Find the orthogonal projection of 𝐯=(1,2,-1,2)^T onto the span of (1,-1,2,5)^T and (2,1,0,-1)^T

Find the orthogonal projection of 𝐯=(1,2,-1,2)^T onto the...

Key Concepts

Subspace and Span

The span of a set of vectors constitutes a subspace composed of all possible linear combinations of those vectors. Understanding the concept of a span is critical to the projection process, as it defines the target space onto which the original vector is projected.

Solving Projection Systems

Finding the orthogonal projection typically involves setting up a system of linear equations based on the requirement that the residual vector (the difference between the original vector and its projection) must be orthogonal to every vector in the subspace. This system incorporates the properties of the inner product, and solving it yields the appropriate coefficients for the projection.

Orthogonal Projection

An orthogonal projection of a vector onto a subspace is the unique point in that subspace that minimizes the distance to the original vector. This is achieved by ensuring that the difference between the vector and its projection is orthogonal to the subspace with respect to the given inner product, meaning no component of the error lies within the subspace.

Weighted Inner Product

A weighted inner product is a generalization of the standard inner product where each component of the vectors is multiplied by a predefined weight. This definition modifies measurements of length, angle, and orthogonality in the space, and is essential for accurately computing projections or other operations in settings where uniform scaling does not apply.

Transcript

00:02 We're given a subspace w, which is the set of all polynomials of degree at most 2, p2, with an interproduct of f and g, is the integral from 0 to 1 of f of t times g of t.

00:32 And we're asked to find the projection of the function f of t equals t cubed, onto the space w.

00:46 As a hint, we're told, to use the orthogonal polynomials, 1, 2t minus 1, and 6t squared minus 60 plus 1.

01:07 So we found out these were orthogonal in exercise 722.

01:18 Now, because these are orthogonal, we can just calculate the fourier code.

01:22 To find a projection of f onto this space.

01:29 These are an orthogonal basis for w.

01:40 Okay, so the 4a coefficient c1 is the inner product of t cubed with 1 over the inner product of 1 with itself.

02:00 So this is the integral from 0 to 1 of t cubed d t over the integral from 0 to 1 of 1 d t.

02:11 And this is one -fourth over one, which is one -fourth.

02:22 The second fourier coefficient c2 is the interproduct of t -cubed with 2t -minus 1 over the inner -product of 2t -1 with itself.

02:37 This is the integral from 0 to 1 of well t -cubed times 2t minus 1 d t.

02:46 Over the integral from 0 to 1 of 2t minus 1 squared d t.

02:54 This is equal to the integral from 0 to 1 of 2 t to 4th minus t cubed t over the integral from 0 to 1 of 4 t squared plus minus 4 t plus 1 d2...

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