How to compute the norm of a linear mapping? Is there hints for this exercise from a topology book of mine? First there is a notation that if $A:E\to F$ is a linear mapping between norm spaces then we denote $$|A|=\sup\{|Ax|:|x|\leq 1\}.$$ Now, suppose that $A:\mathbb R^2\to \mathbb R^2$ is a linear mapping such that $$A=\begin{pmatrix}1 & 3\\2 & 4\end{pmatrix}. $$ Compute $|A|$. I had difficulties even to start. Do I need eigenvectors? I was given the answer $|A|=\sqrt{15+\sqrt{221}}$.
 A: If $|\cdot|$ is the Euclidean norm on both of $E $ and $F $, then $|A|$ is the biggest singular value, i.e. the square root of the largest eigenvalue of $A^TA $.
A: Disclaimer: The other answers are more concise and the OP has already accepted one, but I had already started writing mine, so I went with it.

Note that the norm of $A$ is defined as the supremum of a particular set, namely the image of the closed unit ball under $|A|$ (which is a composition of $A$ with $|\cdot|$), $|A|(B_1(0))$. Note that $|A|(B_1(0))$ is a subset of $\mathbb{R}$.
Then all we need more is the definition of supremum of a subset of $\mathbb{R}$. If $S(\neq\emptyset)\subseteq\mathbb{R}$ is bounded from above, (by the supremum axiom) it has a least upper bound $\sup{A}$. This means that $A$ has no member greater than this number, and any number smaller than this number cannot be an upper bound for $A$. There is a practical way of characterizing $\sup{A}$: $\alpha=\sup{A}$ if and only if the following two conditions hold:


*

*$\forall a\in A:a\leq\alpha$

*$\forall \varepsilon>0,\exists a_\varepsilon\in A: \alpha-\varepsilon<a_\varepsilon\leq\alpha.$

