0
$\begingroup$

Let $a$, $c$, $d$, $v \in \mathbb{R}^n$ are vectors, and $A, B \in \mathbb{R}^{n \times n}$ are matrices. Suppose that $ v = Ac-d $, and $a = ABc- \|B\| d$ where $\| B \|$ is the maximum value of the norms of the eigenvalues of $B$.

Is it true that $\| a \| \leq \|B\| \| v\|$?

Improve this question
$\endgroup$
2
  • $\begingroup$ Do you really mean $\lVert B\rVert$ is the maximum eigenvalue, or the maximum of the absolute values of the eigenvalues (which would be the operator norm)? E.g., if $B=-I$ (where $I$=identity), is $\lVert B\rVert=1$ or $\lVert B\rVert=-1$. $\endgroup$
    – Dominique Unruh
    Commented May 26, 2019 at 9:04
  • $\begingroup$ Oops, I modified it. Thank you. $\endgroup$
    – livehhh
    Commented May 26, 2019 at 9:06

1 Answer 1

Reset to default
0
$\begingroup$

No, it is not true in general.

Example: $A:=I$ (identity). $B:=-I$. $c\neq 0$ arbitrary. $d:=c$.

Then $v=Ac-d=0$ and $a=ABc-\lVert B\rVert d=-c-1d=-2c\neq 0$. Then $\lVert a\rVert > 0$ and $\lVert B\rVert\lVert v\rVert=0$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ So, you mean that $B = -I$, right? I understand that there exists an example. Thank you. $\endgroup$
    – livehhh
    Commented May 26, 2019 at 9:38
  • $\begingroup$ Yes, the "$-$" got lost accidentally. I edited it now. $\endgroup$
    – Dominique Unruh
    Commented May 26, 2019 at 9:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .