Show that the solutions for the linear system of equations:
$$\begin{aligned} 0 + x_2 +3x_3 – x_4 + 2x_5 &= 0 \\ 2x_1 + 3x_2 + x_3 + 3x_4 &= 0 \\ x_1 + x_2 – x_3 + 2x_4 – x_5 &= 0 \end{aligned}$$
is a subspace of $\mathbb R^5$. What is the dimension of the subspace and determine a basis for the subspace?
I really don't know how to solve this problem. I have achieved this augmented matrix through Gaussian elimination:
$$ \begin{bmatrix}
1& 0& -4& 3& -3& 0 \\
0& 1& 3& -1& 2& 0 \\
0& 0& 0& 0& 0& 0
\end{bmatrix} $$
Any hints or some steps I've missed?
Edit
My professor says the dimension is $3$.
Best Answer
You're almost there. Now your free variables are $x_3=s$,$x_4=t$ and $x_5=u$. Using backward substitution we get $$ x_1=4s-3t+3u \\ x_2=-3s+t-2u\\ x_3=s \\ x_4=t \\ x_5=u $$
Therefore we can write every solution as $$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix}=s\begin{bmatrix} 4 \\ -3 \\ 1 \\ 0 \\ 0 \end{bmatrix}+t\begin{bmatrix} -3 \\ 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}+u\begin{bmatrix} 3 \\ -2 \\ 0 \\ 0 \\ 1 \end{bmatrix} $$ with $s,t,u \in \mathbb{R}$.
Thus the subspace has dimension $3$ and a basis is given by $\begin{bmatrix} 4 \\ -3 \\ 1 \\ 0 \\ 0 \end{bmatrix},\begin{bmatrix} -1 \\ 1 \\ 0 \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix} 3 \\ -2 \\ 0 \\ 0 \\ 1 \end{bmatrix}$
To prove is a subspace you need:
$0$ belongs to it: this is clear by taking $s=t=u=0$
It's closed under sums: if $(s,t,u)$ and $(s',t',u')$ gives us two different solutions, the sum of them is given by $(s+s',t+t',u+u')$
It's closed under scalar multiplication: if $(s,t,u)$ gives us a solution and we multiply it by $k \in \mathbb{R}$, then we still have a solution given by $(ks,kt,ku)$.
Therefore it is a subspace of $\mathbb{R}^5$