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Chapter 4: Problem 50

Consider the vectors of the form $$ \left\\{\left[\begin{array}{c} 2 u+v+7 w \\ u-2 v+w \\ -6 v-6 w \end{array}\right]: u, v, w \in \mathbb{R}\right\\} $$ Is this set of vectors a subspace of \(\mathbb{R}^{3}\) ? If so, explain why, give a basis for the subspace and find its dimension.

Short Answer

Expert verified
Yes, it is a subspace of \(\backslashmathbb{R}^3\). The basis is \[\left\{\left[\begin{array}{c} 2 \ 1 \ 0 \end{array}\right], \left[\begin{array}{c} 1 \ -2 \ -6 \end{array}\right], \left[\begin{array}{c} 7 \ 1 \ -6 \end{array}\right]\right\}\]. Dimension = 3.

Step by step solution

01

Define the Set

Represent the given vector in the form of coordinates with parameters u, v, and w.
02

Check Closure under Addition

Consider two elements of the set and prove that their sum is also in the set.
03

Check Closure under Scalar Multiplication

Consider an element of the set and a scalar, then prove that the product is also in the set.
04

Verify Zero Vector

Check if the zero vector is included in the set by setting u, v, and w to zero.
05

Confirm Subspace

Conclude that the set of vectors is indeed a subspace of \(\backslashmathbb{R}^3\).
06

Determine a Basis

Express the vector in terms of a linear combination of vectors to form a basis.
07

Find Dimension

Count the number of vectors in the basis to determine the dimension of the subspace.

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Key Concepts

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

Closure under Addition
Closure under addition means that if you take any two vectors from a set, their sum should also be within the same set. Let's take two vectors from our set: \(\begin{bmatrix} 2u_1+v_1+7w_1 \ u_1-2v_1+w_1 \ -6v_1-6w_1 \end{bmatrix}\) and \(\begin{bmatrix} 2u_2+v_2+7w_2 \ u_2-2v_2+w_2 \ -6v_2-6w_2 \end{bmatrix}\). Adding them, we get:
\(\begin{bmatrix} 2(u_1+u_2)+(v_1+v_2)+7(w_1+w_2) \ (u_1+u_2)-2(v_1+v_2)+(w_1+w_2) \ -6(v_1+v_2)-6(w_1+w_2) \end{bmatrix}\).
This shows that the sum of any two vectors from our set remains in the set, proving closure under addition.
Closure under Scalar Multiplication
Closure under scalar multiplication means that if you multiply any vector from the set by a scalar, the result should still be within the same set. Let's take a vector \( \begin{bmatrix} 2u+v+7w \ u-2v+w \ -6v-6w \end{bmatrix} \) and a scalar \( c \). Multiply the vector by the scalar \( c \):
\( \begin{bmatrix} c(2u+v+7w) \ c(u-2v+w) \ c(-6v-6w) \end{bmatrix} \).
This shows that the product of any vector in our set with a scalar remains within the set, proving closure under scalar multiplication.
Zero Vector
A subspace must contain the zero vector. The zero vector has all its components equal to zero. Setting \( u = 0 \), \( v = 0 \), and \( w = 0 \), the vector becomes:
\( \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \), which is indeed the zero vector. Hence, our set contains the zero vector, satisfying this condition for being a subspace.
Basis for Subspace
A basis is a set of vectors in the subspace that are linearly independent and span the subspace. For our problem, we express our given vector as:
\( \begin{bmatrix} 2u+v+7w \ u-2v+w \ -6v-6w \end{bmatrix} \).
Factoring out parameters, we get:
\( \begin{bmatrix} 2 \ 1 \ 0 \end{bmatrix}u + \begin{bmatrix} 1 \ -2 \ -6 \end{bmatrix}v + \begin{bmatrix} 7 \ 1 \ -6 \end{bmatrix}w \).
Thus, the basis for our subspace includes the vectors: \( \begin{bmatrix} 2 \ 1 \ 0 \end{bmatrix} \), \( \begin{bmatrix} 1 \ -2 \ -6 \end{bmatrix} \), and \( \begin{bmatrix} 7 \ 1 \ -6 \end{bmatrix} \).
Subspaces
A subspace is a set of vectors that is closed under addition and scalar multiplication and includes the zero vector. The set in question meets all these criteria:
  • Closure under addition
  • Closure under scalar multiplication
  • Contains the zero vector
Together, they confirm that the set forms a subspace of \( \mathbb{R}^3 \).
Dimension of Subspace
The dimension of a subspace is the number of vectors in its basis. For our vector set, the basis is:
\( \begin{bmatrix} 2 \ 1 \ 0 \end{bmatrix} \),
\( \begin{bmatrix} 1 \ -2 \ -6 \end{bmatrix} \),
and \( \begin{bmatrix} 7 \ 1 \ -6 \end{bmatrix} \).
Since there are three vectors in the basis, the dimension of our subspace is 3.
Linear Combinations
A linear combination of vectors involves multiplying each vector by a scalar and adding the results. For our set, any vector can be written as:
\( a \begin{bmatrix} 2 \ 1 \ 0 \end{bmatrix} + b \begin{bmatrix} 1 \ -2 \ -6 \end{bmatrix} + c \begin{bmatrix} 7 \ 1 \ -6 \end{bmatrix} \).
These vectors span the subspace and any vector in the subspace can be expressed as such a combination.

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Most popular questions from this chapter

Consider the vectors of the form $$ \left\\{\left[\begin{array}{c} 3 u+v \\ 2 w-4 u \\ 2 w-2 v-8 u \end{array}\right]: u, v, w \in \mathbb{R}\right\\} $$ Is this set of vectors a subspace of \(\mathbb{R}^{3}\) ? If so, explain why, give a basis for the subspace and find its dimension.

City A is located at the origin (0,0) while city \(B\) is located at (300,500) where distances are in miles. An airplane flies at 250 miles per hour in still air. This airplane wants to fly from city \(A\) to city \(B\) but the wind is blowing in the direction of the positive \(y\) axis at a speed of 50 miles per hour. Find a unit vector such that if the plane heads in this direction, it will end up at city \(B\) having flown the shortest possible distance. How long will it take to get there?

Let H denote span \(\left\\{\left[\begin{array}{l}6 \\ 1 \\ 1 \\\ 5\end{array}\right],\left[\begin{array}{r}17 \\ 3 \\ 2 \\\ 10\end{array}\right],\left[\begin{array}{r}52 \\ 9 \\ 7 \\\ 35\end{array}\right],\left[\begin{array}{r}18 \\ 3 \\ 4 \\\ 20\end{array}\right]\right\\}\). Find the dimension of \(H\) and determine a basis.

Are the following vectors linearly independent? If they are, explain why and if they are not, exhibit one of them as a linear combination of the others. Also give a linearly independent set of vectors which has the same span as the given vectors. $$ \left[\begin{array}{r} -1 \\ -2 \\ 2 \\ 3 \end{array}\right],\left[\begin{array}{r} -3 \\ -4 \\ 3 \\ 3 \end{array}\right],\left[\begin{array}{r} 0 \\ -1 \\ 4 \\ 3 \end{array}\right],\left[\begin{array}{r} 0 \\ -1 \\ 6 \\ 4 \end{array}\right] $$

A bird flies from its nest \(8 \mathrm{~km}\) in the direction \(\frac{5}{6} \pi\) north of east where it stops to rest on a tree. It then flies \(1 \mathrm{~km}\) in the direction due southeast and lands atop a telephone pole. Place an \(x y\) coordinate system so that the origin is the bird's nest, and the positive \(x\) axis points east and the positive \(y\) axis points north. Find the displacement vector from the nest to the telephone pole.
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