Weighted Frobenius norm's inner product Let $W$ be a symmetric and positive definite real matrix. We know that the Frobenius norm $$\Vert A\Vert_F^2=\mathrm{trace}(A^TA)=\sum_{i,j}|a_{i,j}|^2$$ is induced by Frobenius inner product $$\langle A,B\rangle_F^2=\mathrm{trace}(A^TB)=\sum_{i,j}a_{i,j}b_{i,j}.$$ Is it also true for weighted Frobenius norm $$\Vert A\Vert_W^2=\Vert W^{\frac 12}AW^{\frac 12}\Vert_F=\mathrm{trace}(W^{\frac 12}A^TWAW^{\frac 12})?$$ Would $\mathrm{trace}(W^{\frac 12}A^TWBW^{\frac 12})$ be the inner product in this case? Or maybe $W$ would need to satisfy additional conditions? What would those be?
 A: Yes, assuming that $W$ is symmetric and Positive definitie (SPD). 
In general, given a vector space $V$, an inner product $<,>$ and a SPD map $L:V\to V$, the function $<L^{1/2}v,L^{1/2}w>$ is also an inner product (I leave verification of this fact as an exercise). Now, we simply need to verify that $L(A)= WAW$ is SPD with respect to the Frobenius inner product (henceforth denoted <,>). 
To check symmetry, note that $<LA,LB>=Tr((LA)^TLB)$, and the fact that the the trace does not change under taking the traspose implies that this is equal to $<LB,LA>$. To check positive-definiteness, we have to check that $<LA,A>$ is positive provided $A\ne 0$. But $<LA,A>=Tr((LA)^TA)=Tr(WA^TWA)$. Using the property $Tr(BC)=Tr(CB)$, we write this $Tr(W^{1/2} A^T WA W^{1/2})=|W^{1/2}AW^{1/2}|^2$. If we can shown that $W^{1/2}AW^{1/2}$ is not identically zero, then we will be finished,by the positive-definiteness of the Frobenius norm. However, since $W$ is positive-definite it follows that $W^{1/2}$ is invertible, so if $W^{1/2} A W^{1/2}=0$, then multiply on the left and right by $W^{-1/2}$ implies that $A=0$, a contradiction. This shows that $L$ is SPD with respect to the Frobenius norn, and thus that $<L^{1/2}A,L^{1/2}B>$ is an inner product. Finally, note that $L^{1/2}(A)=W^{1/2}AW^{1/2}$/ 
