Let $A$ be an n x n matrix, where the trace is the sum of the diagonal elements. Show that $tr(A^{T}A)$=$\|A\|_2^2$.
Question/Concerns: I was given this question. However, I'm concerned about what norm they are talking about in the question. Is it the induced Matrix 2-norm squared? Or do you think that it is the Frobenius Norm squared? I know that:$\lambda_{\max}(A^TA)=\|A\|_2^2$ where this notation represents the induced Matrix 2-norm squared. However, it is not always true that the trace of $A^TA$ is equal to the maximum eigenvalue of $A^TA$. Example:(Taken from another stack exchange post): Consider $$A = \begin{pmatrix} \frac{1}{2} \ \ 0 \\ 0 \ \ \frac{1}{3}\end{pmatrix}$$Then $$\hbox{tr}(A^TA) = \hbox{tr}\begin{pmatrix} \frac{1}{4} \ \ 0 \\ 0 \ \ \frac{1}{9}\end{pmatrix} = \frac{13}{36} < \frac{1}{4} = \lambda_{max} \ \mbox {greatest eigenvalue of $A^{T}$*A}$$.
Thus, they must be talking about the Frobenius Norm? I'm just confused because the Frobenius Norm is typically denoted: $\|A\|_F$. Thanks for the help!
$$ (A^T A)_{kl} = \sum_{r}a^T_{rl} a_{kr} = \sum_{r}a_{lr} a_{kr} $$ so, taking the trace $$trace(A^T A) == \sum_{k} (A^T A)_{kk} =\sum_{kr}a_{kr} a_{kr} $$
The formula should tell you which norm (squared) this is.