# EBounds for sum of squared eigenvalues

Given a matrix $A$ of order $n$, which is symmetric semidefinite positive and at least one of its eigenvalues is zero, I'd like to find an upper bound for the sum of the square of its eigenvalues, expressed in terms of the trace of the matrix.

In literature I could only find the following one: $\mbox{tr}(A^2)\le [\mbox{tr}(A)]^2$

But in my case it is too loose. Is there anything better around? I believe that positive semidefiniteness should play a preminent role here.

EDIT: As correctly pointed out, the elements I have given are insufficient for a tighter bound. I know more about this matrix $A$. In particular $A$ is given by: $P^{1/2}BP^{1/2}$ where $P$ is psd, symmetric, with all elements on the diagonal equal to one (is a correlation matrix) and $B$ is a symmetric idempotent matrix.

EDIT 2: To clarify: if the trace of $A$ is $n-1$ (its maximum possible value), then the trace of $A^2$ is still $n-1$, rather than $(n-1)^2$, whatever the elements of $P$ are. If the trace of $A$ is very close to $n-1$, then the trace of $A^2$ is still very close to $n-1$... That's why I'm expecting a tighter bound on the trace of $A$ rather than simply $[tr(A)]^2$.

See the simulation example in the image... when $trace(A)$ (x-axis) is approaching $2$, the simulation shows clearly that something happens, and the trace of $A^2$ (y-axis) is clearly bounded by another (unknown) function of the trace. I believe that this bound should have something to do with this http://mathoverflow.net/questions/107434/upper-bounds-on-eigenvalues-of-psd-matrix. • Edit helps. The previous version was an x-y problem: meta.stackexchange.com/questions/66377/what-is-the-xy-problem Apr 8, 2016 at 14:06
• What does your bound have to be in terms of? This bound is still tight if it is just in terms of the trace of A (take B to be a rank 1 projection matrix onto the space spanned by your favorite eigenvector of P). Apr 8, 2016 at 14:09
• @Batman, I've added an edit which shows why I'm expecting tighter bounds in terms of the trace alone to exist.
– xanz
Apr 8, 2016 at 16:35

Consider the $2 \times 2$ diagonal matrix $$\begin{bmatrix} u & 0 \\ 0 & 0 \end{bmatrix}$$ with $u \ge 0$ This is symmetric positive semidefinite. Its trace is $u$; the trace of its square is $u^2$; that shows that the bound you've cited is tight. So unless you know more about your matrix than you've told us, there's no tighter bound.

• Yeah, indeed I know more about it, I'll add some more details to see if that helps.
– xanz
Apr 8, 2016 at 13:50

You can't improve this in general for psd matrices.

If you take $A$ to be a rank 1 p.s.d. matrix (i.e. $A = \lambda u u^T$ for some $\lambda>0$ and $u$ to be a unit vector), $tr(A^2) = \lambda^2 = (tr(A))^2$.

• You're right. I've added some more details on this matrix but I don't know if the problem is simple to solve.
– xanz
Apr 8, 2016 at 13:55