TITLE: How to show Von Neumann Trace inequality $ \text{Tr}(AB) \leq \sum_{i=1}^n \sigma_{A,i}\sigma_{B,i} $?
QUESTION [6 upvotes]: Let $A,B$ have the appropriate size. How can we show Von Neumann Trace inequality $ \text{Tr}(AB) \leq \sum_{i=1}^n \sigma_{A,i}\sigma_{B,i} $?
Also, what is the intuition behind this inequality?

REPLY [2 votes]: A different (non-standard) proof:
Consider the SVD of $A$, with singular values $\sigma_i(A)$ and associated orthonormal bases $e_i$, $\tilde{e}_i$; similarly for $B$, with bases $f_j$, $\tilde{f}_j$. Then \begin{align}
\mathrm{tr}(A^*B)&=\mathrm{tr}\langle Ae_i,Be_i\rangle\\
&=\sum_i\sigma_i(A)\langle\tilde{e}_i,Be_i\rangle\\
&=\sum_{ij}\sigma_i(A)\langle\tilde{e}_i,\tilde{f}_j\rangle\langle \tilde{f}_j,Be_i\rangle \\
&=\sum_{ij}\sigma_i(A)\sigma_j(B)\langle\tilde{e}_i\tilde{f}_j\rangle\langle f_j,e_i\rangle\\
|\mathrm{tr}(A^*B)|&\le\sum_{ij}\sigma_i(A)\sigma_j(B)(|\langle\tilde{e}_i,\tilde{f}_j\rangle|^2+|\langle f_j,e_i\rangle|^2)/2\\
&=\vec{\sigma}(A)^*\,T\,\vec{\sigma}(B)
\end{align}
$T$ is a doubly stochastic matrix since $$\sum_iT_{ij}=\tfrac{1}{2}\sum_i|\langle\tilde{e}_i,\tilde{f}_j\rangle|^2+\tfrac{1}{2}\sum_i|\langle f_j,e_i\rangle|^2|=\tfrac{1}{2}\|\tilde{f}_j\|^2+\tfrac{1}{2}\|f_j\|^2=1$$ (and similarly for columns).
But by Birkhoff's theorem, every doubly stochastic matrix is the convex sum of permutation matrices, $$T=\sum_k\alpha_kP_k,\qquad\qquad \sum_k\alpha_k=1,$$ hence $$|\mathrm{tr}(A^*B)|\le\sum_k\alpha_k(\vec{\sigma}(A)^*P_k\vec{\sigma}(B))\le\max_k\vec{\sigma}(A)^*P_k\vec{\sigma}(B)$$ By the rearrangement inequality, the largest sum of $(\sigma_i(A))$ dotted with the permuted values of $(\sigma_i(B))$ is the one in which the values align from biggest to smallest. This is precisely the von Neumann trace inequality.