Connection between Trace of $A^TA$ and eigenvalues of $A$

Question

I am visiting a course about linear algebra. Our professor proved the equality between the Frobeniusnorm and the Schatten-2-norm. For this he used the following equality for a real matrix A.

$\sqrt{(\lambda_1^2+...+\lambda_n^2)}=\sqrt{tr(A^TA)}$

where $\lambda_1, ..., \lambda_n$ are the eigenvalues of $A^TA$ and $tr(A^TA)$ denotes the trace of $A^TA$. May someone please explain, why this equality holds?

Thank you very much!

Answer 1

It is well known that the trace of any matrix is the sum of its eigenvalues (counted with algebraic multiplicity). This is equal valid for $A^TA$ as for any other matrix.

So it really ought to be $$ \lambda_1+\cdots+\lambda_n=\operatorname{tr}(A^TA) $$

Answer 2

What You do have though is $$\sqrt{tr(A^{\perp}A)}=\sqrt{\sum_{j=1}^n\sigma_j²}$$ where $\sigma_j,j=1,...,n$ are the singular values of $A$, i.e. the eigenvalues of $\sqrt{A^{\perp}A}$, $\sigma_j^2$ are the eigenvalues of $A^{\perp}A$ and the right hand side is the $2$-Schatten norm of $A$ by definition. Now choose the standard basis $e_j,j=1,...,n$ and calculate: $$tr(A^{\perp}A)=\sum_{j=1}^n\langle A^{\perp}Ae_j,e_j\rangle=\sum_{j=1}^n\langle Ae_jAe_j\rangle=\sum_{j=1}^n|Ae_j|^2=$$ $$\sum_{j=1}^n\sum_{k=1}^n|\langle Ae_j,e_k\rangle|^2=\sum_{j=1}^n\sum_{k=1}^n|a_{jk}|^2$$ using Parseval's equality and the right hand side is the square of the Frobenius-norm $||.||_F$ of the matrix $A$. If $||.||_2$ denotes the $2$-Schatten-norm this shows: $$||A||_2=||A||_F.$$ An analogous result holds for compact linear operators on an infinite dimensional Hilbert-space that belong to the Hilbert-Schmidt-class.