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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):

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

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  • $\begingroup$ One of the way to do it would be to figure out the dimension of the vector space. In which case it suffices to find that many linearly independent vectors to prove that they are basis. $\endgroup$ – Jack Yoon May 28 '15 at 13:06
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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 $W$.

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  • $\begingroup$ I see, but why is my answer to (b) wrong? Since it's a linearly independent set? $\endgroup$ – nTuply May 28 '15 at 13:27
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    $\begingroup$ @nTuply I did not solve the example (b), but what you have written down, is a basis for the whole space $\Bbb R^5$. You should find a basis for the subspace $W \subset \Bbb R^5$. In case of example (b), you will see that $W \neq \Bbb R^5$. $\endgroup$ – aexl May 28 '15 at 13:32
  • $\begingroup$ Oh I see it now. Thanks. I'll try to solve it. $\endgroup$ – nTuply May 28 '15 at 13:37
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(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.

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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.

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