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I need your expertise in solving the following problem:

Let $A,B \in \mathbb{R}^{n \times n}$ be two inverse matrices such that $\left| B \right| \geq \left| A\right|$ then how can we show that for every $x \in \mathbb{R}^n$ such that the following applies: $$ \left\| Bx\right\|_2 \geq \left\| Ax\right\|_2$$

Is it possible?

Please advise and thanks in advance.

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  • $\begingroup$ What is the definition of $|A|$ for an $n\times n$ matrix $A$? Is it the $sup \|Ax\|$ over unit vectors $x\in \mathbb{R}^n$? $\endgroup$ – ChocolateAndCheese Dec 21 '16 at 20:42
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In general we do not have $\left\| Bx\right\|_2 \geq \left\| Ax\right\|_2$.

Example: $x \in ker(B)$ and $x \notin ker(A)$

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  • $\begingroup$ What kind of conditions must be applied so that we can the required inequality? $\endgroup$ – user3492773 Dec 21 '16 at 19:46
  • $\begingroup$ @Fred In this situation $A$ and $B$ are invertible so the $x$ in your counterexample does not exist. $\endgroup$ – ChocolateAndCheese Dec 21 '16 at 20:45
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This is not in general true.
For example, let $A= \begin{pmatrix} 2 & 0 \\ 0 & \frac{1}{3}\end{pmatrix}$ and $B = \begin{pmatrix} \frac{1}{2} & 0 \\ 0 & 3\end{pmatrix}$. Then $A$ and $B$ are inverses with $\|A\| = 2$ and $\|B\| = 3$. However, for the vector $x = \begin{pmatrix}1 \\ 0\end{pmatrix}$, we have $\|Ax\|_2 = 2$ and $\|Bx\|_2 = \frac{1}{2}$.

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