Relations between matrix norm and determinant 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?
$$|Ax|_2 \leq \|A\|_{\text{op}} |x|_2 \leq C_n |\det A| \ |x|_2 \leq \cdots$$
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! :)
 A: 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 $t\to0$.
A: Consider the matrices of the form $\left(\begin{smallmatrix}1&x\\0&1\end{smallmatrix}\right)$, with $x\in(0,+\infty)$. All of them have determinant equal to $1$, but their norms are arbitrarily large. Does this answer your question?
A: I'm not sure how to prove it, but you might find this exact formula for the norm from Hubbard & Hubbard's Vector Calculus interesting:

If A is a real 2x2 matrix, then
$$||A|| = \left(\frac{|A|^2+\sqrt{|A|^4-4(\det A)^2}}{2}\right)^{1/2}$$
where |A| is the Frobenius norm of $A$.

José Carlos Santos's answer shows that the determinant alone does not carry enough information to determine the norm. For the matrix $\begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix}$ that he considered in his example, we see that as $x \to \infty$, while $\det A \to 1$, the above formula is still able to work because $|A| \to \infty$.
