Linked Questions

2
votes
2answers
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 $M^2<N^2$?
1
vote
0answers
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 ...
1
vote
0answers
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.
0
votes
0answers
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 ...
1
vote
0answers
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.
0
votes
0answers
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?
3
votes
2answers
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)$\sqrt[k]{B}-\sqrt[k]{A}$...
3
votes
2answers
253 views

Is it possible to show that exp(A)-exp(B) is negative definite provided $A-B$ is negative definite? [closed]

If two symmetric square matrices $A$ and $B$ are such that $A-B$ is negative definite, I am to prove that $\exp(A)-\exp(B)$ is negative definite. I initially used the following exponential series $e^...
1
vote
2answers
526 views

If $A, B, A^2-B^2$ are positive definite prove that $A-B$ is positive definite [closed]

If $A, B, A^2-B^2$ are real symmetric positive definite matrices, prove that $A-B$ is also positive definite .
3
votes
1answer
144 views

How to prove positive semidefiniteness of square root and square of two matrices

[1] The similar question I post is here. Let matrices $X$ and $Y$ be positive semidefinite (PSD). (1) $X \succeq Y$ implies $X^{1/2} \succeq Y^{1/2}$. (2) $X \succeq Y$ does not imply $X^{2} \...
1
vote
1answer
94 views

Squares of positive semidefinite matrices

Suppose $L_1 \succeq L_2$, where $L_1,L_2$ are positive semidefinite matrices (actually combinatorial Laplacians). Is the following inequality true, and if no, under which conditions? $$L_1^2 \...
1
vote
1answer
148 views

Feasibility of Matrix Inequality

I need to show if the following inequality is true $$ (A + B)^{-1}M (A + B)^{-1} - A^{-1} M A^{-1} \preceq 0$$ given that $(A,B)=(A^T,B^T) \succ 0$ and $M = M^T \succeq 0$ also we have that $A + B \...
4
votes
0answers
121 views

Squaring matrices and positive semidefiniteness

Suppose that $A$ and $B$ are positive definite, symmetric, $n\times n$ real matrices such that $A-B$ is positive semidefinite. Write $A\succeq B$. When is it true that $A^2\succeq B^2$? I know given ...
3
votes
1answer
42 views

Do we have $\|A\|_F\leq \|B\|_F$ if $-B\preceq A\preceq B$?

In the question, $A$ is a symmetric matrix, and $B$ is a positive semi-definite matrix. $A\preceq B$ means that $B-A$ is a positive semi-definite matrix. $\|\|_F$ means the Frobenius norm.
1
vote
1answer
54 views

Can $A^2\preceq \gamma^2 B^2$ lead to that $A\preceq \gamma B$?

In the question, $A$ and $B$ are positive semi-definite matrices, $\gamma\geq 0$ is a constant, and $A\preceq \gamma B$ means that $\gamma B-A$ is positive semi-definite. We have known another fact ...