# 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?

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 $$$$ is also an inner product (I leave verification of this fact as an exercise). Now, we simply need to verify that $$L(A)= WAW$$ is SPD with respect to the Frobenius inner product (henceforth denoted <,>).
To check symmetry, note that $$=Tr((LA)^TLB)$$, and the fact that the the trace does not change under taking the traspose implies that this is equal to $$$$. To check positive-definiteness, we have to check that $$$$ is positive provided $$A\ne 0$$. But $$=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$$ is SPD with respect to the Frobenius norn, and thus that $$$$ is an inner product. Finally, note that $$L^{1/2}(A)=W^{1/2}AW^{1/2}$$/