We have an $(n \times n)$ real symmetric positive semi-definite matrix A such that $A\neq 0$. Does this imply that $tr(A)>0$?
Comments: It follows from the p.s.d of the matrix that all eigenvalues, $\{\lambda_i\}_{i=1}^n$, are non-negative: Hence, $tr(A)=\sum_i^{n}\lambda_i\geq 0$. However I don't see how the fact that $A$ is symmetric and $A\neq 0$ helps us to show that $tr(A)\neq 0$
Thanks to @ChongxuRen, I found a correct answer.
From p.s.d of matrix $A$ we have that $tr(A)\geq 0$, Hence it remains to show that $tr(A)\neq 0$.
Lets prove it from contradiction. Lets $tr(A)=0$, this implies that all eigenvalues are zero (since $0\leq tr(A)$). In general, if all eigenvalues are zero it is not necessary that the matrix is zero. However, if matrix $A$ is a real and symmetric than we have $A=\sum_{i=1}^n \lambda_i v_i v_i'$, where $\{\lambda_i\}$ are eigenvectors and $\{v_i\}$ is associated orthonormal eigenvectors. Hence, for real symmetric matrix if all eigenvalues are zero than $A=\sum_{i=1}^n \lambda_i v_i v_i'=0$ and this contradicts initial condition $A\neq 0$.