If the set W is a vector space, find a set S of vectors that spans it. Otherwise, state that W is not a vector space. W is the set of all vectors of the form [a - 4b 5 4a + b -a - b], where a and bare arbitrary real numbers.
a. [1 5 4 -1], [-4 0 1 -1]
b. [1 0 4 -1], [-4 5 1 -1]
c. [1 0 4 -1], [-4 0 1 -1], [0 5 0 0]
d. Not a vector space

Choice d. The set of vectors: [tex]\displaystyle \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex] isn't a vector space over [tex]\mathbb{R}[/tex].

Explanation:

Let a set of vectors [tex]V[/tex] to be a vector field over some field [tex]\mathbb{F}[/tex] (for this question, that "field" is the set of all real number.) The following must be true:

The set of vectors [tex]V[/tex] includes the identity element [tex]\mathbf{0}[/tex]. In other words, there exists a vector [tex]\mathbf{0} \in V[/tex] such that for all [tex]\mathbf{v} \in V[/tex], [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex].
[tex]V[/tex] should be closed under vector addition. In other words, for all [tex]\mathbf{u},\, \mathbf{v} \in V[/tex], [tex]\mathbf{u} + \mathbf{v} \in V[/tex].
[tex]V[/tex] should also be closed under scalar multiplication. In other words, for all [tex]\mathbf{v} \in V[/tex] and all "scalar" [tex]m \in \mathbb{F}[/tex] (in this question, the "field" is the set of all real numbers, so [tex]m[/tex] can be any real number,) [tex]a\,\mathbf{v} \in V[/tex].

Note that in the general form of a vector in [tex]V[/tex], the second component is a always non-zero. Because of that non-zero component,

Assume by contradiction that [tex]V[/tex] is indeed a vector field. Therefore, it should contain a zero vector. Let [tex]\mathbf{0}[/tex] denote that zero vector. For all [tex]\mathbf{v} \in V[/tex], [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex].

Using the definition of set [tex]V[/tex]: [tex]\displaystyle \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex], there exist real numbers [tex]a[/tex] and [tex]b[/tex], such that:

[tex]\displaystyle \mathbf{v} = \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex].

Hence, [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex] is equivalent to:

[tex]\displaystyle \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} + \mathbf{0} = \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex].

Apply the third property that [tex]V[/tex] is closed under scalar multiplication. [tex]-1[/tex] is indeed a real number. Therefore, if [tex]\mathbf{v}[/tex] is in

Therefore:

[tex]\displaystyle -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} \in V[/tex].

Apply the second property and add [tex]\displaystyle - \mathbf{v} = -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix}[/tex] to both sides of [tex]\mathbf{v} + \mathbf{0} = \mathbf{v}[/tex]. The left-hand side becomes:

[tex]\mathbf{v} - \mathbf{v} + \mathbf{0} = \mathbf{0}[/tex].

The right-hand side becomes:

[tex]\displaystyle \begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} -\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix} = \begin{bmatrix}a - 4\, b - (a - 4\, b) \\ 5 - 5 \\ 4\, a+ b-(4\, a+ b)\\ -a -b - (-a -b)\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex].

Therefore:

[tex]\displaystyle \mathbf{0} = \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex].

However, [tex]\mathbf{0} = \displaystyle \begin{bmatrix}0 \\ 0 \\ 0 \\0\end{bmatrix}[/tex] isn't a member of the set [tex]\displaystyle V = \left\{\begin{bmatrix}a - 4\, b \\ 5 \\ 4\, a+ b\\ -a -b\end{bmatrix},\; a,\, b\in \mathbb{R} \right\}[/tex]. That's a contradiction, because [tex]\mathbf{0}[/tex] was supposed to be part of [tex]V[/tex].

Hence, [tex]V[/tex] isn't a vector space by contradiction.