(\beta\,I - \alpha\,X^{-2}\Big)^T:dX \cr \frac{\partial\phi}{\partial X} &= \frac{\;\big(\beta\,I - \alpha\,X^{-2}\big)^T}{{\cal P}} \cr \cr }$$ Setting the gradient to zero reveals $X$ to be a scalar multiple of the identity matrix, i.e. $$\eqalign{ X^{-2} &= \bigg(\frac{\beta}{\alpha}\bigg)\,I = \sigma^{-2}I \quad\implies X &= \sigma I \cr }$$ The constrained matrix can be evaluated without knowing the precise value of the $\sigma$ scalar $$\eqalign{ t &= {\rm Tr}(I) = I:I \cr B &= \bigg(\frac{{\cal P}}{I:\sigma I}\bigg)\sigma I = \bigg(\frac{{\cal P}}{t}\bigg) I \cr }$$ which is the paper's result.