Proving $||A^{-1} v || \geq ||A^{-1} u ||$ implies $||v|| \geq ||u||$.

81 Views Asked by At

I hope to show that $||A^{-1} v || \geq ||A^{-1} u || \implies ||v|| \geq ||u||$ .

$A$ is a positive definite matrix of size $n \times n$ and $v, u$ is a vector of size $n \times 1$.

$||\cdot||$ can be any norm (e.g., 1-norm or 2-norm). That is, if $||\cdot||$ should be 1-norm to satisfy the equation, it is ok, and I would like to know that condition.

How can I prove this? If I can't, which conditions are additionally needed? Or, I wonder if this is absolutely impossible.

I would really appreciate your help!!

1

There are 1 best solutions below

1
On BEST ANSWER

This is not possible. Let $A^{-1}v = \lambda v$ and $A^{-1}u = \mu u$. Then what you are trying to prove is $\lambda \lVert v \rVert \geq \mu \lVert u \rVert$ implies $\lVert v \rVert \geq \lVert u \rVert$, which is obviously not true for any norm.