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4.6: Other Norms

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    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 one is a diagram with a vertical axis labeled x_2 and horizontal axis labeled x_1. A large square centered at the origin with its diagonals on the axes is drawn, and around this is a circle centered at the origin and of the same diameter as the diagonals of the square. In addition, a square centered at the origin with the same width as the diagonal of the first square is the outermost shape of the diagram. The side of the smaller square that lies in the fourth quadrant is labeled locus of x such that  ||x||_1 = 1. The section of the circle that lies in the first quadrant is labeled locus of x such that ||x||_2 = 1. The section of the larger square that lies in the second quadrant is labeled locus of x such that  ||x||_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 two is a diagram of metroville. North points upward. There are five streets that travel north-south, labeled from left to right, street -2, street -1, street 0, street 1, and street 2. There are also five streets that run east-west labeled from top to bottom, avenue 2, avenue 1, avenue 0, avenue -1, avenue -2. At the intersection of avenue 0 and street -2 is point A. At the intersection of street 0 and avenue 0 is point C. At the intersection of avenue -2 and street 1 is point B.
    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)

    This page titled 4.6: Other Norms is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Louis Scharf (OpenStax CNX) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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