Is there relationship between magnitude of matrix-vector multiplication and determinant of that matrix?

Question

If I have a matrix $A$ and vector $x$ is there such a relationship or something similar involving determinants? $$\|Ax\| \leq |\det A|\|x\|$$ where the absolute values indicate the usual Euclidean norm?

A bit confusing. Use $|x|$ (\|x\|) to denote a norm, and $|x|$ (|x|) to denote the absolute value. — J. M. ain't a mathematician, Apr 20 '12 at 11:04
Look up operator norm for the thing you need to replace $\det$ with to make this true. — Chris Eagle, Apr 20 '12 at 11:20
The determinant should be replaced by something like the spectral radius. — Fabian, Apr 20 '12 at 11:25
Thanks. So can I write: $|Ax|{R^n} \leq | A |{R} |x| \leq C|A|{\infty}|x|$ where $|\cdot |{R}$ is some norm on $\mathbb{R}^n$ and $|A|_{\infty}$ is the biggest absolute-valued element in the matrix? I got this by equivalence of norms and so I need a constant $C$. — AIOM, Apr 20 '12 at 12:53

score 8 · Answer 1 · answered Sep 12 '17 at 13:06

AGTortorella showed that it cannot be true when $det(A) = 0$, but for $det(A) \ne 0$ this inequality also doesn't hold. Consider:

$A=\left(\begin{array}{cc} 1 & a\\ 0 & 1 \end{array} \right)$

$det(A) = 1$ for every $a$

Let $x = \left(\begin{array}{cc} 1 \\ 1 \end{array} \right)$

Then: $Ax = \left(\begin{array}{cc} 1 \cdot 1 + a \cdot 1 \\ 0 \cdot 1 + 1 \cdot 1 \end{array} \right) = \left(\begin{array}{cc} a + 1 \\ 1 \end{array} \right)$

So clearly:

$\|Ax\| \to \infty$ when $a \to \infty$

but

$|det(A)|\|x\|=1 \cdot \sqrt 2$ for every $a$.

score 6 · Accepted Answer · answered Apr 20 '12 at 11:14

6

No. Because if $\det A=0$ then $|\det A|\|x\|=0,$ for any $x$, while $\|Ax\|$ can take any value varying $x$.
Consider for example $$A=\left(\begin{array}{cc} 1 & 0\\ 0 & 0 \end{array} \right).$$

answered Apr 20 '12 at 11:14

agt

4,772

What if $detA\not=0$? – The way of life May 27 '17 at 17:18

score 3 · Answer 3 · answered Nov 28 '14 at 00:35

If $A$ is an invertible operator of an n dimensional vector space to itself then

$$\frac{1}{det(A^{-1})^{1/n}} \leq |A|\leq \frac{|A^{-1}|^{n-1}}{det(A^{-1})}$$

This you can get by the relation

$$s_1...s_n = det(A^{-1})$$

where $s_i$ are singular values of $A^{-1}$ which satisfy

$$\frac{1}{|A|}=s_1 \leq ... \leq s_n = |A^{-1}|$$

And you get the relation immediately.

Is there relationship between magnitude of matrix-vector multiplication and determinant of that matrix?

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