Specifying a base for the vectorspace $\{(x_1,x_2,x_3,x_4)\in\mathbb{R}^4:x_1+3x_2+2x_4=0,2x_1+x_2+x_3=0\}$ over $\mathbb{R}$

Question

Specifying a base for the vectorspace $\{(x_1,x_2,x_3,x_4)\in\mathbb{R}^4:x_1+3x_2+2x_4=0,\quad2x_1+x_2+x_3=0\}$ over $\mathbb{R}$

How many vectors do I need to specify a base for this vectorspace? (I guess 4) Or could I just solve the equation system with Gauß:

\begin{bmatrix} 1 & 2 &| 0 \\ 3 & 1 &| 0 \\ 0 & 1 &| 0 \\ 2 & 0 &| 0 \\ \end{bmatrix} Which would be: \begin{bmatrix} 1 & 2 &| 0 \\ 0 & -5&| 0 \\ 0 & 0 &| 0 \\ 0 & 0 &| 0 \\ \end{bmatrix} Therefore the Base $\mathcal{B}$ is $\mathcal{B}:=\left\{\begin{pmatrix}1 \\ 3 \\ 0 \\2 \end{pmatrix},\begin{pmatrix}2 \\ 1 \\ 1 \\ 0\end{pmatrix}\mid \in \mathbb{R}\right\}$

Is this wrong? Do I need to choose vectors in order to specify a base?

Answer 1

You have two linearly independent (check this, one is not a multiple of the other) constraints on the components of the vectors in $\mathbb{R}^4$, so this amounts to $4-2$ degrees of freedom for the vectors in the space. You will therefore require a basis of $2$ vectors.

If you form an augmented matrix the other way up to how you have (the coefficients in your equations should correspond to the rows), you can solve your equations to obtain a basis:

$$\left\{\begin{pmatrix}-3\\1\\5\\0\end{pmatrix}, \begin{pmatrix}2\\-4\\0\\5\end{pmatrix}\ \right\}$$

Answer 2

There are two independent constraints, therefore the dimension of the sub-space will be $2$.

$2x_1+x_2+x_3=0 \Rightarrow x_3 = -2x_1-x_2\\ x_1+3x_2+2x_4=0 \Rightarrow x_4 = -\frac 1 2 x_1 - \frac 3 2 x_2$

So if you know the values of $x_1$ and $x_2$ then that also determines the values of $x_3$ and $x_4$.

One possible basis is therefore

$\{(1,0,-2,-\frac 1 2),(0,1,-1,-\frac 3 2)\}$