# Relation between norm and determinant of a linear operator [closed]

Let $A$ be a $n\times n$ matrix and define $T:\mathbb R^n\to \mathbb R^n$ by $T(X)=AX$. Is there a formula that can present the norm $\|T\|$ as the determinant $\det(A)$?

## closed as unclear what you're asking by PhoemueX, Ahaan S. Rungta, Najib Idrissi, Bruno Joyal, Rory DaultonDec 28 '14 at 19:50

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• Do you mean can you write the norm as a function of the determinant? If so, then no. – copper.hat Dec 28 '14 at 18:33
• Saying the obvious, the matrix $\begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}$ has determinant $1$ and it does not preserve norms as the identity matrix does. What do you mean by "present the norm as the determinant"? – Ian Mateus Dec 28 '14 at 18:44
• To elaborate on the other comment, note that (for $n\geq 2$) there is a matrix $A\neq 0$ with $\det(A)=0$, but if you could write the norm as a function of the determinant, then $0= \Vert 0\Vert = f(\det 0)=f(0)=f(\det A)=\Vert A\Vert \neq 0$, contradiction. – PhoemueX Dec 28 '14 at 18:47
• Thanks to you all. I mean write the question as a function of the determinant and I got the answer. – mac Jan 1 '15 at 13:23