# Trying to find the basis of a subspace given components satisfying a condition?

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

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 $$\color{red}{\langle2,1,-1,0\rangle,\langle0,0,0,1\rangle}$$

• Thanks, this was a huge help! – DesPhantomes Oct 21 '18 at 23:55

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.

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