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Let $A$ and $B$ be two positive semi-definite $n\times n$ matrices. We say $A\geq B$ if $A-B$ is positive semi-definite.

Let $A^{\frac{1}{2}}$ be the square root of the positive semi-definite matrix $A$.

How to prove if $A\geq B$ then $A^{\frac{1}{2}}\geq B^{\frac{1}{2}}$?

I could prove this for $B=I$, (i.e., $A\geq I \implies A^{\frac{1}{2}}\geq I$. But I am not able to prove it for the general case.


1) All positive semi-definite matrices are symmetric for me.

2)Square root of a matrix:- Let $A$ be a symmetric matrix, then there exists an orthogonal matrix $P$ such that $P^TAP=diag(\lambda_1,\lambda_2,\cdot, \lambda_n)$, where $\lambda_1,\lambda_2,\cdot, \lambda_n$ are eigenvalues of $A$.

Define $A^{\frac{1}{2}}:=Pdiag(\lambda_1^{\frac{1}{2}},\lambda_2^{\frac{1}{2}},\cdot, \lambda_n^{\frac{1}{2}})P^T$ for a positive semi-definite matrix $A$. (Notice, $\lambda_i^{\frac{1}{2}}$ are well-defined as $\lambda_i$'s are non-negative (since A is positive semi-definite).


marked as duplicate by user1551 linear-algebra Aug 9 '16 at 3:38

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  • $\begingroup$ It can be shown that for a positive semi-definite matrix A, there exists precisely one positive semi-definite matrix B s.t BB=A, and we call B "the" square root (sometimes principal square root) of A. $\endgroup$ – mb- Aug 8 '16 at 22:59
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    $\begingroup$ By continuity, it suffices to consider only the case where $B$ is positive definite. $\endgroup$ – user1551 Aug 9 '16 at 3:36