MAC 2103. Module 11 lnner Product Spaces II. Learning Objectives. Upon completing this module, you should be able to: Construct an orthonormal set of vectors from an orthogonal set of vectors . Find the coordinate vector with respect to a given orthonormal basis.
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Module 11
lnner Product Spaces II
Upon completing this module, you should be able to:
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The major topics in this module:
Orthogonal Bases, Gram-Schmidt Process,
Least Squares and Best Approximation
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We have learned from the previous module that two vectors u and v in an inner product space V are orthogonal to each other iff <u,v> = 0.
To obtain an orthonormal set, we will normalize each of the vectors in the orthogonal set.
How to normalize the vectors? This can be done by dividing each of them by their respective norm and making each of them a unit vector.
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Example 1: Find the orthonormal set of vectors from the following set of vectors: Let S ={v1, v2} where v1 = (5,0) and v2= (0,-3).
Step 1: Verify that the set of vectors are mutually orthogonal with respect to the Euclidean inner product on ℜ².
Step 2: Find the norm for both vectors.
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Step 3: Normalize the vectors in the orthogonal set.
Step 4: Verify that the set S is orthonormal by showing that
and
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Orthonormal Set: An orthogonal set in which each vector is a unit vector.
Orthonormal basis: A basis consisting of orthonormal vectors in an inner product space. Example: The standard basis for ℜⁿ.
Orthogonal basis: A basis consisting of orthogonal vectors in an inner product space.
Note that if S is an orthogonal set, then S is a linearly independent set.
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If S ={v1, v2 , … , vn} is an orthogonal basis of W, then for any w∈ W,
where
are called the Fourier coefficients.
So the coordinate vector of w,
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If S ={q1, q2 , … , qn} is an orthonormal basis of W, then for any w∈ W,
where are called the Fourier coefficients.
So the coordinate vector of w,
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Example 2: Compute the coefficients and determine the coordinate vectors in Example 1 for u = (10,3).
From Example 1, we have v1 = (5,0), v2= (0,-3) and
In this case, the coefficients are:
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So the coordinate vector of u,
We can see that a nice advantage of working with an orthogonal basis is that the coefficients in any basis representation for a vector are immediately known; they are called Fourier coefficients.
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Example 3: Find the coordinates of w = (2,3) relative to the orthonormal basis for ℜ², B = {v1, v2}, where
Since B is orthonormal, we have
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Let S = {u1, u2, …, um} with nonzero ui∈ℜⁿ for i = 1, 2, … , m. S does not have to be a linearly independent set. It might be that A = [u1u2 … um] is an n x m matrix and the source of S.
The Gram-Schmidt Algorithm: S = {u1, u2, …, um}
Let
For k = 2, 3, … , m, let
If we discard it since it is linearly dependent.
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We then have r ≤ m orthogonal and linearly independent vectors in B = {v1, v2, …, vr}.
span(S) = span(B) = W, a r dimensional subspace of ℜⁿ and B is an orthogonal basis for W. W = col(A) if A = [u1u2 … um].
An orthogonal basis for W is {q1, q2, …, qr} where
for i =1, 2, … , r.
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Let ℜⁿ be the usual Euclidean inner product space of dimension n. Let {u1, u2, …, un} be a nonstandard basis in ℜⁿ.
Step 1: Use the Gram-Schmidt method to construct an orthogonal basis {v1, v2, …, vn} from the basis vectors {u1, u2, …, un}.
Step 2: Normalize the orthogonal basis vectors to obtain the orthonormal basis {q1, q2,…, qn}.
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Example 4:
The set is a nonstandard basis in ℜ³.
Step 1:
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form an orthogonal basis for ℜ³.
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Step 2: Normalize the orthogonal basis vectors to obtain the orthonormal basis B = {q1, q2, q3}.
The norms of these vectors are:
So an orthonormal basis is B where
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The linear system Ax = b always has the associated normal system ATAx = ATb which is consistent and has one or more solutions. Any solution of this system is a least squares solution of Ax = b. Moreover, the orthogonal projection of b on W = col(A) is where x is a least squares solution.
Example 5: Find the least squares solution of the linear system Ax = b given by
Observe that A has two linearly independent column vectors, so ATA is invertible, and there is a unique solution to ATAx = ATb, which will be our least squares solution to Ax = b.
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We have
So the normal system ATAx = ATb in this case is
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By Gauss Elimination, the row-echelon form of
is
The solution to the normal system ATAx = ATb in this case is
is our unique least square solution to Ax = b.
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is the orthogonal projection of b on W = col(A). The is the Best Approximation to b in W since the distance between b and is the minimum for all vectors in W,
Example 6: From the previous example, the orthogonal projection of b on W = col(A) is
Thus, we have obtained the best approximation to b in W.
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We have learned to :
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Some of these slides have been adapted/modified in part/whole from the following textbook:
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