In [1]:

```
import numpy as np
import numpy.linalg as la
import matplotlib.pyplot as pt
```

Here's a matrix of which we're trying to compute the norm:

In [14]:

```
n = 2
A = np.random.randn(n, n)
A
```

Out[14]:

Recall:

$$||A||=\max_{\|x\|=1} \|Ax\|,$$where the vector norm must be specified, and the value of the matrix norm $\|A\|$ depends on the choice of vector norm.

For instance, for the $p$-norms, we often write:

$$||A||_2=\max_{\|x\|=1} \|Ax\|_2,$$and similarly for different values of $p$.

We can approximate this by just producing very many random vectors and evaluating the formula:

In [3]:

```
xs = np.random.randn(n, 1000)
```

First, we need to bring all those vectors to have norm 1. First, compute the norms:

In [4]:

```
p = 2
norm_xs = np.sum(np.abs(xs)**p, axis=0)**(1/p)
norm_xs.shape
```

Out[4]:

Then, divide by the norms and assign to `normalized_xs`

:

Then check the norm of a randomly chosen vector.

In [16]:

```
normalized_xs = xs/norm_xs
la.norm(normalized_xs[:, 316], p)
```

Out[16]:

Let's take a look:

In [17]:

```
pt.plot(normalized_xs[0], normalized_xs[1], "o")
pt.gca().set_aspect("equal")
```

Now apply $A$ to these normalized vectors:

In [7]:

```
A_nxs = A.dot(normalized_xs)
```

Let's take a look again:

In [8]:

```
pt.plot(normalized_xs[0], normalized_xs[1], "o", label="x")
pt.plot(A_nxs[0], A_nxs[1], "o", label="Ax")
pt.legend()
pt.gca().set_aspect("equal")
```

Next, compute norms of the $Ax$ vectors:

In [9]:

```
norm_Axs = np.sum(np.abs(A_nxs)**p, axis=0)**(1/p)
norm_Axs.shape
```

Out[9]:

What's the biggest one?

In [10]:

```
np.max(norm_Axs)
```

Out[10]:

Compare that with what `numpy`

thinks the matrix norm is:

In [11]:

```
la.norm(A, p)
```

Out[11]:

In [ ]:

```
```