Is this the proper method for finding the basis and dimension of a vector space?

Question

I'm a bit new at linear algebra, so please bear with me.

I'm trying to figure out how to find the basis and dimension of this vector space:

U = { $(x,y,z,t)^T$| $x,y,z,t \in R$, x + 2y = 0, z − 3t = 0}

This has been my process:

First I made this co-effecient matrix:

[ 1 2 0 0]

[ 0 0 -3 1]

And turned into REF form:

[1 2 0 0]

[0 0 1 -1/3]

That gave me y and t into variables, and keeping that in mind, I made these equations:

x+ 2y = 0 and z- (t/3) = 0

Which turns into: x = -2y and z = t/3

I put all this into this vector:

$(-2y,y,t/3,t)^T$

which equals: y$(-2,1,0,0)^T$+t$(0,0,1/3,1)^T$

This means that U = span{$(-2,1,0,0)^T$,$(0,0,1/3,1)^T$}

Since they are linearly independent this means they are a basis, and the dimension would be 2. This math looks right to me, but I'm not confident enough to assume that, so I'm just wondering if I went about this correctly. Thanks in advance!

Answer 1

I post this as an answer because it may be too much long for a comment.

You correctly proved that $U = span(e_1, e_2)$, where $e_1 = (-2,1,0,0)^{\text{T}}$ and $e_2 = (0,0,\frac{1}{3},1)^{\text{T}}$. As pointed out in the comments section, in order to prove linear independence of $e_1$, $e_2$ (and thus obtaining a basis for $U$) you can use the definition:

1. let $t_1, t_2 \in \mathbb{R}$ and assume that $t_1 e_1 + t_2 e_2 = 0$ (as vectors);
2. prove that under this assumption it follows that $t_1 = 0 = t_2$.

To do this just make explicit what $t_1 e_1 + t_2 e_2 = 0$ means.

P.S. : if it is the case that you can compute the ${rank}$ of a matrix, you can prove that $\{e_1,e_2\}$ is a basis by showing that $rank(e_1^{T}, e_2^{T})$ equals 2.