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Chapter 6: Problem 12

In Exercises 11 and \(12,\) find the closest point to \(y\) in the subspace \(W\) spanned by \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2} .\) $$ \mathbf{y}=\left[\begin{array}{r}{3} \\ {-1} \\ {1} \\\ {13}\end{array}\right], \mathbf{v}_{1}=\left[\begin{array}{r}{1} \\ {-2} \\\ {-1} \\ {2}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{-4} \\\ {1} \\ {0} \\ {3}\end{array}\right] $$

Short Answer

Expert verified
The closest point is \( \begin{bmatrix} -6 \\ 5 \\ -2 \\ -1 \end{bmatrix} \).

Step by step solution

01

Understand the Problem

We need to find the orthogonal projection of the vector \( \mathbf{y} \) onto the subspace \( W \) spanned by the vectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \). This projection will be the closest point to \( \mathbf{y} \) in \( W \).
02

Verify Linear Independence

First, we check that \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are linearly independent. Compute the determinant of their coefficient matrix or check that neither is a scalar multiple of the other. Here they are linearly independent as no scalar \( c \) satisfies \( \mathbf{v}_1 = c \mathbf{v}_2 \ \).
03

Find the Gram Matrix

Construct the Gram matrix \( G \) using \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \): \[ G = \begin{bmatrix} \mathbf{v}_1 \cdot \mathbf{v}_1 & \mathbf{v}_1 \cdot \mathbf{v}_2 \ \mathbf{v}_2 \cdot \mathbf{v}_1 & \mathbf{v}_2 \cdot \mathbf{v}_2 \end{bmatrix} \] Calculate: \( \mathbf{v}_1 \cdot \mathbf{v}_1 = 10 \), \( \mathbf{v}_1 \cdot \mathbf{v}_2 = -13 \), \( \mathbf{v}_2 \cdot \mathbf{v}_2 = 26 \). So, \( G = \begin{bmatrix} 10 & -13 \ -13 & 26 \end{bmatrix} \).
04

Calculate Projection Coefficients

Find the vector \( \mathbf{b} \) as follows: \[ \mathbf{b} = \begin{bmatrix} \mathbf{v}_1 \cdot \mathbf{y} \ \mathbf{v}_2 \cdot \mathbf{y} \end{bmatrix} = \begin{bmatrix} -18 \ 48 \end{bmatrix} \] Solve the equation \( G \mathbf{c} = \mathbf{b} \) for \( \mathbf{c} \), where \( \mathbf{c} = \begin{bmatrix} c_1 \ c_2 \end{bmatrix} \).
05

Solve Linear System for c

By solving the system: \[ 10c_1 - 13c_2 = -18 \ -13c_1 + 26c_2 = 48 \] Using techniques such as substitution or elimination, we find \ \( c_1 = -2, \ c_2 = 1 \) .
06

Compute the Projection

The projection \( \mathbf{p} \) is given by: \[ \mathbf{p} = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 = -2 \begin{bmatrix} 1 \ -2 \ -1 \ 2 \end{bmatrix} + 1 \begin{bmatrix} -4 \ 1 \ 0 \ 3 \end{bmatrix} \] Calculate this to get \[ \mathbf{p} = \begin{bmatrix} -6 \ 5 \ -2 \ -1 \end{bmatrix} \].
07

Verify Result

To ensure accuracy, verify \( \mathbf{p} \) is in \( W \) by checking it equals \( c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 \). Calculations show it matches, confirming \( \mathbf{p} = \begin{bmatrix} -6 \ 5 \ -2 \ -1 \end{bmatrix} \).

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Key Concepts

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

Linear Independence
Linear independence is an important concept in determining the uniqueness of a set of vectors. When vectors are linearly independent, it essentially means that no vector in the set can be expressed as a linear combination of the others.
In our problem, vectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) need to be checked for linear independence to ensure the subspace they span doesn't dimish its dimension.
  • Calculate the determinant of their coefficient matrix.
  • Check that neither vector is a scalar multiple of the other.
For \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \), they are not scalar multiples of each other, confirming their independence. This property is crucial, as only then do we know the subspace has the right form and that the Gram matrix used later will be non-singular.
Gram Matrix
The Gram matrix is derived from a set of vectors and contains invaluable information about their geometry, such as angles and lengths. It helps in computing projections. For our vectors \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \), the elements of the Gram matrix \( G \) are constructed by computing the dot products of each vector with itself and with each other.
The Gram matrix \( G \) for this exercise is as follows:
  • \( \mathbf{v}_1 \cdot \mathbf{v}_1 = 10 \)
  • \( \mathbf{v}_1 \cdot \mathbf{v}_2 = -13 \)
  • \( \mathbf{v}_2 \cdot \mathbf{v}_2 = 26 \)
So, the matrix \( G \) becomes:\[ G = \begin{bmatrix} 10 & -13 \ -13 & 26 \end{bmatrix} \]This matrix helps in finding the coefficients for the projection of a vector onto the subspace defined by \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \).
Projection Coefficients
Projection coefficients tell us how much each basis vector contributes to the projection of a vector onto a subspace. To determine these coefficients, we first calculate the vector \( \mathbf{b} \) formed by dot products of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) with the vector \( \mathbf{y} \).
  • \( \mathbf{v}_1 \cdot \mathbf{y} = -18 \)
  • \( \mathbf{v}_2 \cdot \mathbf{y} = 48 \)
Thus, \( \mathbf{b} = \begin{bmatrix} -18 \ 48 \end{bmatrix} \).
We can then find \( \mathbf{c} \), the projection coefficients, by solving \( G \mathbf{c} = \mathbf{b} \). Essentially, \( \mathbf{c} \) gives the linear combination of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) that represents the projection.
Linear System Solving
Solving the linear system \( G \mathbf{c} = \mathbf{b} \) is key to finding the projection coefficients. This involves using algebraic methods such as substitution or elimination to solve for \( \mathbf{c} \).
Our system is:\[\begin{align*}10c_1 - 13c_2 &= -18 \ -13c_1 + 26c_2 &= 48\end{align*}\]Using algebraic techniques, we find:
  • \( c_1 = -2 \)
  • \( c_2 = 1 \)
With these coefficients, we create the linear combination of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \), leading to the final projection vector \( \mathbf{p} \). This projection is the closest point in the subspace to the vector \( \mathbf{y} \).

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Most popular questions from this chapter

Find the equation \(y=\beta_{0}+\beta_{1} x\) of the least-squares line that best fits the given data points. \((1,0),(2,1),(4,2),(5,3)\)

Let \(\mathbf{y}=\left[\begin{array}{r}{5} \\ {-9} \\ {5}\end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{r}{-3} \\ {-5} \\ {1}\end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{r}{-3} \\ {2} \\ {1}\end{array}\right] .\) Find the distance from y to the plane in \(\mathbb{R}^{3}\) spanned by \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\)

Let \(\mathbf{u}_{1}, \ldots, \mathbf{u}_{p}\) be an orthogonal basis for a subspace \(W\) of \(\mathbb{R}^{n},\) and let \(T : \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) be defined by \(T(\mathbf{x})=\operatorname{proj}_{W} \mathbf{x}\) Show that \(T\) is a linear transformation.

[M] Use the Gram-Schmidt process as in Example 2 to produce an orthogonal basis for the column space of $$ A=\left[\begin{array}{rrrr}{-10} & {13} & {7} & {-11} \\ {2} & {1} & {-5} & {3} \\ {-6} & {3} & {13} & {-3} \\ {16} & {-16} & {-2} & {5} \\ {2} & {1} & {-5} & {-7}\end{array}\right] $$

In Excrcises \(1-6,\) the given set is a basis for a subspace \(W .\) Use the Gram-Schmidt process to produce an orthogonal basis for \(W\) . $$ \left[\begin{array}{r}{1} \\ {-4} \\ {0} \\\ {1}\end{array}\right],\left[\begin{array}{r}{7} \\ {-7} \\ {-4} \\\ {1}\end{array}\right] $$
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