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Chapter 7: Problem 38

Find the standard matrix of the orthogonal projection onto the subspace \(W\). Then use this matrix to find the orthogonal projection of \(\mathbf{v}\) onto \(W\). $$W=\operatorname{span}\left(\left[\begin{array}{r} 1 \\ -2 \end{array}\right]\right), \mathbf{v}=\left[\begin{array}{l} 1 \\ 1 \end{array}\right]$$

Short Answer

Expert verified
The orthogonal projection of \( \mathbf{v} \) is \( \begin{bmatrix} -0.2 \\ 0.4 \end{bmatrix} \).

Step by step solution

01

Understand the Problem

We are asked to find the standard matrix of the orthogonal projection onto the subspace spanned by a single vector and then use it to find the orthogonal projection of another vector onto this subspace.
02

Write Down the Basis of Subspace W

Since \(W\) is the span of the vector \(\begin{bmatrix} 1 \ -2 \end{bmatrix}\), it means any vector in \(W\) can be written as \( k \begin{bmatrix} 1 \ -2 \end{bmatrix}\) where \(k\) is a scalar.
03

Use the Formula for Projection Matrix

The formula for the projection matrix \(P\) onto a subspace spanned by a single vector \(\begin{bmatrix} a \ b \end{bmatrix}\) is \P = \frac{1}{{a^2 + b^2}} \begin{bmatrix} a^2 & ab \ ab & b^2 \end{bmatrix}\. Our vector is \(\begin{bmatrix} 1 \ -2 \end{bmatrix}\), so we plug it into the formula.
04

Calculate the Projection Matrix

Substitute \(a=1\) and \(b=-2\) into the formula: \P = \frac{1}{{1^2 + (-2)^2}} \begin{bmatrix} 1^2 & 1(-2) \ 1(-2) & (-2)^2 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 1 & -2 \ -2 & 4 \end{bmatrix}.\So the projection matrix \(P\) is \(\begin{bmatrix} 0.2 & -0.4 \ -0.4 & 0.8 \end{bmatrix}\).
05

Use Projection Matrix to Find Orthogonal Projection

To find the orthogonal projection of \(\mathbf{v} = \begin{bmatrix} 1 \ 1 \end{bmatrix}\) onto \(W\), we calculate P \mathbf{v} = \begin{bmatrix} 0.2 & -0.4 \ -0.4 & 0.8 \end{bmatrix} \begin{bmatrix} 1 \ 1 \end{bmatrix}.\This results in calculating elements as: \(0.2 \times 1 + (-0.4) \times 1 = -0.2\) and \(-0.4 \times 1 + 0.8 \times 1 = 0.4\).\Thus, the projection is \(\begin{bmatrix} -0.2 \ 0.4 \end{bmatrix}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Matrix
The standard matrix of a linear transformation, like the orthogonal projection, gives us a clear way to represent and apply the transformation using matrix multiplication. For an orthogonal projection onto a subspace, this matrix depends on the vector(s) that span the subspace.
The purpose of a standard matrix is to allow us to transform vectors from the general space into the subspace in a straightforward manner.
  • It provides a structured way to compute projections for any vector.
  • For the problem at hand, we derived the projection matrix using the formulas specific to the span of a single vector.
  • This matrix is particularly useful as it enables us to find the orthogonal projections quickly and efficiently.
Subspace
In the context of linear algebra, a subspace is a set of vectors that is closed under addition and scalar multiplication. It is like a "plane" within the larger vector space where all vectors reside.
Recognizing the subspace is essential in finding the projections or reductions in the complexity of a problem.
  • In our exercise, the subspace is defined by the span of \[\begin{bmatrix} 1 \ -2 \end{bmatrix}\].
  • Understanding this basis allows us to delve into constructing meaningful components like projection matrices.
A subspace must pass through the origin and possess linearity, ensuring every linear combination of its vectors stays within the subspace.
Projection Matrix
A projection matrix is a tool used to project vectors onto a subspace. It essentially "drops" a given vector onto the closest point within the specified subspace, reducing dimensions while preserving the structure.
For a subspace spanned by a single vector \[\begin{bmatrix} a \ b \end{bmatrix}\], the projection matrix can be constructed simply.
  • It involves utilizing the components of the vector in the formula for the matrix: \[P = \frac{1}{{a^2 + b^2}} \begin{bmatrix} a^2 & ab \ ab & b^2 \end{bmatrix}\]
  • The matrix operates by "aligning" vectors onto the subspace, effectively filtering out components not in line with the span.
  • This process ensures optimal proximity between a vector and its projection.
Span
The concept of span involves all the vectors that can be formed using linear combinations of a given set of vectors. Particularly important in defining a subspace, the span sets the foundation of what a subspace contains.
When vectors are said to "span" a space, they essentially form the "building blocks" of that space.
  • For \[W=\text{span}\left(\begin{bmatrix} 1 \ -2 \end{bmatrix}\right)\], each vector in the subspace can be represented as a scalar multiple of this basis vector.
  • This allows us to rigorously define what's in the subspace and subsequently project onto it.
  • Span is useful for determining minimal representations of subspaces, especially in orthogonal projections.
Basis of a Subspace
A basis of a subspace is a minimal set of vectors that span the entire subspace. These vectors are independent and provide a means to express every other vector within the subspace through linear combinations.
Understanding the basis helps you grasp the structure and dimension of the subspace.
  • In the exercise, the vector \([1, -2]^T\) essentially forms the basis for this one-dimensional subspace.
  • Establishing the basis is crucial in forming projection matrices, as it determines the "direction" vectors are projected onto.
  • Each unique basis ensures no redundancy, providing the simplest yet complete description of the subspace.
Identifying a basis helps in efficiently managing transformations and operations over the subspace, a vital part of solving linear algebra problems.

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Most popular questions from this chapter

Find \(\operatorname{cond}_{1}(A)\) and \(\operatorname{cond}_{\infty}(A)\). State whether the given matrix is ill-conditioned. $$A=\left[\begin{array}{rr}1 & -2 \\\\-3 & 6\end{array}\right]$$

Show that if \(A\) is an invertible matrix, then cond \((A) \geq 1\) with respect to any matrix norm.

Compute the pseudoinverse of \(A\). $$A=\left[\begin{array}{ll} 1 & 3 \\ 3 & 1 \\ 2 & 2 \end{array}\right]$$

Find a polar decomposition of the matrices $$A=\left[\begin{array}{rrr} 4 & 2 & -3 \\ -2 & 2 & 6 \\ 4 & -1 & 6 \end{array}\right]$$

Let \(Q\) be an orthogonal matrix such that \(Q A\) makes sense. Show that \(A\) and \(Q A\) have the same singular values.
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