Operator norm of product of matrices

Question

Could someone explain why the following is true? $$\max_{A: Tr(A)=1} \langle B,A \rangle = \|B\|_{\infty}.$$ Where the infinity norm is the standard operator norm of a matrix, and $\langle A,B\rangle =Tr(A^*B)$ is the standard inner product. I am not able to formally prove it, but can intuitively see it being true.

Answer 1

Of course, you need absolute value: $$ \max\left\{|\langle B,A\rangle|\,:\ \text{Tr}(A)=1\right\}=\|B\|_\infty. $$ Also, as stated, the equality is false: consider $$ B=\begin{bmatrix}1&1\\0&1\end{bmatrix}, \ \ \ A=\begin{bmatrix}1&n\\ 1&0\end{bmatrix}. $$ Then $\text{Tr}(A)=1$, and $$\langle B,A\rangle=\text{Tr}(B^*A)=n+1,$$ for every $n$. The right requirement is that the one-norm of $A$ is one: $$\tag{1} \max\left\{|\langle B,A\rangle|\,:\ \text{Tr}(|A|)=1\right\}=\|B\|_\infty. $$ So we can simply assume that $A$ is positive and $\text{Tr}(A)=1$. By using the inequality $|\text{Tr}(X)|=\text{Tr}(|X|)$, we get $$ |\langle B,A\rangle|=|\text{Tr}(B^*A)|\leq\text{Tr}(|B^*A|) =\text{Tr}((A^*BB^*A)^{1/2})\leq\|B\|_\infty\,\text{Tr}((A^*A)^{1/2})=\|B\|_\infty. $$ Above we are using the inequalities, $BB^*\leq\|B\|^2\,I$, and that the square root preserves order. We have shown that $$\tag{2} \max\left\{|\langle B,A\rangle|\,:\ \text{Tr}(|A|)=1\right\}\leq\|B\|_\infty. $$ Now fix $\varepsilon>0$. Then there exist $x,y\in H$ with $\|x\|=\|y\|=1$ and $\langle Bx,y\rangle>\|B\|_\infty-\varepsilon.$ Let $\{e_n\}$ be an orthonormal basis of $H$ where $e_1=x$, and write $E_{x,y}$ for the rank-one operator that maps $x$ to $y$ (i.e., $E_{x,y}=\langle \cdot,x\rangle y$). Then $|E_{x,y}|=(E_{y,x}E_{x,y})^{1/2}=E_{y,y}^{1/2}=E_{y,y}$, so $\text{Tr}(|E_{x,y})=1$; and \begin{align} |\text{Tr}(B^*E_{y,x})|&=|\text{Tr}(BE_{x,y})|=|\text{Tr}(E_{x,y}BE_{x,x})|=\left|\sum_{n=1}^\infty\langle E_{x,y}BE_{x,x}e_n,e_n\rangle\right|\\ \ \\ &=\left|\sum_{n=1}^\infty\langle BE_{x,x}e_n,E_{y,x}e_n\rangle\right| =\langle Bx,y\rangle>\|B\|_\infty-\varepsilon. \end{align} Thus, as $\varepsilon$ is arbitrary, $$\tag{3} \max\left\{|\langle B,A\rangle|\,:\ \text{Tr}(|A|)=1\right\}\geq\|B\|_\infty. $$ By combining $(2)$ and $(3)$, we get $(1)$.