norm(c(1,1), type="2")     # 1.414214
norm(c(1, 1, 1), type="2")  # 1.732051

OK let's see if this helps you. Suppose you have two functions $f,g:[a,b]\to \mathbb{R}$. If someone asks you what is distance between $f(x)$ and $g(x)$ it is easy you would say $|f(x)-g(x)|$. But if I ask what is the distance between $f$ and $g$, this question is kind of absurd. But I can ask what is the distance between $f$ and $g$ on average? Then it is $$ \dfrac{1}{b-a}\int_a^b |f(x)-g(x)|dx=\dfrac{||f-g||_1}{b-a} $$ which gives the $L^1$-norm. But this is just one of the many different ways you can do the averaging: Another way would be related to the integral $$ \left[\int_a^b|f(x)-g(x)|^p dx\right]^{1/p}:=||f-g||_{p} $$ which is the $L^p$-norm in general.

Let us investigate the norm of $f(x)=x^n$ in $[0,1]$ for different $L_p$ norms. I suggest you draw the graphs of $x^{p}$ for a few $p$ to see how higher $p$ makes $x^{p}$ flatter near the origin and how the integral therefore favors the vicinity of $x=1$ more and more as $p$ becomes bigger. $$ ||x||_p=\left[\int_0^1 x^{p}dx\right]^{1/p}=\frac{1}{(p+1)^{1/p}} $$ The $L^p$ norm is smaller than $L^m$ norm if $m>p$ because the behavior near more points is downplayed in $m$ in comparison to $p$. So depending on what you want to capture in your averaging and how you want to define `the distance' between functions, you utilize different $L^p$ norms.

This also motivates why the $L^\infty$ norm is nothing but the essential supremum of $f$; i.e. you filter everything out other than the highest values of $f(x)$ as you let $p\to \infty$.

The first point is proven as follows.

From the SVD of $A = UDV^T$ we can see that eigenvalues of $A^TA = VD^2V^T$ are just squared ones from $A$. At the same time the columns of $V$ are the eigenvectors of $A^TA$. So, exploiting orthogonality of eigenvectors

$$\|Ax\|_2^2 = \|UDVx\|_2^2 = \|D\left(Vx\right)\|_2^2 = \|De_{\lambda}\|x\|\|_2^2 = \|\sqrt{\lambda}\|x\|\|_2^2= \lambda\|x\|^2$$

The proof is based on the property of the second matrix norm

$${\displaystyle \|A\|_{2}={\sqrt {\lambda _{\max }(A^{^{T}}A)}}}$$. $${\displaystyle \|A^{^{T}}A\|_{2}={\sqrt {\lambda _{\max }(A^{^{T}}AA^{^{T}}A)}}}.$$

The same reasoning from the first point says that the eigenvalues of $A^{^{T}}AA^{^{T}}A$ are just the ones of $A^{^{T}}A$ being squared. So $${\displaystyle \|A\|_{2}={\sqrt {\lambda _{\max }(A^{^{T}}A)}}}$$ $${\displaystyle \|A^{^{T}}A\|_{2}={\sqrt {\lambda _{\max }(A^{^{T}}AA^{^{T}}A)}}} = {\sqrt {\lambda _{\max }^2(A^{^{T}}A)}} = \vert\lambda _{\max }(A^{^{T}}A)\vert = \lambda_{\max}(A^TA)=\max(\lambda_{\min}^2(A),\lambda_{\max}^2(A)) \ge \|A\|_{2}^2 $$

Edit

The inequality becomes after the notion that $\lambda_{\max}(A^TA)=\max(\lambda_{\min}^2(A),\lambda_{\max}^2(A))$, because eigenvalues of $A$ could be negative.

This example with norm and random data

A = randn(2000,2000);
tic; 
n1 = norm(A) 
toc;

gives

n1 =  89.298
Elapsed time is 2.16777 seconds.

You can try eigs to find only one (the largest) eigenvalue of the symmetric matrix A'*A (or A*A' if it is smaller for A rectangular). It uses a Lanczos iteration method.

tic; 
B = A'*A; % symmetric positive-definite. B = A*A' if it is smaller
n2 = sqrt(eigs(B, 1)), 
toc

it outputs:

n2 =  89.298
Elapsed time is 0.311942 seconds.

If you don't want to use norm or eigs, and your matrix A has good properties (singular values properly separated), you can try to approximate it with a power iteration method:

tic; 
B = A'*A; % or B = A*A' if it is smaller
x = B(:,1); % example of starting point, x will have the largest eigenvector
x = x/norm(x); 
for i = 1:200 
   y = B*x; 
   y = y/norm(y);
   % norm(x - y); % <- residual, you can try to use it to stop iteration
   x = y; 
end; 
n3 = sqrt(mean(B*x./x)) % translate eigenvalue of B to singular value of A
toc

which for the same random matrix (not particularly good properties) gives a ~0.1% accurate solution:

n3 =  89.420
Elapsed time is 0.428032 seconds.