trace of symmetric matrix problems I have the two problems below from a practice exam. I can prove them on their own but am not exactly sure if/how to show that they only hold for symmetric matrices and for '3)' showing that it only holds for a matrix with only positive eigenvalues. I know that if the eigenvalues are all positive the determinant will be positive and the trace but cant see how that affects whether '3)' is true or not.


*Show that $\operatorname{Tr}(A^2) \leq \operatorname{Tr}(A)^2$ holds for any symmetric matrix $A$ whose eigenvalues are all non-negative.

*Show that $\operatorname{Tr}(AB)^2 \le \operatorname{Tr}(A^2)\operatorname{Tr}(B^2)$ holds for any symmetric matrices $A$ and $B$.

 A: Question No.3 is more related to the fact that for given numbers $x_1,\dots,x_n$, the following inequality $$x_1^2+\dots+x_n^2\leq \,(x_1+\dots+x_n)^2$$ holds only if $x_i$ are non-negative. Let $A$ be any diagonalizable matrix so that $A=T\Lambda T^{-1}$ and $A^2=T\Lambda^2 T^{-1}$. Thus, if $x_1,\dots,x_n$ are the eigenvalues, then $\mathrm{trace}(A^2)=x_1^2+\dots+x_n^2 $ and $\mathrm{trace}(A)^2=(x_1+\dots+x_n)^2 $. Note that symmetric matrices are readily diagonalizable since they are normal. 
Question No.4 is more related to the fact that trace is an inner product in the space of symmetric matrices. In fact, that inequality you have given is Cauchy-Schwartz indeed. 
A: About question  $\#4$:
Notation: Let $C= AB$ and $K^2_{ii}$ denote the element at the $(i,i)$ position of the matrix $K^2$. 
Firstly, due to symmetry of the $n\times n$ matrices $A,B$, it is easy to prove that:
$$c_{ii} \le \left(A^2_{ii}\right)^{1/2} \cdot \left(B^2_{ii}\right)^{1/2},\quad i = 1,\ldots, n.\tag 1$$

Proof of $(1)$
$c_{ii} =^\color{red}{\star\star} a_{i1} \cdot b_{i1} + \cdots +a_{in}\cdot b_{in}= \sum\limits_{j=1}^{n}a_{ij}\cdot b_{ij}\color{blue}{\le^\star} \left(\sum\limits_{j=1}^{n}a^2_{ij}\right)^{1/2}\cdot \left(\sum\limits_{j=1}^nb_{ij}^2\right)^{1/2}=\left(A^2_{ii}\right)^{1/2}\cdot \left(B^2_{ii}\right)^{1/2}\quad \text{QED}$
Thus, we have:
$$\big[\operatorname{trace} (AB)\big]^2=\left(\sum_{i=1}^n c_{ii}\right)^2\le\left[ \sum_{i=1}^n \left(A^2_{ii}\right)^{1/2}\cdot \left(B^2_{ii}\right)^{1/2}\right]^2\color{blue}{\le^\star}\sum_{i=1}^n A^2_{ii} \cdot \sum_{i=1}^n B^2_{ii}=\operatorname{trace} (A^2) \cdot \operatorname{trace} (B^2)$$

$^\color{blue}{\star}$ We have applied the Cauchy-Schwarz inequality.
$^\color{red}{\star\star}$ Normally, it is:
$c_{ii} = a_{i1}b_{1i} + a_{i2}b_{2i} +\cdots + a_{in}b_{ni},$
but notice that $b_{k\ell} = b_{\ell k}$.
