I have a naive question about the trace of a real, symmetric positive semi-definite matrix:

Does the trace of a real, symmetric positive semi-definite matrix have to be larger than $1$?

I know that all the eigenvalues of the matrix must be strictly non-negative, but does the sum of all the non-negative eigenvalues must be larger than $1$?

  • 2
    $\begingroup$ the trace of the zeros matrix is zero. $\endgroup$ – user251257 Jul 19 '15 at 23:05
  • $\begingroup$ Just a clarification. From the context, I can understand that you are talking about real symmetric positive semi - definite matrices, right? $\endgroup$ – thanasissdr Jul 19 '15 at 23:15
  • $\begingroup$ @thanasissdr Yes of course, sorry I forgot to mention that. I have a follow-up question: what if you know that the entries of the matrix are larger than 1? does that change anything? $\endgroup$ – Jack Shi Jul 19 '15 at 23:19
  • $\begingroup$ @thanasissdr Thanks a lot for your help man! really appreciate it ! $\endgroup$ – Jack Shi Jul 19 '15 at 23:38
  • $\begingroup$ If $I_n$ is the $n\times n$ identity matrix, the trace of $\epsilon I_n$ for $\epsilon>0$ is $n\epsilon$, which can be made arbitrarily small (given a fixed $n$) as $\epsilon\downarrow 0$. $\endgroup$ – Kim Jong Un Jul 19 '15 at 23:58

No, that's not necessary. A non-trivial counterexample:

Let $$A = \begin{bmatrix} 0.26 & 0.05 \\ 0.05 & 0.01 \end{bmatrix},$$ which is clearly a real symmetric positive semi-definite matrix (actually, it's positive definite).

This can be verified by computing its eigenvalues, which are: $\lambda_1 = 0.0004$ and $\lambda_2 = 0.2696$. Clearly, the trace of $A$ is smaller than $1$.

  • $\begingroup$ got it. thanks a lot! $\endgroup$ – Jack Shi Jul 19 '15 at 23:12
  • $\begingroup$ @JackShi No problem. You're welcome! $\endgroup$ – thanasissdr Jul 19 '15 at 23:13

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