# 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.

1. Show that $\operatorname{Tr}(A^2) \leq \operatorname{Tr}(A)^2$ holds for any symmetric matrix $A$ whose eigenvalues are all non-negative.
2. Show that $\operatorname{Tr}(AB)^2 \le \operatorname{Tr}(A^2)\operatorname{Tr}(B^2)$ holds for any symmetric matrices $A$ and $B$.
• If $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $A$, what are the eigenvalues of $A^2$? And how the trace of a matrix is related to the eigenvalues? Oct 28, 2015 at 9:08
• for both questions i just worked through an example which proved the questions right. I understand the eigenvalues for $A^2$ are the same as for $A$ but squared and when you add the trace you should get the same value as the eigenvalues added. I just wasn't sure in the second question as it says about all non-negative eigenvalues if just working through an example that proves it right really proves it for all cases and therefore had to go about it a different way Oct 28, 2015 at 12:25
• for the fourth question I cant really prove it as much as I just show that it is true through a worked example, im guessing this is wrong but im not sure how else to do it Oct 28, 2015 at 14:49

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.

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 $b_{k\ell} = b_{\ell k}$.