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I am reading a textbook which states

for $A, B$ symmetric, $A \leq B$ iff $B-A$ is non-negative definite (equiv. positive semi-definite). This is easily seen to be an ordering.

This reads like the following statement for real numbers, applied to matrices: "$a \leq b$ iff $b-a$ is non-negative."

It's not intuitively clear to me why ordering for matrices is defined using positive definiteness. For instance, we could define a "positive" matrix as one that has all positive entries, but we don't do that. What is special about the $x^TAx \geq 0$ definition for positive semi-definite matrices that allows the notion of positiveness of real numbers to generalize to matrices in a useful manner?

mai
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  • https://stats.stackexchange.com/questions/224005/why-are-symmetric-positive-definite-spd-matrices-so-important – saulspatz Sep 19 '18 at 18:09
  • That thread has a lot of useful facts about psd matrices. I think the one that comes closest to replicating the idea of "positiveness" in numbers is that a convex function has psd Hessian (similar to how it has nonnegative second derivative in one dimension). The reasoning behind that using the Taylor expansion isn't very intuitive, though. Maybe that's the best explanation there is. – mai Sep 20 '18 at 01:56

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I see it as follows.

Assuming here that we are working over real numbers.

Now what is a $1\times 1$ symmetric matrix. It is simply $[b]$ where $b$ is a real number. Let us call this matrix $B$. Now $B$ being positive semidefinite means $$x^t Bx \geq 0 $$ for all $ x \in \mathbb{R} $

This will happen if and only if $b\geq 0$ which conveys nothing but positiveness of real number $b$.

So it makes sense to me. In short think of real numbers as $1\times 1$ matrix. What you have written follows.

Shweta Aggrawal
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