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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?

2 Answers2

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

OldGodzilla
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  • The question is why all eigenvalues are nonnegative $\text{Tr}(Y-X)\leq 0$. Please read the statement carefully and confer to the provided links. –  Jan 15 '19 at 17:18
  • Sorry, I took it as the definition of positive semidefinite that all the eigenvalues are nonnegative (this is often the case). – OldGodzilla Jan 15 '19 at 17:28
  • @Saeed: Please delete the answer because it does not follow the statement. The problem is not as easy as it seems. Or you can provide a good answer. –  Jan 15 '19 at 17:30
  • What is your definition of positive semidefinite? Is it that $x^* A x \geq 0$ for all vectors $x \neq 0$? If so, because $A$ is Hermitian, then it is unitarily diagonalizable. Hence, by a change of variables, if $A = UDU^$, where $U$ is unitary and $D$ is diagonal. Then $0 \leq x^ A x = (Ux)^* D Ux = y^* D y$ for all vectors $y \neq 0$ (since unitary matrices are bijective). Hence, taking $y$ to be a standard basis vector gives that each entry of $D$ is nonnegative, or that the eigenvalues of $A$ are nonnegative. So my definition and yours are equivalent. – OldGodzilla Jan 15 '19 at 17:50
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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*}

Tom Chen
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  • still using summation! This is exactly what is in https://math.stackexchange.com/questions/3074592/how-to-show-texttrab-leq-texttrac-where-b-preceq-c which I said. Can you show it without summation? –  Jan 15 '19 at 20:29
  • Ah, that may be very difficult. The trace itself, by definition, is a sum, and any direct proof I can imagine (avoiding eigenvalues, for example), would involve quadratic forms which, in itself, are sums (of squares). – Tom Chen Jan 16 '19 at 05:29