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

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, $$||I_n||^2==trace(I_n^{*}I_n)=n$$, 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.

• You should mention which norm you use on $\mathbb{C}^{n\times n}$. Mar 11, 2021 at 13:52

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.
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.