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So I have two problems built off very similar premises which I have no idea how to solve:

Let W be the Subspace of $\mathbb{R}^4$ consisting of vectors of the form $ x = \{x_1, x_2, x_3, x_4\}$. Find a basis for W when the components of x satisfy the given conditions:

  1. $x_1 - x_2 = 0$
    $x_2-2x_3 = 0$
    $x_3-x_4 = 0$
  2. $-x_1 + 2x_2 = 0 $
    $x_2 + x_3 = 0$

How should I approach solving these? I have no lecturer or anything right now so any help is greatly appreciated

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For the first case, $x_1=x_2$ and $x_3=x_4$. Moreover, $2x_3 = x_2$. This ensures that all vectors that satisfy the three equations are of the form $$\langle 2x,2x,x,x\rangle\;\forall x\in\mathbb R$$

For the second case, $x_4$ can be anything, so all vectors that satisfy those equations are of the form $$\langle2x,x,-x,y\rangle\;\forall x,y\in\mathbb R$$

This means that for the first example, the basis is $\color{red}{\langle2,2,1,1\rangle}$ while for the second, a basis (there are infinitely many valid bases) that works is $\color{red}{\langle2,1,-1,0\rangle,\langle0,0,0,1\rangle}$

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    $\begingroup$ Thanks, this was a huge help! $\endgroup$ – DesPhantomes Oct 21 '18 at 23:55
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For number 6. we have $x_3=x_4$ and $x_2=2x_3= 2x_4$, $x_1=x_2$ therefore the vector $x$ in $W$ is described as $x= x_4(2,2,1,1)$ which is simply a scalar multiple of $(2,2,1,1)$,

Thus the basis for the subspace is just $(2,2,1,1)$ which make it one dimensional.

For number 7. we have $x_1=2x_2, x_3=-x_2,x_4=x_4$

Thus $$x= (2x_2, x_2, -x_2,x_4) = x_2(2,1,-1,0) + x_4(0,0,0,1)$$

Therefore the subspace in spanned by two vectors, $(2,1,-1,0)$ and $(0,0,0,1)$ which makes it two dimensional.

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  • $\begingroup$ Thank you, that's super helpful! $\endgroup$ – DesPhantomes Oct 21 '18 at 23:55

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