How to show the trace inequality of two P.S.D matrices $\text{Tr(X)}\leq\text{Tr(Y)}$ when $X \preceq Y$?

Question

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?

Answer 1

Let $e_i$ be a the basis vector with 1 in element $i$, and 0's elsewhere. Then \begin{align*} e_i^\intercal (Y - X) e_i = Y_{ii} - X_{ii} \ge 0 \end{align*} Summing over $i$, \begin{align*} \text{tr}(Y-X) = \sum_i(Y_{ii} - X_{ii}) \ge 0 \end{align*}

Answer 2

Since $Y - X$ is positive semidefinite, all its eigenvalues are nonnegative, so $Tr(Y-X) \geq 0$. Now, by linearity of the trace operator, $Tr(Y) - Tr(X) \geq 0.$