Let $A=\begin{bmatrix} 0&1 &2 &1 &1 \\ 1&0 &1 &1 &1 \\ -1&2 &3 &1 &1 \end{bmatrix}$ with the homogeneous system $A \mathbf{x}=\mathbf0$.
Problem 1
Find the parametric form for the solution set $N_{A}\subseteq \mathbb{R}^5 $ (null)
Solution so far
After row operations to find the augmented matrix in reduced echelon form, I get that the parametric form with $s=x_3, t=x_4, k=x_5$ is \begin{bmatrix} -s-t-k\\ -2s-t-k\\ s\\ t\\ k\\ \end{bmatrix} or $\mathbf{v}=\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\\ x_5\\ \end{bmatrix}=s\begin{bmatrix} -1\\ -2\\ 1\\ 0\\ 0\\ \end{bmatrix}+t\begin{bmatrix} -1\\ -1\\ 0\\ 1\\ 0\\ \end{bmatrix}+k\begin{bmatrix} -1\\ -1\\ 0\\ 0\\ 1\\ \end{bmatrix}$
Problem 2 (Main)
Specify a basis for the subspace $N_{A}\subseteq \mathbb{R}^5 $. What is the dimension of $N_{A}\subseteq \mathbb{R}^5 $?
My Solution so far
I am thinking the basis would just be the 3 vectors in my parametric form i.e
$\begin{bmatrix} -1\\ -2\\ 1\\ 0\\ 0\\ \end{bmatrix},\begin{bmatrix} -1\\ -1\\ 0\\ 1\\ 0\\ \end{bmatrix},\begin{bmatrix} -1\\ -1\\ 0\\ 0\\ 1\\ \end{bmatrix}$ and then I am thinking the dimension is just the cardinality of all the vectors in the specified solution space so 3?
I am certain this is wrong, though and I would like the approach for the main problem cleared up! Do I have to extend my solutions with the standard basis to get 5 vectors? Thanks!
edit: I have been taught that if i.e $\mathbb{R}^5$ then I would need to extend this system to 5 vectors. Is this not correct?
Yes, those three vectors span $N_A$ and, since they are linearly independent, then are a basis of $N_A$. Therefore $\dim N_A=3$.