Math / Algebra

QuestionLet $u, w, v_{1}, v_{2}$ , and $v_{3}$ be the vectors in $\mathbb{R}^{4}$ defined by $u=\left[\begin{array}{l} 17 \\ 12 \\ 16 \\ 10 \end{array}\right] \quad w=\left[\begin{array}{c} 29 \\ 12 \\ -29 \\ 14 \end{array}\right] \quad v_{1}=\left[\begin{array}{c} 7 \\ -2 \\ 6 \\ -19 \end{array}\right] \quad v_{2}=\left[\begin{array}{c} -2 \\ -19 \\ 1 \\ -18 \end{array}\right] \quad v_{3}=\left[\begin{array}{c} 6 \\ -8 \\ -11 \\ -1 \end{array}\right]$ (a) Is $u \in \operatorname{span}\left\{v_{1}, v_{2}, v_{3}\right\}$ ? Write all zeros if it is not in the span or write zero as a non-trivial (not all zero coefficients) linear combination of $u_{1}, v_{1}, v_{2}$ , and $v_{3}$ if $u$ is in the span. $0=\square u+\square v_{1}+\square v_{2}+\square v_{3}$ (b) Is $w \in \operatorname{span}\left\{v_{1}, v_{2}, v_{3}\right\}$ ? Write all zeros if it is not or if it is in the span write zero as a non-trivial (not all zero coefficients) linear combination of $w$ , $v_{1}, v_{2}$ , and $v_{3}$ if $w$ is in the span. $0=\square w+\square v_{1}+\square v_{2}+\square v_{3}$ (c) Type the dimension of $\operatorname{span}\left\{v_{1}, v_{2}, v_{3}, u\right\}$ : $\square$

Studdy Solution

STEP 1

What is this asking? Can we make the vectors $u$ and $w$ using a mix of the vectors $v_1$ , $v_2$ , and $v_3$ ?
Also, what's the biggest number of independent vectors we can get from the set $\{v_1, v_2, v_3, u\}$ ? Watch out! Don't mix up linear dependence and independence!
If vectors are linearly dependent, it means one can be built from the others.

STEP 2

1. Check if $u$ is in the span of $v_1$ , $v_2$ , and $v_3$ .
2. Check if $w$ is in the span of $v_1$ , $v_2$ , and $v_3$ .
3. Find the dimension of the span of $\{v_1, v_2, v_3, u\}$ .

STEP 3

Set up the equation.
We want to see if we can find numbers $a$ , $b$ , and $c$ such that $u = a v_1 + b v_2 + c v_3$ .

STEP 4

Create an augmented matrix.
This equation gives us a system of four equations with three unknowns.
We can represent this system with an augmented matrix: $\begin{bmatrix} 7 & -2 & 6 & 17 \\ -2 & -19 & -8 & 12 \\ 6 & 1 & -11 & 16 \\ -19 & -18 & -1 & 10 \end{bmatrix}$

STEP 5

Row reduce the matrix.
After row reducing this matrix, we get: $\begin{bmatrix} 1 & 0 & -1 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

STEP 6

Interpret the result.
This tells us that $a - c = 2$ and $b = -1$ .
Let's set $c = 0$ .
Then $a = 2$ .
So, we have a solution: $a=2$ , $b=-1$ , and $c=0$ .

STEP 7

Set up the equation.
Similar to before, we want to find $a$ , $b$ , and $c$ such that $w = a v_1 + b v_2 + c v_3$ .

STEP 8

Create an augmented matrix.
This gives us the augmented matrix: $\begin{bmatrix} 7 & -2 & 6 & 29 \\ -2 & -19 & -8 & 12 \\ 6 & 1 & -11 & -29 \\ -19 & -18 & -1 & 14 \end{bmatrix}$

STEP 9

Row reduce the matrix.
Row reducing, we find: $\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

STEP 10

Interpret the result.
The third row shows us that $0 = 1$ , which is impossible!
So, $w$ is not in the span of $v_1$ , $v_2$ , and $v_3$ .

STEP 11

Consider the relationship between $u$ and the $v_i$ vectors.
We found that $u$ is a linear combination of $v_1$ , $v_2$ , and $v_3$ : $u = 2v_1 - v_2$ .

STEP 12

Determine the dimension.
Since $u$ depends on $v_1$ and $v_2$ , adding $u$ to the set $\{v_1, v_2, v_3\}$ doesn't add any new dimensions to the span.
The row-reduced matrix in step 2.1.3. shows that $v_1$ and $v_2$ are linearly independent, and $v_3$ can be written as $v_3 = v_1 - 2u = v_1 - 2(2v_1 - v_2) = -3v_1 + 2v_2$ .
Thus, the dimension of the span of $\{v_1, v_2, v_3, u\}$ is 2.

STEP 13

(a) 0 = \(-1 $u$ $+2$ $v_1$ $-1$ $v_2$ $+0$ $v_3$ . (b) $0 = 0w + 0v_1 + 0v_2 + 0v_3$ (since $w$ is not in the span). (c) The dimension is 2.

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Check if $u$ is in the span of $v_1$ , $v_2$ , and $v_3$ .
2. Check if $w$ is in the span of $v_1$ , $v_2$ , and $v_3$ .
3. Find the dimension of the span of $\{v_1, v_2, v_3, u\}$ .

STEP 3

Set up the equation.
We want to see if we can find numbers $a$ , $b$ , and $c$ such that $u = a v_1 + b v_2 + c v_3$ .

STEP 4

Create an augmented matrix.
This equation gives us a system of four equations with three unknowns.
We can represent this system with an augmented matrix: $\begin{bmatrix} 7 & -2 & 6 & 17 \\ -2 & -19 & -8 & 12 \\ 6 & 1 & -11 & 16 \\ -19 & -18 & -1 & 10 \end{bmatrix}$

STEP 5

Row reduce the matrix.
After row reducing this matrix, we get: $\begin{bmatrix} 1 & 0 & -1 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

STEP 6

Interpret the result.
This tells us that $a - c = 2$ and $b = -1$ .
Let's set $c = 0$ .
Then $a = 2$ .
So, we have a solution: $a=2$ , $b=-1$ , and $c=0$ .

STEP 7

Set up the equation.
Similar to before, we want to find $a$ , $b$ , and $c$ such that $w = a v_1 + b v_2 + c v_3$ .

STEP 8

Create an augmented matrix.
This gives us the augmented matrix: $\begin{bmatrix} 7 & -2 & 6 & 29 \\ -2 & -19 & -8 & 12 \\ 6 & 1 & -11 & -29 \\ -19 & -18 & -1 & 14 \end{bmatrix}$

STEP 9

Row reduce the matrix.
Row reducing, we find: $\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

STEP 10

Interpret the result.
The third row shows us that $0 = 1$ , which is impossible!
So, $w$ is not in the span of $v_1$ , $v_2$ , and $v_3$ .

STEP 11

Consider the relationship between $u$ and the $v_i$ vectors.
We found that $u$ is a linear combination of $v_1$ , $v_2$ , and $v_3$ : $u = 2v_1 - v_2$ .

STEP 12

STEP 13

(a) 0 = \(-1 $u$ $+2$ $v_1$ $-1$ $v_2$ $+0$ $v_3$ . (b) $0 = 0w + 0v_1 + 0v_2 + 0v_3$ (since $w$ is not in the span). (c) The dimension is 2.

Get Studdy

Studdy solves anything!

Start learning now

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

Studdy Solution

STEP 1

STEP 2

1. Check if uuu is in the span of v1v_1v1​, v2v_2v2​, and v3v_3v3​.2. Check if www is in the span of v1v_1v1​, v2v_2v2​, and v3v_3v3​.3. Find the dimension of the span of {v1,v2,v3,u}\{v_1, v_2, v_3, u\}{v1​,v2​,v3​,u}.

STEP 3

**Set up the equation**.We want to see if we can find numbers aaa, bbb, and ccc such that u=av1+bv2+cv3u = a v_1 + b v_2 + c v_3u=av1​+bv2​+cv3​.

STEP 4

STEP 5

**Row reduce the matrix**.After row reducing this matrix, we get: [10−12010−100000000]\begin{bmatrix} 1 & 0 & -1 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}⎣⎡​1000​0100​−1000​2−100​⎦⎤​

STEP 6

**Interpret the result**.This tells us that a−c=2a - c = 2a−c=2 and b=−1b = -1b=−1.Let's set c=0c = 0c=0.Then a=2a = 2a=2.So, we have a solution: a=2a=2a=2, b=−1b=-1b=−1, and c=0c=0c=0.

STEP 7

**Set up the equation**.Similar to before, we want to find aaa, bbb, and ccc such that w=av1+bv2+cv3w = a v_1 + b v_2 + c v_3w=av1​+bv2​+cv3​.

STEP 8

STEP 9

**Row reduce the matrix**.Row reducing, we find: [10−10010000010000]\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}⎣⎡​1000​0100​−1000​0010​⎦⎤​

STEP 10

**Interpret the result**.The third row shows us that 0=10 = 10=1, which is impossible!So, www is *not* in the span of v1v_1v1​, v2v_2v2​, and v3v_3v3​.

STEP 11

**Consider the relationship between uuu and the viv_ivi​ vectors**.We found that uuu is a linear combination of v1v_1v1​, v2v_2v2​, and v3v_3v3​: u=2v1−v2u = 2v_1 - v_2u=2v1​−v2​.

STEP 12

STEP 13

(a) 0 = \(-1 uuu +2+2+2 v1v_1v1​ −1 -1−1 v2v_2v2​ +0+0+0 v3v_3v3​. (b) 0=0w+0v1+0v2+0v30 = 0w + 0v_1 + 0v_2 + 0v_30=0w+0v1​+0v2​+0v3​ (since www is not in the span). (c) The dimension is 2.

Get Studdy

Studdy solves anything!

Start learning now

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

1. Check if $u$ is in the span of $v_1$ , $v_2$ , and $v_3$ .
2. Check if $w$ is in the span of $v_1$ , $v_2$ , and $v_3$ .
3. Find the dimension of the span of $\{v_1, v_2, v_3, u\}$ .

Set up the equation.
We want to see if we can find numbers $a$ , $b$ , and $c$ such that $u = a v_1 + b v_2 + c v_3$ .

Row reduce the matrix.
After row reducing this matrix, we get: $\begin{bmatrix} 1 & 0 & -1 & 2 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

Interpret the result.
This tells us that $a - c = 2$ and $b = -1$ .
Let's set $c = 0$ .
Then $a = 2$ .
So, we have a solution: $a=2$ , $b=-1$ , and $c=0$ .

Set up the equation.
Similar to before, we want to find $a$ , $b$ , and $c$ such that $w = a v_1 + b v_2 + c v_3$ .

Row reduce the matrix.
Row reducing, we find: $\begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

Interpret the result.
The third row shows us that $0 = 1$ , which is impossible!
So, $w$ is not in the span of $v_1$ , $v_2$ , and $v_3$ .

Consider the relationship between $u$ and the $v_i$ vectors.
We found that $u$ is a linear combination of $v_1$ , $v_2$ , and $v_3$ : $u = 2v_1 - v_2$ .

(a) 0 = \(-1 $u$ $+2$ $v_1$ $-1$ $v_2$ $+0$ $v_3$ . (b) $0 = 0w + 0v_1 + 0v_2 + 0v_3$ (since $w$ is not in the span). (c) The dimension is 2.