Is the Tr $({\bf A}) < 0$ for ${\bf A}$ a negative definite matrix.

Question

Where Tr$({\bf A})$ denotes the sum of the diagonal of the square matrix ${\bf A}$ of size n. ${\bf A}$ comprised of all real elements, with real eigenvalues $\lambda_n<0$.

Answer 1

Look back in your textbook for the definition of a negative definite matrix. Using that, we can say that if $e_1,\dots,e_n$ is the standard basis of $\Bbb R^n$, we have $e_k^TAe_k < 0$ for all $k$, which means that $$ \operatorname{tr}(A) = \sum_{k=1}^n e_k^TAe_k < 0 $$

Answer 2

Note that the trace of a matrix $A$ is also equal to the sum of the eigenvalues of $A$.