4.6: Other Norms

Sometimes we find it useful to use a different definition of distance, corresponding to an alternate norm for vectors. For example, consider the l-norm defined as

\[||x||_1=(|x_1|+|x_2|+⋯+|x_n|) \nonumber \]

where \(|x_i|\) is the magnitude of component \(x_i\). There is also the sup-norm, the “supremum” or maximum of the components \(x_1,...,x_n\) :

\[||x||_{\mathrm{sup}}=\mathrm{max}(|x_1|,|x_2|,...,|x_n|) \nonumber \]

The following examples illustrate what the Euclidean norm, the l-norm, and the sup-norm look like for typical vectors

Example \(\PageIndex{1}\)

Consider the vector \(x=\begin{bmatrix} −3\\1\\2 \end{bmatrix}\). Then

1. \(||x||=[(−3)^2+(1)^2+(2)^2]^{1/2}=(14)^{1/2}\)
2. \(||x||_1=(|−3|+|1|+|2|)=6\)
3. \(||x||_{\mathrm{sup}}=\mathrm{max}(|−3|,|1|,|2|)=3\)

Example \(\PageIndex{2}\)

Figure \(\PageIndex{1}\) shows the locus of two-component vectors \(x=\begin{bmatrix}x_1\\x_2\end{bmatrix}\) with the property that \(||x||=1\),\(||x||_1=1\), or \(||x||_{\mathrm{sup}}=1\).

Figure \(\PageIndex{1}\): Locus of Two-Dimensional Vectors Whose Various Norms Are 1 (Copyright; author via source)

The next example shows how the l-norm is an important part of city life.

Example \(\PageIndex{3}\)

The city of Metroville was laid out by mathematicians as shown in Figure \(\PageIndex{2}\). A person at the intersection of Avenue 0 and Street −2 (point A) is clearly two blocks from the center of town (point C). This is consistent with both the Euclidean norm

\[||A||=\sqrt{0^2+(−2)^2}=\sqrt{4}=2 \nonumber \]

and the l-norm

\[||A||_1=(|0|+|−2|)=2 \nonumber \]

But how far from the center of town is point B at the intersection of Avenue-2 and Street 1? According to the Euclidean norm, the distance is

\[||B||=\sqrt{(−2)^2+(1)^2}=\sqrt{5} \nonumber \]

Figure \(\PageIndex{2}\): Metroville, U.S.A. (Copyright; author via source)

While it is true that point B is \(\sqrt{5}\) blocks from C, it is also clear that the trip would be three blocks by any of the three shortest routes on roads. The appropriate norm is the l-norm:

\[∣1^B||_1=(|−2|+|1|)=3 \nonumber \]

Even more generally, we can define a norm for each value of p from 1 to infinity. The so-called p-norm is

\[|Ix||_p=(|x_1∣∣^p+|x_2∣∣^p+⋯+|x_n|^p)^{1/p} \nonumber \]

Exercise \(\PageIndex{1}\)

Show that the Euclidean norm is the same as the p-norm with p=2 and that the 1-norm is the p-norm with p=1. (It can also be shown that the sup-norm is like a p-norm with p=∞.)

DEMO 4.1 (MATLAB)

From the command level of MATLAB, type the following lines:

>> x = [1;3;-2;4]
>> y = [0;1;2;-0.5]
>> x - y

Check to see whether the answer agrees with the definition of vector subtraction. Now type

>> a = -1.5
>> a * x

Check the answer to see whether it agrees with the definition of scalar multiplication. Now type

>> x' * y

This is how MATLAB does the inner product. Check the result. Type

>> norm(y)
>> sqrt(y' * y)

>> norm(y,1)
>> norm(y' * y)

Now type your own MATLAB expression to find the cosine of the angle between vectors x and y. Put the result in variable t. Then find the angle θ by typing

>> theta = acos(t)

The angle θ is in radians. You may convert it to degrees if you wish by multiplying it by \(180/π\):

>> theta = theta * (180/pi)