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?
Is there relationship between magnitude of matrix-vector multiplication and determinant of that matrix?
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1A bit confusing. Use $\|x\|$ (`\|x\|`) to denote a norm, and $|x|$ (`|x|`) to denote the absolute value. – 2012-04-20
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0Look up [operator norm](http://en.wikipedia.org/wiki/Operator_norm) for the thing you need to replace $\det$ with to make this true. – 2012-04-20
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0The determinant should be replaced by something like the [spectral radius](http://en.wikipedia.org/wiki/Spectral_radius). – 2012-04-20
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0Thanks. 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 $C$. – 2012-04-20
3 Answers
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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0What if $detA\not=0$? – 2017-05-27
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 $a$.
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.