# generalized ordering of positive semi-definite matrix by eigenvalues

I know positive semi-definite matrices are generalizations of non-negative numbers. So "ordering" of the two systems should be pretty much like each other. How to prove the following theorem?

For two symmetric $X$ and $Y$, if $X \geq Y$, then $\lambda_i(X) \geq \lambda_i(Y)$, for every $i$. $\lambda_i(\cdot)$ denotes the $i$-th largest eigenvalue.

And what about the converse statement? Is it true?

Thanks a lot.

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The converse is not true. Try to find an easy $2$-by-$2$ counterexample. –  Jonas Meyer Jan 23 '12 at 6:18
That case $i=1$ is straightforward: $Y\leq X\leq \lambda_1(X)I\implies \lambda_1(Y)\leq \lambda_1(X)$. Considering $-X\leq -Y$, it becomes similarly straightforward to show the $i=n$ case (if $X$ and $Y$ are $n$-by-$n$). The other cases aren't as straightforward unless $X$ and $Y$ commute. –  Jonas Meyer Jan 23 '12 at 6:41

Courant-Fischer Theorem $\Rightarrow$ Weyl Theorem $\Rightarrow$ Monotonicity Theorem
And the converse statement is not true. Here is a simple example: $A = \bigl(\begin{smallmatrix} 2&0\\ 0&4 \end{smallmatrix} \bigr)$ $B = \bigl(\begin{smallmatrix} 3&0\\ 0&1 \end{smallmatrix} \bigr)$