Solved on Dec 03, 2023

Find the closest vector $\mathbf{x}$ in the subspace $W$ spanned by $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ to the given vector $\mathbf{y}$ , where $\mathbf{y}=\left[\begin{array}{r}1 \\ -1 \\ 1 \\ 19\end{array}\right]$ , $\mathbf{v}_{1}=\left[\begin{array}{r}1 \\ -1 \\ -1 \\ 1\end{array}\right]$ , and $\mathbf{v}_{2}=\left[\begin{array}{r}-1 \\ 1 \\ 0 \\ 2\end{array}\right]$ .

STEP 1

Assumptions
1. The vector $\mathbf{y}$ is given by $\left[\begin{array}{r} 1 \\ -1 \\ 1 \\ 19 \end{array}\right]$ .
2. The subspace $W$ is spanned by vectors $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ , where $\mathbf{v}_{1}=\left[\begin{array}{r} 1 \\ -1 \\ -1 \\ 1 \end{array}\right]$ and $\mathbf{v}_{2}=\left[\begin{array}{r} -1 \\ 1 \\ 0 \\ 2 \end{array}\right]$ .
3. We are looking for the closest point to $\mathbf{y}$ in the subspace $W$ .

STEP 2

The closest point to $\mathbf{y}$ in the subspace $W$ is the orthogonal projection of $\mathbf{y}$ onto $W$ . This point, say $\mathbf{p}$ , can be expressed as a linear combination of the vectors spanning $W$ , i.e., $\mathbf{p} = c_1\mathbf{v}_{1} + c_2\mathbf{v}_{2}$ for some scalars $c_1$ and $c_2$ .

STEP 3

The orthogonal projection of $\mathbf{y}$ onto $W$ is given by the formula:
$\mathbf{p} = \frac{\mathbf{y} \cdot \mathbf{v}_{1}}{\mathbf{v}_{1} \cdot \mathbf{v}_{1}}\mathbf{v}_{1} + \frac{\mathbf{y} \cdot \mathbf{v}_{2}}{\mathbf{v}_{2} \cdot \mathbf{v}_{2}}\mathbf{v}_{2}$

STEP 4

First, calculate the dot products $\mathbf{y} \cdot \mathbf{v}_{1}$ , $\mathbf{y} \cdot \mathbf{v}_{2}$ , $\mathbf{v}_{1} \cdot \mathbf{v}_{1}$ , and $\mathbf{v}_{2} \cdot \mathbf{v}_{2}$ .

STEP 5

Calculate $\mathbf{y} \cdot \mathbf{v}_{1}$ :
$\mathbf{y} \cdot \mathbf{v}_{1} = (1)(1) + (-1)(-1) + (1)(-1) + (19)(1) = 20$

STEP 6

Calculate $\mathbf{y} \cdot \mathbf{v}_{2}$ :
$\mathbf{y} \cdot \mathbf{v}_{2} = (1)(-1) + (-1)(1) + (1)(0) + (19)(2) = 35$

STEP 7

Calculate $\mathbf{v}_{1} \cdot \mathbf{v}_{1}$ :
$\mathbf{v}_{1} \cdot \mathbf{v}_{1} = (1)(1) + (-1)(-1) + (-1)(-1) + (1)(1) = 4$

STEP 8

Calculate $\mathbf{v}_{2} \cdot \mathbf{v}_{2}$ :
$\mathbf{v}_{2} \cdot \mathbf{v}_{2} = (-1)(-1) + (1)(1) + (0)(0) + (2)(2) = 6$

STEP 9

Substitute these values into the formula for $\mathbf{p}$ :
$\mathbf{p} = \frac{20}{4}\mathbf{v}_{1} + \frac{35}{6}\mathbf{v}_{2}$

STEP 10

Simplify to get $\mathbf{p}$ :
$\mathbf{p} = 5\mathbf{v}_{1} + \frac{35}{6}\mathbf{v}_{2}$

STEP 11

Substitute the values of $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ into the equation for $\mathbf{p}$ :
$\mathbf{p} = 5\left[\begin{array}{r} 1 \\ -1 \\ -1 \\ 1 \end{array}\right] + \frac{35}{6}\left[\begin{array}{r} -1 \\ 1 \\ 0 \\ 2 \end{array}\right]$

STEP 12

Perform the scalar multiplication:
$\mathbf{p} = \left[\begin{array}{r} 5 \\ -5 \\ -5 \\ 5 \end{array}\right] + \left[\begin{array}{r} -\frac{35}{6} \\ \frac{35}{6} \\ 0 \\ \frac{70}{3} \end{array}\right]$

STEP 13

Add the vectors to get $\mathbf{p}$ :
$\mathbf{p} = \left[\begin{array}{r} 5 - \frac{35}{6} \\ -5 + \frac{35}{6} \\ -5 \\ 5 + \frac{70}{3} \end{array}\right] = \left[\begin{array}{r} -\frac{5}{6} \\ \frac{5}{6} \\ -5 \\ \frac{85}{3} \end{array}\right]$
So, the closest point to $\mathbf{y}$ in the subspace $W$ is the vector $\mathbf{p} = \left[\begin{array}{r} -\frac{5}{6} \\ \frac{5}{6} \\ -5 \\ \frac{85}{3} \end{array}\right]$ .

Was this helpful?

STEP 1

STEP 2

STEP 3

The orthogonal projection of $\mathbf{y}$ onto $W$ is given by the formula:
$\mathbf{p} = \frac{\mathbf{y} \cdot \mathbf{v}_{1}}{\mathbf{v}_{1} \cdot \mathbf{v}_{1}}\mathbf{v}_{1} + \frac{\mathbf{y} \cdot \mathbf{v}_{2}}{\mathbf{v}_{2} \cdot \mathbf{v}_{2}}\mathbf{v}_{2}$

STEP 4

First, calculate the dot products $\mathbf{y} \cdot \mathbf{v}_{1}$ , $\mathbf{y} \cdot \mathbf{v}_{2}$ , $\mathbf{v}_{1} \cdot \mathbf{v}_{1}$ , and $\mathbf{v}_{2} \cdot \mathbf{v}_{2}$ .

STEP 5

Calculate $\mathbf{y} \cdot \mathbf{v}_{1}$ :
$\mathbf{y} \cdot \mathbf{v}_{1} = (1)(1) + (-1)(-1) + (1)(-1) + (19)(1) = 20$

STEP 6

Calculate $\mathbf{y} \cdot \mathbf{v}_{2}$ :
$\mathbf{y} \cdot \mathbf{v}_{2} = (1)(-1) + (-1)(1) + (1)(0) + (19)(2) = 35$

STEP 7

Calculate $\mathbf{v}_{1} \cdot \mathbf{v}_{1}$ :
$\mathbf{v}_{1} \cdot \mathbf{v}_{1} = (1)(1) + (-1)(-1) + (-1)(-1) + (1)(1) = 4$

STEP 8

Calculate $\mathbf{v}_{2} \cdot \mathbf{v}_{2}$ :
$\mathbf{v}_{2} \cdot \mathbf{v}_{2} = (-1)(-1) + (1)(1) + (0)(0) + (2)(2) = 6$

STEP 9

Substitute these values into the formula for $\mathbf{p}$ :
$\mathbf{p} = \frac{20}{4}\mathbf{v}_{1} + \frac{35}{6}\mathbf{v}_{2}$

STEP 10

Simplify to get $\mathbf{p}$ :
$\mathbf{p} = 5\mathbf{v}_{1} + \frac{35}{6}\mathbf{v}_{2}$

STEP 11

Substitute the values of $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ into the equation for $\mathbf{p}$ :
$\mathbf{p} = 5\left[\begin{array}{r} 1 \\ -1 \\ -1 \\ 1 \end{array}\right] + \frac{35}{6}\left[\begin{array}{r} -1 \\ 1 \\ 0 \\ 2 \end{array}\right]$

STEP 12

Perform the scalar multiplication:
$\mathbf{p} = \left[\begin{array}{r} 5 \\ -5 \\ -5 \\ 5 \end{array}\right] + \left[\begin{array}{r} -\frac{35}{6} \\ \frac{35}{6} \\ 0 \\ \frac{70}{3} \end{array}\right]$

STEP 13

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

STEP 1

STEP 2

STEP 3

STEP 4

First, calculate the dot products y⋅v1\mathbf{y} \cdot \mathbf{v}_{1}y⋅v1​, y⋅v2\mathbf{y} \cdot \mathbf{v}_{2}y⋅v2​, v1⋅v1\mathbf{v}_{1} \cdot \mathbf{v}_{1}v1​⋅v1​, and v2⋅v2\mathbf{v}_{2} \cdot \mathbf{v}_{2}v2​⋅v2​.

STEP 5

Calculate y⋅v1\mathbf{y} \cdot \mathbf{v}_{1}y⋅v1​:y⋅v1=(1)(1)+(−1)(−1)+(1)(−1)+(19)(1)=20\mathbf{y} \cdot \mathbf{v}_{1} = (1)(1) + (-1)(-1) + (1)(-1) + (19)(1) = 20y⋅v1​=(1)(1)+(−1)(−1)+(1)(−1)+(19)(1)=20

STEP 6

Calculate y⋅v2\mathbf{y} \cdot \mathbf{v}_{2}y⋅v2​:y⋅v2=(1)(−1)+(−1)(1)+(1)(0)+(19)(2)=35\mathbf{y} \cdot \mathbf{v}_{2} = (1)(-1) + (-1)(1) + (1)(0) + (19)(2) = 35y⋅v2​=(1)(−1)+(−1)(1)+(1)(0)+(19)(2)=35

STEP 7

Calculate v1⋅v1\mathbf{v}_{1} \cdot \mathbf{v}_{1}v1​⋅v1​:v1⋅v1=(1)(1)+(−1)(−1)+(−1)(−1)+(1)(1)=4\mathbf{v}_{1} \cdot \mathbf{v}_{1} = (1)(1) + (-1)(-1) + (-1)(-1) + (1)(1) = 4v1​⋅v1​=(1)(1)+(−1)(−1)+(−1)(−1)+(1)(1)=4

STEP 8

Calculate v2⋅v2\mathbf{v}_{2} \cdot \mathbf{v}_{2}v2​⋅v2​:v2⋅v2=(−1)(−1)+(1)(1)+(0)(0)+(2)(2)=6\mathbf{v}_{2} \cdot \mathbf{v}_{2} = (-1)(-1) + (1)(1) + (0)(0) + (2)(2) = 6v2​⋅v2​=(−1)(−1)+(1)(1)+(0)(0)+(2)(2)=6

STEP 9

Substitute these values into the formula for p\mathbf{p}p:p=204v1+356v2\mathbf{p} = \frac{20}{4}\mathbf{v}_{1} + \frac{35}{6}\mathbf{v}_{2}p=420​v1​+635​v2​

STEP 10

Simplify to get p\mathbf{p}p:p=5v1+356v2\mathbf{p} = 5\mathbf{v}_{1} + \frac{35}{6}\mathbf{v}_{2}p=5v1​+635​v2​

STEP 11

STEP 12

STEP 13

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

First, calculate the dot products $\mathbf{y} \cdot \mathbf{v}_{1}$ , $\mathbf{y} \cdot \mathbf{v}_{2}$ , $\mathbf{v}_{1} \cdot \mathbf{v}_{1}$ , and $\mathbf{v}_{2} \cdot \mathbf{v}_{2}$ .

Calculate $\mathbf{y} \cdot \mathbf{v}_{1}$ :
$\mathbf{y} \cdot \mathbf{v}_{1} = (1)(1) + (-1)(-1) + (1)(-1) + (19)(1) = 20$

Calculate $\mathbf{y} \cdot \mathbf{v}_{2}$ :
$\mathbf{y} \cdot \mathbf{v}_{2} = (1)(-1) + (-1)(1) + (1)(0) + (19)(2) = 35$

Calculate $\mathbf{v}_{1} \cdot \mathbf{v}_{1}$ :
$\mathbf{v}_{1} \cdot \mathbf{v}_{1} = (1)(1) + (-1)(-1) + (-1)(-1) + (1)(1) = 4$

Calculate $\mathbf{v}_{2} \cdot \mathbf{v}_{2}$ :
$\mathbf{v}_{2} \cdot \mathbf{v}_{2} = (-1)(-1) + (1)(1) + (0)(0) + (2)(2) = 6$

Substitute these values into the formula for $\mathbf{p}$ :
$\mathbf{p} = \frac{20}{4}\mathbf{v}_{1} + \frac{35}{6}\mathbf{v}_{2}$

Simplify to get $\mathbf{p}$ :
$\mathbf{p} = 5\mathbf{v}_{1} + \frac{35}{6}\mathbf{v}_{2}$