How can I show that: $\text{Tr}(A^*A)=\sum_{\lambda\in\sigma(A^*A)}\lambda$?

Question

Let $A\in \mathbb{K}^{n\times n}$, how can we show that: $$ ||A||^2_F=\text{Tr}(A^*A)=\sum_{\lambda\in\sigma(A^*A)}\lambda$$

Where $A^*=\bar A^T.$

Thanks in advance for any help!

Answer 1

The trace of a matrix is the sum of the eigenvalues, up to multiplicity. As $A^*A$ is Hermitian, it is diagonalisable, hence there is an invertible matrix $P$ and diagonal matrix $D$, whose diagonal entries are the eigenvalues of $A^*A$, up to multiplicity, such that $$A^*A = P^{-1} D P.$$ Using the fact that $\operatorname{Tr}(CD) = \operatorname{Tr}(DC)$, we get $$\operatorname{Tr}(A^*A) = \operatorname{Tr}(P^{-1} D P) = \operatorname{Tr}(DPP^{-1}) = \operatorname{Tr}(D),$$ which is the sum of the diagonal entries of $D$, i.e. the sum of the eigenvalues of $A^*A$ up to multiplicity.

Here are two things worth noting:

1. The notation $\sum_{\lambda \in \sigma(A^* A)} \lambda$ could be misleading, in that it doesn't make it clear that eigenvalues must be summed up to multiplicity. For example, if $A = I_{2 \times 2}$, then $A^*A = I_{2 \times 2}$, which has one eigenvalue of $1$, up to a multiplicity of $2$, i.e. $\sigma(A^* A) = \{1\}$. One could reasonably interpret the sum as simply $1$, when the trace of $A^*A$ is actually $1 + 1 = 2$.

2. The fact that $\operatorname{Tr}(A)$ is the sum of the eigenvalues up to multiplicity is true for any matrix, even non-diagonalisable matrices. We may not be able to get a diagonal matrix $D$, but we can get an upper-triangular matrix $D$ (e.g. the Jordan Normal Form).