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.
Best Answer
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.