How to find the basis of the following vector spaces?

Question

I'm trying, in vain, to find the basis of the following vector spaces:

(a) $W = \{x = (x_1 , x_2 , x_3 ) ∈ \Bbb R^3 : x_1 − 2x_2 + x_3 = 0, 2x_1 − 3x_2 + x_3 = 0\}$

(b) $W = \{x = (x_1 , x_2 , x_3 , x_4 , x_5 ) ∈ \Bbb R^5 : x_1 − x_3 − x_4 = 0\}$

(c) $W = \{x = (x_1 , x_2 , x_3 , x_4 , x_5 ) ∈ \Bbb R^5 : x_2 = x_3 = x_4 , x_1 + x_5 = 0\}$

I understand that if I have a vector space $V$, then the basis $\mathcal B$ for that vector is the set of vectors which is linearly independent and spans all $V$.

However, how do I apply this to solve the questions above?

Thanks.

Edit: I tried to solve the second one which seems easier, but the problem I have is I could get a basis of 5 linearly independent vectors for it, but it's not the right answer. I'm not sure what I'm doing wrong. Here is what I got for (b):

$\mathcal B(V) = \{(1,0,0,0,0), (0,1,0,0,0), (0,0,1,0,0), (0,0,0,1,0), (0,0,0,0,1)\}$

Answer 1

Let's consider the first example. The vector space $W$ can be described as the solutions of this system of linear equations: $$\underbrace{\begin{bmatrix} 1 & -2 & 1 \\ 2 & -3 & 1 \end{bmatrix}}_{=:A} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

By elementary row operations we get $A$ into the form $$ B = \begin{bmatrix} 1 & -2 & 1 \\ 0 & 1 & -1 \end{bmatrix} \; $$ Now set $x_3 = t \in \Bbb R$ arbitrary. From the second row, we get $x_2 = t$, and from the first row $$ x_1 = 2x_2 - x_3 = 2t - t = t \; ,$$ so we find $$ W = \{ (t,t,t) \, : \, t \in \Bbb R \} .$$ Now we see, that $ \mathcal B = \{ (1,1,1) \}$ is a basis for $W$, because it's clearly linearly independant and the vector $(1,1,1)$ spans the whole vector space $W$.

Answer 2

(a) Here we have two equations and three unknowns.Hence it have only one free variable. So we have 1 linearly independent solution to this. Hence dimension of $W$ is . Let $x_3$ be free variable put $x_3=1$ then we get $x_1=1~ \&~ x_2=1$. So a basis for vector space $W$ is $\{(1,1,1)\}$. Similarly you can solve (b) & (c) also.

Answer 3

For the first one eliminate $x_3$ from the conditions (ie subtract the second equation from the first)$$x_1-2x_2+x_3-2x_1+3x_2-x_3 = 0 \Rightarrow x_1=x_2$$ Subbing back into the first equation gives $$x_1-2x_1+x_3=0 \Rightarrow x_1=x_3$$ So for any $x \in \mathbb{R}^3$ we have $(x_1,x_2,x_3)=(x_1,x_1,x_1)=x_1(1,1,1)$ So $Span{(1,1,1)}$ is a basis for first. The basis for the next two I leave to you. Hope this helps.