[Math] How to show Von Neumann Trace inequality $ \text{Tr}(AB) \leq \sum_{i=1}^n \sigma_{A,i}\sigma_{B,i} $

Question

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?

Answer 1

One may view the trace inequality as a pile of Cauchy-Schwarz inequalities.

Let us tackle the (more general) complex case here. Using the tracial property and singular value decomposition, one can reduce the inequality $|\operatorname{tr}(AB)|\le\sum_i\sigma_i(A)\sigma_i(B)$ to $$ |\operatorname{tr}(DUSV^\ast)|\le\operatorname{tr}(DS)\tag{1} $$ where $U,V$ are two unitary matrices and $D=\operatorname{diag}(d_1,\ldots,d_n),\,S=\operatorname{diag}(s_1,\ldots,s_n)$ are two diagonal matrices with nonnegative and decreasing diagonal entries. Let $P_k$ denotes the orthogonal projection matrix $I_k\oplus0_{n-k}$. Note that $D$ is a non-negatively weighted combinations of the $P_i$s. In fact, $$ D=(d_1-d_2)P_1+\cdots+(d_{n-1}-d_n)P_{n-1}+d_nP_n $$ and similarly for $S$. For convenience, let us write $D=\sum_ka_kP_k$ and $S=\sum_lb_lP_l$, where the $a_k$s and $b_l$s are nonnegative. The inequality $(1)$ thus becomes $$ \left|\sum_{k,l}a_kb_l\operatorname{tr}(P_kUP_lV^\ast)\right| \le\sum_{k,l}a_kb_l\operatorname{tr}(P_kP_l).\tag{2} $$ So, by triangle inequality, it suffices to prove that $$ \left|\operatorname{tr}(P_kUP_lV^\ast)\right| \le\operatorname{tr}(P_kP_l)\tag{3} $$ for each pair of $k$ and $l$. Assume that $k\ge l$, or else interchange the roles of $k$ and $l$. As $P_kUP_l=[P_ku_1,\,\ldots,\,P_ku_l,\,0,\ldots,0]$, the inequality $(3)$ is equivalent to $$ \left|\sum_{i=1}^l \langle P_ku_i,\,v_i\rangle\right|\le l.\tag{4} $$ Since $P_k$ is an orthogonal projection and the columns of $U$ are unit vectors, $\|P_ku_i\|_2\le1$. Therefore $(4)$ follows from Cauchy-Schwarz inequality.