Base of a subspace of R4?

Question

I have $F = \{(a, b, c, d) \in\mathbb R^4| a-2b-c+d = 0 , 2a-b+c-2d = 0\}$. The exercise asks me to demonstrate it's a subspace of $\mathbb R^4$, and up to there, no problem. The second part, however, asks me to find a base for $F$ and give its dimension. I've tried to look through the existing posts, but still I can't make it work. Could anyone help me out?

Answer 1

You can write the problem in the following form: \begin{equation}\begin{bmatrix} 1 & -2 & -1 & 1\\ 2& -1& 1 & -2 \end{bmatrix}\begin{bmatrix} x\\ y\\ z\\ t \end{bmatrix} =\begin{bmatrix} 0\\ 0 \end{bmatrix} \end{equation} Now that we have reformulated the problem to linear algebra, we seen that basically we have to find the nullspace of that matrix. Please learn to do this by hand, but Wolfram Alpha gives us: \begin{equation}b_1 = \begin{bmatrix}-1\\ -1\\ 1\\ 0 \end{bmatrix} , b_2 = \begin{bmatrix}5\\ 4\\ 0\\ 3 \end{bmatrix} \end{equation} source: https://www.wolframalpha.com/input/?i=nullspace+of+%7B%7B1%2C-2%2C-1%2C1%7D%2C%7B2%2C-1%2C1%2C-2%7D%7D

Answer 2

The two equations $x−2y−z+t=0$ and $2x−y+z−2t=0$ are independent, and $4−2=2$,

so $F$ is two-dimensional.

$x−2y−z+t=0$ and $2x−y+z−2t=0\implies3x−3y−t=0.$

Two independent solutions of that are $(x,y,t)=(1,1,0)$ and $(1,0,3)$.

If you find the $z$ values to satisfy the equations, you will then have a basis for $F$.