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In the finite-dimensional framework you have a normed space $X$, which may be assumed to be $R^n$ equipped with some norm, and you are looking at the set of linear transformations $T:X\to <!--\; \to \; -->X$, on which you can define a norm as follows
$\Vert <!--\; \Vert \; -->T\Vert <!--\; \Vert \; -->=max\{\Vert <!--\; \Vert \; -->Tx\Vert <!--\; \Vert \; -->:\Vert <!--\; \Vert \; -->x\Vert <!--\; \Vert \; -->\le <!--\; \le \; -->1\}$
$\Vert <!--\; \Vert \; -->T\Vert <!--\; \Vert \; -->=max\{\Vert <!--\; \Vert \; -->Tx\Vert <!--\; \Vert \; -->:\Vert <!--\; \Vert \; -->x\Vert <!--\; \Vert \; -->\le <!--\; \le \; -->1\}$
Thus the norm of a linear map is the maximal "stretch" in any direction, measured in terms of the underlying norm of $X$. Since all linear maps of a finite dimensional space are continuous, the maximum is always attained. Moreover, it is always attained on the boundary of the unit ball of $X$, namely at some $x\in <!--\; \in \; -->X$ for which $\Vert <!--\; \Vert \; -->x\Vert <!--\; \Vert \; -->=1$. Invoking convexity arguments, it can be shown that the maximum is always attained at some extreme point of the unit ball of $X$. These are points in the unit ball of $X$ that cannot be expressed as a convex combination of other points in the unit ball. This simple fact provides relatively simple formulas in case the underlying norm's unit ball does indeed possess a finite number of extreme points. For example, if the underlying unit ball is a polytope, then the norm of a linear map is the maximum of ∥$\Vert <!--\; \Vert \; -->Tx\Vert <!--\; \Vert \; -->$ taken over the extreme points only, and since this a finite set, it is perhaps easier to compute the norm. For example, if $X={\mathrm{\ell <!--\; \ell \; -->}}_1^n$, then the extreme points of the unit ball of $X$ are just $\pm <!--\; \pm \; -->e_i$, where $e_i$ are the standard unit vectors in $R^n$, and so the norm of a linear map is given by
and if we think of $T$ as being represented by an $n\times <!--\; \times \; -->n$ matrix relative to the basis $e_i$, then $T{e}_i$ is the i'th column of the matrix, so we just need to pick the column whose ${\mathrm{\ell <!--\; \ell \; -->}}_1^n$ norm is maximal. As another example, if $X={\mathrm{\ell <!--\; \ell \; -->}}_{\mathrm{\infty <!--\; \infty \; -->}}^n$, then the extreme points of the unit ball are the $2^n$ sign-vectors, namely $\mathrm{ϵ<!--\; ϵ\; -->}=(\pm <!--\; \pm \; -->1,\pm <!--\; \pm \; -->1,\dots <!--\; \dots \; -->,\pm <!--\; \pm \; -->1)$, and so the norm of a linear map $T:{\mathrm{\ell <!--\; \ell \; -->}}_{\mathrm{\infty <!--\; \infty \; -->}}^n\to <!--\; \to \; -->{\mathrm{\ell <!--\; \ell \; -->}}_{\mathrm{\infty <!--\; \infty \; -->}}^n$ is the maximum of $\Vert <!--\; \Vert \; -->T\mathrm{ϵ<!--\; ϵ\; -->}{\Vert <!--\; \Vert \; -->}_{\mathrm{\infty <!--\; \infty \; -->}}$ over all sign vectors.The situation for the Euclidean norm gets a bit more complicated, because every point of the unit ball's boundary of the Euclidean norm is an extreme point, so that in principle we have no preferred points on which we can compute the norm. I am not aware of a simple formula that connects the norm of a linear map $T:{\mathrm{\ell <!--\; \ell \; -->}}_2^n\to <!--\; \to \; -->{\mathrm{\ell <!--\; \ell \; -->}}_2^n$ to the entries of the matrix. (Here when I say "norm" I mean the norm defined by the general formula above, namely, the maximum of $\Vert <!--\; \Vert \; -->Tx\Vert <!--\; \Vert \; -->$ over $x$ varying in the underlying unit ball. Of course there are many other norms, for example, the Frobenius norm, for which there is a simple formula in terms of the matrix entries.)Even though there is no simple formula for the norm in terms of the matrix entries, the norm is intimately related to the eigenvalues of the matrix, provided these exist. This comes as no surprise, because eigenvalues measure the stretching in various directions. Thus, if $A$ is a symmetric matrix, then we know that there exists an orthonormal basis of eigenvectors of $A$, hence there exists an orthonormal matrix $U$ such that $UA{U}^{-<!--\; -\; -->1}$ is a diagonal matrix, and the diagonal consists of the eigenvalues of $A$. Then we can compute
$\Vert <!--\; \Vert \; -->A\Vert <!--\; \Vert \; -->=max\{\Vert <!--\; \Vert \; -->Ax{\Vert <!--\; \Vert \; -->}_2:\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}_2=1\}=max\{\Vert <!--\; \Vert \; -->UAx{\Vert <!--\; \Vert \; -->}_2:\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}_2=1\}=max\{\Vert <!--\; \Vert \; -->UA{U}^{-<!--\; -\; -->1}x\Vert <!--\; \Vert \; -->:\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}_2=1\}$
here we used the fact that $U$ preserves lengths as well as the fact that the set of all $U^{-<!--\; -\; -->1}x$ with $\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}_2=1$ coincides with the Euclidean unit ball. Now the action of the diagonal matrix on any $x$ is simply multiplying its coordinates by the diagonal entries, so we deduce that the norm of $A$ equals
$max{\left(\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{d}_i^2{x}_i^2\right)}^{1/2}$
where the maximum is taken over all $x$ such that $\Vert <!--\; \Vert \; -->x{\Vert <!--\; \Vert \; -->}_2=1$. It is a simple exercise to prove that this is equal to $\underset{i}{max}|d_i|$. In other words,
The norm of a symmetric matrix is equal to the maximal absolute value of its eigenvalues.
In general however, there is no simple formula.