Showing that $M_n(\mathbb{C})$ is a unital C*-algebra

Question

I am trying to show that the set of $n \times n$ complex matrices $M_n(\mathbb{C})$ is a unital C*-algebra. Showing that it is a unital *-algebra is no problem, but I am struggling to show that $$ \| A^*A \| = \| A \|^2. $$ I am using the norm on $M_n(\mathbb{C})$ defined by the inner product $$ \langle A, B \rangle = \operatorname{Tr}(A^*B) $$ for all $A, B \in M_n(\mathbb{C})$. That is, $$ \|A \| = \sqrt{\langle A,A \rangle} = \sqrt{\operatorname{Tr}(A^*A)}. $$ Then $$ \|A^*A\| = \sqrt{\operatorname{Tr}([A^*A]^*A^*A)} = \sqrt{\operatorname{Tr}(A^*AA^*A)}. $$ But $$ \|A\|^2 = \operatorname{Tr}(A^*A) = \sqrt{\operatorname{Tr}(A^*A)\operatorname{Tr}(A^*A)}.$$ So I need to show that $$\operatorname{Tr}(A^*A A^*A) = \operatorname{Tr}(A^*A)\operatorname{Tr}(A^*A).$$ I am aware of the trace property $$\operatorname{Tr}(A \otimes B) = \operatorname{Tr}(A)\operatorname{Tr}(B).$$ But is $\operatorname{Tr}(A^*A A^*A) = \operatorname{Tr}(A^*A \otimes A^*A)$? The Kronecker product gives $A^*A \otimes A^*A$ as an $n^2 \times n^2$ matrix, whereas $A^*A A^*A$ is an $n \times n$ matrix.

Answer 1

If you apply your formula to $A$ diagonal, you are trying to show that $$ (|a_{11}|^4+\cdots+|a_{nn}|^4)^{1/2}=|a_{11}|^2+\cdots+|a_{nn}|^2, $$ which is easily seen to fail in most examples.

A C$^*$-algebra has a unique norm, since it is not too hard to see that in any C$^*$-algebra $$ \|A\|^2=\|A^*A\|=\max\sigma(A^*A). $$ So the norm you need to use is $$\|A\|=\text{largest singular value}.$$ A more manageable characterization of this norm is that it is the operator norm when you see $A$ as a linear operator on $\mathbb C^n$ with the Euclidean norm. That is $$ A=\sup\{\|Ax\|:\ x\in\mathbb C^n,\ \|x \|=1\}. $$