Multiplication Property of Vector Inequality

Question

I'm a beginner of matrix inequality. Now I have a vector inequality like below

\begin{equation} \begin{aligned} A \lambda \succeq \mu \end{aligned} \end{equation}

where $A$ is a $n \times n$ symmetric positive definite matrix and $\lambda \succeq 0$ is a $n \times 1$ vector. May I ask if I can draw the following conclusion?

\begin{equation} \begin{aligned} \lambda \succeq A^{-1} \mu \end{aligned} \end{equation}

If the equation above is correct, how can I prove it?

Thanks in advance!

If your matrix is semidefinite, its inverse might not be defined. — mathcounterexamples.net, Sep 30 '15 at 06:00
Yes, you are right. But let me assume this inverse matrix exists. How about that? — Greenwicher, Sep 30 '15 at 06:18
@joriki Thanks. My question is if we assume A is invertible, then can we derive the second inequality? — Greenwicher, Sep 30 '15 at 07:18
I know. I'm just saying that you should correct the question to state that the matrix is positive definite if that's the case you're interested in. — joriki, Sep 30 '15 at 07:18
@mathcounterexamples.net I have edited the problem description to let $A$ is symmetric positive definite matrix. — Greenwicher, Sep 30 '15 at 07:19

score 3 · Accepted Answer · answered Sep 30 '15 at 07:17

3

No, you cannot. Choose $\lambda=0$; certainly $0\succeq\mu$ does not imply $0\succeq A^{-1}\mu$, for instance for $\mu=\pmatrix{-1\\0}$ and $A^{-1}=\pmatrix{3&-1\\-1&3}$.

answered Sep 30 '15 at 07:17

joriki

238,052

Thanks for your counterexample. It's very helpful. – Greenwicher Sep 30 '15 at 07:23

Multiplication Property of Vector Inequality

1 Answers1