the norm of the trace as a linear functional on $\mathbb{C}^{n \times n}$

Question

Let $T:\mathbb{C}^{n \times n} \rightarrow \mathbb{C}$ be a linear functional on the space of all $n \times n$ matrices whose entries are complex numbers. I need to show that the norm of the trace $||trace||$ " as a linear functional" is $\sqrt n$. I have proved one direction as follows:

$|trace(A)| \leq ||trace||||A||$

this holds for all $A \in \mathbb{C}^{n \times n}$. So, pick $A=I_n$ identity matrix, we get:

$n \leq ||trace|| \sqrt n, \quad \quad \quad \quad \quad (1)$

(Here, $ ||I_n||^2=<I_n,I_n>=trace(I_n^{*}I_n)=n$, where $I_n^*$ represents the conjugate transpose of $I_n$.)

For the second direction, I know that

$||trace|| = \sup_{A\in \mathbb{C}^{n \times n}}=\{ |trace(A)| : ||A|| \leq 1$, but I don't know how can I get the upper bound $\sqrt n$ of the definition of the norm.

Answer 1

Your norm seems to be induced by the inner product $\langle A,B\rangle = \operatorname{Tr}(B^*A)$ so we can use Cauchy-Schwarz inequality

$$\left|\operatorname{Tr}(A)\right|^2 = \left|\operatorname{Tr}(I^*A)\right|^2 =|\langle A,I\rangle|^2 \le \|A\|^2\|I\|^2 = n\|A\|^2$$ and hence $\left|\operatorname{Tr}(A)\right| \le \sqrt{n}\|A\|$.

Furthermore, CS states that equality holds if and only if $A$ and $I$ are proportional, so the bound $\sqrt{n}$ is attained.

Answer 2

Hint:

Let $e_1,\dots e_n$ be the standard basis of $\mathbb{C}^n$. Then for any $A\in\mathbb{C}^{n\times n}$, we have $$\hbox{trace}A=\sum_{i=1}^n\langle Ae_i,e_i\rangle$$ where $\langle\cdot \rangle$ is the standard inner product. The result follows from the CS inequality.