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Chapter 4: Problem 53

Consider the set of vectors \(S\) given by $$ \left\\{\left[\begin{array}{c} u+v+w \\ 2 u+2 v+4 w \\ u+v+w \\ 0 \end{array}\right]: u, v, w \in \mathbb{R}\right\\} $$ Is S a subspace of \(\mathbb{R}^{4} ?\) If so, explain why, give a basis for the subspace and find its dimension.

Short Answer

Expert verified
The set S is a subspace of \mathbb{R}^{4}\ with basis \{ \begin{bmatrix} 1 \ 2 \ 1 \ 0 \end{bmatrix}\ , \begin{bmatrix} 1 \ 4 \ 1 \ 0 \end{bmatrix}\ \} and dimension 2.

Step by step solution

01

- Check for zero vector

The zero vector must be in the set to be a subspace. Set \(u = 0\), \(v = 0\), and \(w = 0\). This yields: \(\begin{bmatrix} 0 \ 0 \ 0 \ 0 \ \end{bmatrix}\). This must be in every subspace.
02

- Check for closure under addition

Check if the sum of any two vectors in the set is also in the set. Consider two vectors: \(\begin{bmatrix} u_1+v_1+w_1 \ 2u_1+2v_1+4w_1 \ u_1+v_1+w_1 \ 0 \end{bmatrix}\) and \(\begin{bmatrix} u_2+v_2+w_2 \ 2u_2+2v_2+4w_2 \ u_2+v_2+w_2 \ 0 \end{bmatrix}\). Then, their sum is: \(\begin{bmatrix} (u_1+u_2) + (v_1+v_2) + (w_1+w_2) \ 2(u_1+u_2) + 2(v_1+v_2) + 4(w_1+w_2) \ (u_1+u_2) + (v_1+v_2) + (w_1+w_2) \ 0 \end{bmatrix}\), which can also be written in the form of an element in the set.
03

- Closure under scalar multiplication

Check if multiplying a vector in the set by a scalar \(c\) results in a vector that is also in the set. Consider a vector: \(\begin{bmatrix} u+v+w \ 2u+2v+4w \ u+v+w \ 0 \end{bmatrix}\). Multiplying by scalar \(c\) gives: \(\begin{bmatrix} c(u+v+w) \ c(2u+2v+4w) \ c(u+v+w) \ 0 \end{bmatrix}\), which is of the same form and thus remains in the set.
04

- Basis and Dimension

Determine if the set is a space by factoring out common factors. Express the general vector as: \(\begin{bmatrix} u+v+w \ 2(u+v+2w) \ u+v+w \ 0 \end{bmatrix}\). Notice that the vectors can be written as a linear combination: \(c_1\begin{bmatrix} 1 \ 2 \ 1 \ 0 \end{bmatrix} + c_2\begin{bmatrix} 1 \ 2 \ 1 \ 0 \end{bmatrix} + c_3\begin{bmatrix} 1 \ 4 \ 1 \ 0 \end{bmatrix}\). The set of vectors \(\begin{bmatrix} 1 \ 2 \ 1 \ 0 \end{bmatrix}\) and \(\begin{bmatrix} 1 \ 4 \ 1 \ 0 \end{bmatrix}\) forms a basis for the subspace. Since there are two linearly independent vectors, the dimension is 2.

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Key Concepts

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

Basis of Vector Space
A basis of a vector space is a set of vectors that are both linearly independent and span the entire space. This means you can form any vector in the space as a linear combination of the basis vectors. In our problem, we need to express vectors in the form:
\[\begin{pmatrix} u+v+w \ 2u+2v+4w \ u+v+w \ 0 \end{pmatrix} \]
We then hypothesize basis vectors that might span the entire subspace and check for linear independence.
We re-write the general vector using constants. We end up with: \[\begin{pmatrix} 1 \ 2 \ 1 \ 0 \end{pmatrix} \text{and} \begin{pmatrix} 1 \ 4 \ 1 \ 0 \end{pmatrix} \]. Since these vectors are linearly independent and span the subspace, they form a basis for it.
Closure under Addition and Scalar Multiplication
For a set to be a subspace, it must be closed under addition and scalar multiplication.
Closure under addition means if you take any two vectors in your set and add them together, the result must be in the set too. We checked this by taking two vectors: \[\begin{pmatrix} u_1+v_1+w_1 \ 2u_1+2v_1+4w_1 \ u_1+v_1+w_1 \ 0 \end{pmatrix} \] and \[\begin{pmatrix} u_2+v_2+w_2 \ 2u_2+2v_2+4w_2 \ u_2+v_2+w_2 \ 0 \end{pmatrix} \].
Their sum: \[\begin{pmatrix} (u_1+u_2) + (v_1+v_2) + (w_1+w_2) \ 2(u_1+u_2) + 2(v_1+v_2) + 4(w_1+w_2) \ (u_1+u_2) + (v_1+v_2) + (w_1+w_2) \ 0 \end{pmatrix} \] is also of the correct general form and so remains in the set.
Similarly, closure under scalar multiplication means if you multiply any vector in your set by any scalar, the result is still in the set. This was verified by multiplying: \[\begin{pmatrix} u+v+w\ 2u+2v+4w \ u+v+w \ 0 \end{pmatrix} \] by a scalar \(c\), resulting in: \[\begin{pmatrix} c(u+v+w) \ c(2u+2v+4w) \ c(u+v+w) \ 0 \end{pmatrix} \], which still fits the original form.
Dimension of Subspace
The dimension of a subspace is the number of vectors in its basis. In simple terms, it tells us the 'number of directions' in the subspace.
In our problem, we started by identifying vectors that could potentially span the subspace. By expressing the original vectors in the correct form, we defined a set of vectors: \[\begin{pmatrix} 1 \ 2 \ 1 \ 0 \end{pmatrix} \] and \[\begin{pmatrix} 1 \ 4 \ 1 \ 0 \end{pmatrix} \]. Since these two vectors are linearly independent, they can form our basis.
The number of these vectors (in this case, two) gives our dimension. Thus, the dimension of this subspace is 2.

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Most popular questions from this chapter

Three forces act on an object. Two are \(\left[\begin{array}{r}6 \\ -3 \\\ 3\end{array}\right]\) \(\left[\begin{array}{l}2 \\ 1 \\ 3\end{array}\right]\) Newtons. Find the third force if the total force on the object is to be \(\left[\begin{array}{l}7 \\ 1 \\ 3\end{array}\right]\).

Find the rank of the following matrix. Also find a basis for the row and column spaces. $$ \left[\begin{array}{rrrrrr} 1 & 0 & 3 & 0 & 7 & 0 \\ 3 & 1 & 10 & 0 & 23 & 0 \\ 1 & 1 & 4 & 1 & 7 & 0 \\ 1 & -1 & 2 & -2 & 9 & 1 \end{array}\right] $$

Here are some vectors in \(\mathbb{R}^{4}\). $$ \left[\begin{array}{r} 1 \\ 2 \\ -2 \\ 1 \end{array}\right],\left[\begin{array}{r} 1 \\ 3 \\ -3 \\ 1 \end{array}\right],\left[\begin{array}{r} 1 \\ -1 \\ 1 \\ 1 \end{array}\right],\left[\begin{array}{r} 2 \\ -3 \\ 3 \\ 2 \end{array}\right],\left[\begin{array}{r} 1 \\ 3 \\ -2 \\ 1 \end{array}\right] $$ Thse vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Here are some vectors. $$ \left[\begin{array}{r} 1 \\ -1 \\ -2 \end{array}\right],\left[\begin{array}{r} 1 \\ 0 \\ -2 \end{array}\right],\left[\begin{array}{r} 1 \\ -5 \\ -2 \end{array}\right],\left[\begin{array}{r} -1 \\ 5 \\ 2 \end{array}\right] $$ Now here is another vector: $$ \left[\begin{array}{r} 1 \\ 1 \\ -1 \end{array}\right] $$ Is this vector in the span of the first four vectors? If it is, exhibit a linear combination of the first four vectors which equals this vector, using as few vectors as possible in the linear combination.

An object moves 10 meters in the direction of \(\vec{j}+\vec{i}\). There are two forces acting on this object, \(\vec{F}_{1}=\vec{i}+2 \vec{j}+2 \vec{k},\) and \(\vec{F}_{2}=5 \vec{i}+2 \vec{j}-6 \vec{k} .\) Find the total work done on the object by the two forces. Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force. Why?
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