$A \in \mathbb{R}^{n \times n}$ is a symmetric, PSD matrix. I want to write $\|A\|_F$ as a function of the eigenvalues of $A$

Question

Given symmetric and PSD matrix $A \in \mathbb{R}^{nxn}$, I'm trying to express its Frobenius norm in terms of the eigenvalues of $A$ only. My attempt looks like this:

$$\begin{aligned} \|A\|_F &= \sqrt{\mbox{Tr}(A^TA)} \\ &= \sqrt{\mbox{Tr}(A^2)} \\ &= \sqrt{\sum_{i=1}^{n} \lambda_i(A^2)} \end{aligned}$$

At this point I'm close, but I need to write this in terms of the eigenvalues of $A$, not $A^2$. Any suggestions would be appreciated — I'm new to linear algebra, especially matrices, and so a simple/thorough explanation would be nice. Thanks!

Answer 1

If $A = Q \Lambda Q^T$ is the eigendecomposition of $A$, then we can see that

\begin{equation} A^2 = Q \Lambda Q^T Q \Lambda Q^T = Q \Lambda \Lambda Q^T = Q \Lambda^2 Q^T \end{equation}

Because $\Lambda$ is a diagonal matrix, we can deduce that $\lambda_i(A^2) = \lambda_i(A)^2$. Plugging this into your last equation gives us the result:

\begin{equation} \|A\|_F = \sqrt{\sum_i^n \lambda_i(A)^2} \end{equation}