Equality in trace duality

Question

For $A,B\in\mathbb{R}^{n\times m}$ we have the trace duality property $$|\langle A, B \rangle|\leq \|A\|_1 \|B\|_{\infty}$$ where $\|A\|_p$ is the Schatten $p$-norm (i.e. $\|\cdot \|_1$ is the nuclear norm equal to the sum of singular values, and $\|\cdot\|_{\infty}$ is the operator norm equal to the largest singular value) and the inner product is $\langle A, B \rangle = \text{tr}(A^{\top}B)$.

There are at least two methods to prove this inequality. One is using the Fischer-Courant min-max principle (see for example this question), and the other is by the aid of symmetric gauge functions (see Chapter 4 of Matrix Analysis (1997) form Bathia). None of these proofs establish sufficient (or necessary) conditions to get an equality.

Do anyone know a way to get equality?

Answer 1

I developed an answer which coincides with the Proposition of loup blanc. Let $n\geq m$ and $A = U_1D_mV^{\top}$ the reduced SDV of $A$ and $A = UDV^{\top}$ the full SVD of $A$. Here $U=(U_1,U_2)\in\mathbb{R}^{n\times n}$ is orthonormal, $U_1\in\mathbb{R}^{n\times m}$, $V\in\mathbb{R}^{m\times m}$ is orthonormal, $D\in\mathbb{R}^{n\times m}$ and $D_m\in\mathbb{R}^{m\times m}$.

Take $B = U_1V^{\top}\in\mathbb{R}^{n\times m}$. Then $$ \|B\|_{\infty} = \|U_1V^{\top}\|_{\infty} = \|UI_{n\times m}V^{\top}\|_{\infty} = 1, $$ and $$ \text{tr}(A^{\top}B) = \text{tr}(VD_mU_1^{\top}U_1V^{\top}) = \text{tr}(D_m) = \|A\|_1. $$ so $$ |\text{tr}(A^{\top}B)| = \text{tr}(A^{\top}B) = \|A\|_1\|B\|_{\infty}. $$ Observe that $A^{\top}B$ is symmetric and $\geq 0$, wich coincides with loup blanc's Proposition.

If someone finds a mistake, please let me know!

Answer 2

That follows is a case of equality of the relation

$(*)$ $|tr(A^TB)|\leq \sigma_1(B)\sum_i \sigma_i(A)$.

I don't know if there are other such instances (at least, I'm pretty sure there are not when $n=2$).

Let $A,B$ be $n\times n$ symmetric matrices that commute. Then, up to an orthonormal change of basis, we may assume that they are diagonal. Moreover, they have the form

1. $B=diag(\lambda I_k,\lambda_1,\cdots,\lambda_{n-k})$ where, for every $i$, $|\lambda|\geq |\lambda_i|$.

2. $A=diag(\mu_1,\cdots,\mu_k,0_{n-k})$ where, for every $j$, $\mu_j\geq 0$ (or, for every $j$, $\mu_j\leq 0$).

EDIT. $\textbf{Proposition}$. If $(*)$ is an equality, then $A^TB$ is symmetric $\geq 0$ or $\leq 0$.

$\textbf{Proof}$. One always has $|tr(A^TB)|\leq \sum_i\sigma_i(A^TB)\leq \sigma_1(B)\sum_i \sigma_i(A)$.

Then $|tr(A^TB)|= \sum_i\sigma_i(A^TB)$, and consequently, $spectrum(A^TB)\subset [0,+\infty[$ or $]-\infty,0]$, and the SVD of $A^TB$ is a diagonalization via an orthonormal basis. $\square$