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I was wondering whether there is a way to obtain the determinant of a matrix out of its norm (when the matrix is regular otherwise it is not true). If $A$ is a square matrix of dimension $n\geq 1$, and $\det A\neq 0$, do we have something like?

$$|Ax|_2 \leq \|A\|_{op} |x|_2 \leq C_n |\det A| \ |x|_2 \leq \cdots$$ or similar? Or maybe a similar estimate for different norms for $A$ and $x$ if needed since many matrix norms are related to eigen- or singular values which are related to the determinant.

Thanks a lot! :)

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Consider the matrices of the form $\left(\begin{smallmatrix}1&x\\0&1\end{smallmatrix}\right)$, with $x\in(0,+\infty)$. They all have determinant $1$, but their norms are arbitrarily large. Does this answer your question?

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  • $\begingroup$ In some sense yes. So the question should be reformulated. For instance. $|Ax|_2\leq \|A\|_{op}|x|_2\leq C_{n,A} |\det A| \ |x|_2$ (i.e. the constant depends on the values of $A$?) $\endgroup$ – Martingalo Jul 13 '17 at 8:50
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    $\begingroup$ @Martingalo Sure. Take $\left(\begin{smallmatrix}\sqrt{x^2+1}&x\\x&\sqrt{x^2+1}\end{smallmatrix}\right)$. Its determinant is $1$, but its norm is arbitrarily large. $\endgroup$ – José Carlos Santos Jul 13 '17 at 8:55
  • $\begingroup$ Funily enough if $A$ is invertible and its determinant positive we may find some estimate for its inverse because: $$\|A^{-1}\| = \frac{1}{\det A} \|Adj(A^\ast)\|$$ but of course the constant is another matrix. $\endgroup$ – Martingalo Jul 13 '17 at 9:01
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    $\begingroup$ But anyway there is a bound for the determinant: en.wikipedia.org/wiki/Hadamard%27s_inequality So we have something like $|\det A|\lesssim \|A\|^n$ for a $n\times n$-Matrix. $\endgroup$ – tofurind Nov 9 '17 at 16:13
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I don't see any way to obtain an upper bound of $\|Ax\|$ using the determinant, because the norm of a matrix can remain unchanged when the determinant approaches zero. E.g. when $x=(1,0,\ldots,0)$ and $A=\operatorname{diag}(1,t,\ldots,t)$, we have $\|Ax\|_2=1$ but $\det A\to0$ as $t\to0$.

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