Let $X,Y$ be two Positive Semi-Definite matrices. How can we show the following in the most elegant and shortest way? Because I know how to prove it but I think there is a better way?
Alos, MaoWao shows it differently using summation.
$$\text{Tr}(X)\leq\text{Tr}(Y)$$ when $X \preceq Y$, where $Y-X$ is positive semi-definite?
My try is:
$x^T(Y-X)x \geq 0$ so $\text{Tr}(xx^T(Y-X)) \geq 0$.
Then using $\text{tr}(AB) \leq \text{tr(A)} \text{tr(B)}$
$$0 \leq \text{Tr}(xx^T(Y-X)) \leq \text{Tr}(xx^T)\text{Tr}(Y-X)$$
hence the claim.
I do not want to use this $\text{tr}(AB) \leq \text{tr(A)} \text{tr(B)}$ or sumation. Is there any other way to show it shorter?
Best Answer
Since $Y - X$ is positive semidefinite, all its eigenvalues are nonnegative, so $Tr(Y-X) \geq 0$. Now, by linearity of the trace operator, $Tr(Y) - Tr(X) \geq 0.$