Does a positive semidefinite matrix always have a non-negative trace?

Question

If $A$ is a positive semidefinite matrix ($A\succeq 0)$, does it imply that $\mbox{Tr}(A)\geq 0$, where the $\mbox{Tr}(\cdot)$ denotes the trace.

If not, any counter-example? Thanks.

Suppose a matrix has a negative diagonal element. Can you see that the matrix cannot be positive semidefinite? — Daniel Fischer, Sep 11 '14 at 19:03
I know most likely this statement is wrong, but I just could not figure out an counter-example. — MIMIGA, Sep 11 '14 at 19:06
@MIMIGA the statement is correct: if $A \succeq 0$, then $\mathrm{Tr}(A) \geq 0$. — Ben Grossmann, Sep 11 '14 at 19:12
@Omnomnomnom Is the converse true? This is, if $\mbox{Tr}(A)\geq 0$, then $A\succeq 0$. — Diego Fonseca, Aug 31 '16 at 15:19
@Omnomnomnom But $A=\left(\begin{array}{l}2&0\0&-1\end{array}\right)$ is not positive semidefinite matrix. Note that for $x=\left(\begin{array}{l}0 \ 1\end{array}\right)$ we have $x^{T}Ax\leq 0$. — Diego Fonseca, Sep 01 '16 at 02:53
Even if all diagonal elements are $\ge0$, we need not have $A\succeq0$. Let $A=\begin{pmatrix} 1 & 2 \ 2 & 1 \end{pmatrix}$. As $\det A=-3$, $A$ has a negative eigenvalue $s$; hence $x^* Ax=x^*sx=s|x|^2<0$ for a corresponding eigenvector $x$.
(This also shows that $A\succeq0$ implies $\sigma(A)\subset[0,\infty)$. — user3810316, Apr 27 '22 at 11:18

score 39 · Answer 1 · answered Sep 11 '14 at 19:16

Suppose that $A = [a_{ij}]_{i,j=1}^n$ is such that $a_{ii} < 0$ for some $i$. Let $e_i$ be the $i$th standard basis vector; that is, $$ e_i = (\overbrace{0,\cdots,0}^{i-1},1,0,\dots,0) $$ then $e_i^T Ae_i = a_{ii} < 0$, which means that $A$ is not positive semidefinite.

So, if $A$ is positive semidefinite, then all diagonal elements are non-negative, which means that the trace is non-negative.

score 20 · Answer 2 · answered Sep 11 '14 at 19:21

20

Yes. If the matrix is semi-positive definite, all the eigenvalues are non-negative. The trace being equal to the sum of eigenvalues, we conclude that the trace has to be non-negative.

answered Sep 11 '14 at 19:21

Drmanifold

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score 8 · Answer 3 · answered Sep 11 '14 at 19:15

8

We know that positive semi-definite matrices have nonnegative eigenvalues, and the trace of a matrix is equal to the sum of its eigenvalues. Because sum of nonnegative numbers is nonnegative, the trace of a positive semi-definite matrix is nonnegative.

answered Sep 11 '14 at 19:15

user153012

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Does a positive semidefinite matrix always have a non-negative trace?

3 Answers3

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