[This is just a repetition of user1551’s wonderful Math.SE answer. Link to the answer: Link.]

Ref:

• user1551’s Math.SE answer. Link to the answer: Link.

Let ${ A , B \in \mathbb{C} ^{n \times n} . }$ Consider the inner product

\[{ \text{tr}(A ^{\ast} B) = \sum _{1 \leq i, j \leq n} \overline{a _{i,j}} b _{i, j} . }\]

Consider the SVDs

\[{ A = U _A \Sigma _A V _A ^{\ast}, \quad B = U _B \Sigma _B V _B ^{\ast} }\]

of ${ A, B }$ respectively.

Q) Can we relate ${ \text{tr}(A ^{\ast} B) }$ to ${ \text{tr}(\Sigma _A ^{\ast} \Sigma _B) = \sum _{1 = 1} ^{n} \sigma _i (A) \sigma _i (B) }$?

Note that

\[{ \begin{aligned} &\, \text{tr}(A ^{\ast} B) \\ = &\, \text{tr}(V _A \Sigma _A U _A ^{\ast} U _B \Sigma _B V _B ^{\ast}) \\ = &\, \text{tr}( \Sigma _A U _A ^{\ast} U _B \Sigma _B V _B ^{\ast} V _A) \\ = &\, \text{tr}(\Sigma _A U \Sigma _B V ^{\ast}) \end{aligned} }\]

for orthonormal matrices ${ U = U _A ^{\ast} U _B }$ and ${ V = V _A ^{\ast} V _B . }$

We will again see the power of expanding things in the right basis…

Note that we have the basis ${ (P _1 , \ldots, P _n) }$ of the space of diagonal matrices in ${ \mathbb{C} ^{n \times n} , }$ where ${ P _j = \text{diag} (\underbrace{1, \ldots, 1} _{j \text{ many}} , 0, \ldots, 0) . }$

Expressing ${ \Sigma _A, \Sigma _B }$ in this basis:

\[{ {\begin{aligned} &\, \Sigma _A \\ = &\, \sigma _1 (A) P _1 + \sigma _2 (A) (P _2 - P _1) + \ldots + \sigma _n (A) (P _n - P _{n-1}) \\ = &\, (\sigma _1 (A) - \sigma _2 (A)) P _1 + (\sigma _2 (A) - \sigma _3 (A)) P _2 + \ldots + (\sigma _{n-1} (A) - \sigma _n (A)) P _{n-1} + \sigma _n (A) P _n \end{aligned}} }\]

and similarly

\[{ {\begin{aligned} &\, \Sigma _B \\ = &\, \sigma _1 (B) P _1 + \sigma _2 (B) (P _2 - P _1) + \ldots + \sigma _n (B) (P _n - P _{n-1}) \\ = &\, (\sigma _1 (B) - \sigma _2 (B)) P _1 + (\sigma _2 (B) - \sigma _3 (B)) P _2 + \ldots + (\sigma _{n-1} (B) - \sigma _n (B)) P _{n-1} + \sigma _n (A) P _n . \end{aligned}} }\]

Substituting in the expression of interest:

\[{ {\begin{aligned} &\, \text{tr}(\Sigma _A U \Sigma _B V ^{\ast}) \\ = &\, \text{tr} \left( \left( \sum _{i=1} ^{n} \alpha _i P _i \right) U \left( \sum _{i=1} ^{n} \beta _j P _j \right) V ^{\ast} \right) \\ = &\, \sum _{1 \leq i, j \leq n} \alpha _i \beta _j \text{tr} \left( P _i U P _j V ^{\ast} \right) \end{aligned}} }\]

where

\[{ \alpha _i = (\sigma _i (A) - \sigma _{i+1} (A)), \quad \beta _j = (\sigma _j (B) - \sigma _{j+1} (B)) . }\]

Note that each individual term in the expansion has magnitude

\[{ {\begin{aligned} &\, \left\vert \text{tr}((P _i U P _j ) V ^{\ast}) \right\vert \\ = &\, \left\vert \text{tr} \left( \begin{pmatrix} U _{[i,j]} &O \\ O &O \end{pmatrix} V ^{\ast} \right) \right\vert \\ = &\, \begin{cases} &\, \leq i \quad \text{ if } i \leq j, \\ &\, = \left\vert \text{tr} \left( V ^{\ast} \begin{pmatrix} U _{[i,j]} &O \\ O &O \end{pmatrix} \right) \right\vert \leq j \quad \text{ if } i > j \end{cases} \\ \leq &\, \min \lbrace i, j \rbrace . \end{aligned}} }\]

For each case ${ i \leq j }$ and ${ i > j }$, we have used Cauchy-Schwarz ${ \vert w _1 z _1 + \ldots + w _n z _n \vert }$ ${ \leq \vert w _1 \vert \vert z _1 \vert + \ldots + \vert w _n \vert \vert z _n \vert }$ ${ \leq \sqrt{\sum \vert w _i \vert ^2} \sqrt{\sum \vert z _i \vert ^2 } }$ to get the inequalities.

Hence the expression of interest has magnitude

\[{ {\begin{aligned} &\, \vert \text{tr}(\Sigma _A U \Sigma _B V ^{\ast}) \vert \\ = &\, \left\vert \sum _{1 \leq i, j \leq n} \alpha _i \beta _j \text{tr} \left( P _i U P _j V ^{\ast} \right) \right\vert \\ \leq &\, \sum _{1 \leq i, j \leq n} \alpha _i \beta _j \left \vert \text{tr} \left( P _i U P _j V ^{\ast} \right) \right \vert \\ \leq &\, \sum _{1 \leq i, j \leq n} \alpha _i \beta _j \min \lbrace i, j \rbrace \\ = &\, \sum _{1 \leq i, j \leq n} \alpha _i \beta _j \text{tr}(P _i P _j) \\ = &\, \text{tr}\left((\alpha _1 P _1 + \ldots + \alpha _n P _n) (\beta _1 P _1 + \ldots + \beta _n P _n) \right) \\ = &\, \text{tr}( \Sigma _A \Sigma _B) \\ = &\, \sigma _1 (A) \sigma _1 (B) + \ldots + \sigma _n (A) \sigma _n (B) . \end{aligned}} }\]

To summarise:

Thm [Von Neumann’s trace inequality]

Let ${ A , B \in \mathbb{C} ^{n \times n} . }$ Then their inner product is bounded by

\[{ \boxed{ \vert \text{tr}(A ^{\ast} B) \vert \leq \text{tr}(\Sigma _A ^{\ast} \Sigma _B) = \sum _{i = 1} ^{n} \sigma _i (A) \sigma _i (B) } . }\]

For other trace inequalities, see the Wiki page: Trace inequality.