TITLE: Why is the Operator Norm so hard to calculate?
QUESTION [26 upvotes]: I recently took a better look at the operator norm defined on a matrix $\mathbf A \in \Bbb{K}^{n\times n}$ as follows:
$$
\|\mathbf A\|_p=\sup\{\|\mathbf Ax\|_p \mid x\in\Bbb{K}^n\land\|x\|=1\}
$$
The first time I looked at this I thought "ok, lets calculate it for a few example matrices". I started with $n = 3$ and $p = 2$, just to start "simple". Let
$$
\mathbf A = \left[\begin{matrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{matrix}\right]\quad a_{ij}\in\Bbb{K}^n
$$
Now if we're going to minimize $\|\mathbf Ax\|_2$ ($ = \|\mathbf Ax\|$), we might as well make it easy on ourselves and only minimize $\|\mathbf Ax\|^2$ so as to not worry about that annoying radical. We get
$$
\begin{align}
\|\mathbf Ax\|^2 & = (a_{11}x_1 + a_{12}x_2 + a_{13}x_3)^2 + (a_{21}x_1 + a_{22}x_2 + a_{23}x_3)^2 + (a_{31}x_1 + a_{32}x_2 + a_{33}x_3)^2 \\
& = Ax_1^2 + Bx_2^2 + Cx_3^2 + Dx_1x_2 + Ex_1x_3 + Fx_2x_3
\end{align}
$$
where
$$
\begin{align}
A & = a_{11}^2+a_{21}^2+a_{31}^2 \\
B & = a_{12}^2+a_{22}^2+a_{32}^2 \\
C & = a_{13}^2+a_{23}^2+a_{33}^2 \\
D & = 2(a_{11}a_{12} + a_{21}a_{22} + a_{31}a_{33}) \\
E & = 2(a_{11}a_{13} + a_{21}a_{23} + a_{31}a_{33}) \\
F & = 2(a_{12}a_{13} + a_{22}a_{23} + a_{32}a_{33})
\end{align}
$$
Now lets define 
$$
G(x_1,\ x_2,\ x_3) = Ax_1^2+Bx_2^2+Cx_3^2+Dx_1x_2+Ex_1x_3+Fx_2x_3
$$
So if we want to minimize $||\mathbf Ax||^2$, we're either going to have to minimize
$$
N(x_1,\ x_2,\ x_3) = \frac{G(x_1,\ x_2,\ x_3)}{x_1^2+x_2^2+x_3^2}
$$
or simply minimize $G$ with the constraint $g(x_1,\ x_2,\ x_3) = x_1^2 + x_2^2 + x_3^2 = 1$. The latter seemed easier to me, so I gave it a shot using Lagrange multipliers.
As usual, I defined 
$$
\mathcal{L}(x_1,\ x_2,\ x_3,\ \lambda) = G(x_1,\ x_2,\ x_3)-\lambda g(x_1,\ x_2,\ x_3)
$$
setting it's gradient to zero gives
$$
\nabla \mathcal L = 0 \implies \begin{cases}
2(A - \lambda)x_1 + Dx_2 + Ex_3 & = 0 \\
Dx_1 + 2(B - \lambda)x_2 + Fx_3 & = 0 \\
Ex_1 + Fx_2 + 2(C - \lambda)x_3 & = 0 \\
x_1^2 + x_2^2 + x_3^2 - 1 & = 0
\end{cases}
$$
Now this is where I really started to get stuck. I tried solving the first three equations for $x_1,\ x_2,$ and $x_3$ but didn't end up with anything I could use. I tried solving for $x_1$ in terms of $x_2,\ x_3,$ and $\lambda$, then $x_2$ in terms of $x_3$ and $\lambda$, and then subbing that all into the third equation, but ended up with either $x_3 = 0$ or 
$$
4\lambda^3 - 4\lambda^2(A+B+C) + \lambda(4AB+4AC+4BC-D2+E^2+F^2)-4ABC-AF^2-BE^2+CD^2+DEF = 0
$$
which, although technically solvable for $\lambda$ via the cubic equation, would be incredibly messy.
Now, I probably created my own roadblock for this problem, because I didn't want to think about the system of equations logically and just wanted to bash it out. Regardless of my approach, it seems like the operator norm is a very difficult thing to calculate, and I only analyzed the case where $n = 3$ and $p = 2$. What about the general case? What if $n = 75$ and $p = 9/4$? How on earth would you calculate it then?
The questions above are rhetorical, however, and my actual question is as follows:

Why define such an ordinary norm for matrices which is so difficult to calculate in general?

I see the operator norm everywhere, and it seems like the standard norm for a lot of theorems (unless I'm mistaken and ||A|| just means any matrix norm). So why would we define such a standard norm in a way that is so difficult to calculate? What's the point? Is it easy to work with in theorems? I get that it intuitively makes sense as a norm, but it can't possibly be that easy to work with, especially in comparison to things like the Frobenius norm. 
So why do we care about this definition?

REPLY [4 votes]: It depends on the norm which you take to begin with. Some matrix norms are hard to compute, others are not. In your example, for $p=2$, the norm of the matrix $A\in \mathbb{K}^{n\times n}$ is the square root of the maximal eigenvalue of $A^* A$. This computation is not too hard, even in large dimensions, as this is a Hermitian, resp. symmetric eigenvalues problem. 
Similarly, for $p=1$ and $p=\infty$ the matrix norm has simple expressions, as the column sum, resp. row sum norm.
The technical reason why operator norms are great has been pointed out in previous answers. Submultiplicativity is very handy for many types of estimates. For instance, you get that $\|A\| \geq r(A)$, where $r(A)$ is the spectral radius of $\|A\|$. And on top of that the famous Gelfand formula
$$ r(A) = \lim_{k\to \infty} \|A^k\|^{1/k} = \inf_{k\geq 1} \|A^k\|^{1/k},$$
which even holds for bounded linear operators in Banach spaces.