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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?

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A bit confusing. Use $\|x\|$ (\|x\|) to denote a norm, and $|x|$ (|x|) to denote the absolute value. –  Guess who it is. 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

2 Answers 2

up vote 4 down vote accepted

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

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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.

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