41 views

$M^2<N^2$ if $M,N$ are two positive definite matrix [duplicate]

If $M,N$ are two positive definite matrix st. $M<N$, is that true that $M^2<N^2$?
97 views

$A\geq B$ implies $A^{\frac{1}{2}}\geq B^{\frac{1}{2}}$ for positive semi-definite matrices $A$ and $B$. [duplicate]

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 ...
60 views

If $A\ B$ are symmetric positive definite matrices then $A>B$ iff $\sqrt{A}>\sqrt{B}$ [duplicate]

My guess is, it only holds one way i.e. $A>B$ then $\sqrt{A}>\sqrt{B}$ but not otherwise. Any proof or counterexample would be appreciated.
35 views

Can $U\Sigma U^T \preceq UVSV^TU^T$ lead to $( UVS^{-1}V^TU^T )^2 \preceq (U\Sigma^{-1} U^T)^2$? [duplicate]

Can $U\Sigma U^T \preceq UVSV^TU^T$ lead to $( UVS^{-1}V^TU^T )^2 \preceq (U\Sigma^{-1} U^T)^2$? where $UU^T=U^TU=VV^T=V^TV=I$ and $\Sigma, S$ are square matrix only with positive elements in its ...
31 views

If matrices $A,B$ and $(A^2-B^2)$ are all positive semidefinite, is $A-B$ also? [duplicate]

The question is in the title. I have not found a counter example, so I try to prove it. However, contract diagonalization at the same time seems useless. Thanks very much.
15 views

Positive definite squares [duplicate]

Suppose that $A, B$ are real $n\times n$ symmetric positive definite matrices such that $A - B$ is positive semi-definite. Does it follow that $A^2 - B^2$ is positive semi-definite?
479 views

if the matrix such $B-A,A$ is Positive-semidefinite,then $\sqrt{B}-\sqrt{A}$ is Positive-semidefinite

Question: let the matrix $A,B$ such $B-A,A$ is Positive-semidefinite show that: $\sqrt{B}-\sqrt{A}$ is Positive-semidefinite maybe The general is true? question 2: (2)$\sqrt[k]{B}-\sqrt[k]{A}$...