Suppose V=R2V=R2,
A
=
span
2
-
1
,
and
x
=
2
1
.
A
=
span
2
-
1
,
and
x
=
2
1
.
(1)We will want to find x^∈Ax^∈A that minimizes x-x^px-x^p. Since x^∈Ax^∈A, we can write
x
^
=
2
α
-
α
x
^
=
2
α
-
α
(2)and thus
e
=
x
-
x
^
=
2
-
2
α
1
+
α
.
e
=
x
-
x
^
=
2
-
2
α
1
+
α
.
(3)While we can solve for α∈Rα∈R to minimize epep
directly in some cases, a geometric interpretation is also useful. In each
case, on can imagine growing an ℓpℓp ball centered on xx until the ball
intersects with AA. This will be the point x^∈Ax^∈A. that is
closest to xx in the ℓpℓp norm. We first illustrate this for the ℓ2ℓ2 norm below:
In order to calculate x^x^ we can apply the orthogonality principle. Since 〈e,[21]T〉=0〈e,[21]T〉=0 we obtain a solution defined by α=35α=35.
We now observe that in the case of the ℓ∞ℓ∞ norm the picture changes somewhat. The closest point in ℓ∞ℓ∞ is illustrated below:
Note that the error is no longer orthogonal to the subspace AA. In this case we can still calculate x^x^ from the observation that the two terms in the error should be equal, which yields α=13α=13.
The situation is even more different for the case of the ℓ1ℓ1 norm, which is illustrated below:
We now observe that x^x^ corresponds to α=1α=1. Note that in this case the error term is [02]T[02]T. This punctuates a general trend: for large values of pp, the ℓpℓp norm tends to spread error evenly across all terms, while for small values of pp the error is more highly concentrated.