Answer
Basis, $B=\left\{\begin{bmatrix}3\\6\\-9\\-3\end{bmatrix},\begin{bmatrix}6\\-2\\5\\1\end{bmatrix}\right\}$
Since the two vectors are linearly independent, they form the basis for the set and,
the dimension is 2.
Work Step by Step
We are required to find the basis and state the dimension:
Let the given subspace be represented by a vector $\mathbf{\vec{z}}$ such that:
$\mathbf{\vec{z}}=\begin{bmatrix}3a+6b-c\\6a-2b-2c\\-9a+5b+3c\\-3a+b+c\end{bmatrix} $
Writing the given set in parametric form;
$\mathbf{\vec{z}}=\begin{bmatrix}3a+6b-c\\6a-2b-2c\\-9a+5b+3c\\-3a+b+c\end{bmatrix} =a\begin{bmatrix}3\\6\\-9\\-3\end{bmatrix}+b\begin{bmatrix}6\\-2\\5\\1\end{bmatrix}+c\begin{bmatrix}-1\\-2\\3\\1\end{bmatrix}$
The given subspace is a linear combination of three vectors
$v_{1}=\begin{bmatrix}3\\6\\-9\\-3\end{bmatrix},v_{2}=\begin{bmatrix}6\\-2\\5\\1\end{bmatrix},v_{3}=\begin{bmatrix}-1\\-2\\3\\1\end{bmatrix}$
From the set ,it can be seen that $v_{1}$ is a multiple of $v_{3}$.
such that;
$v_{1}=-3v_{3}$
Hence the spanning set is reduced to $v_{1},v_{2}$ and the basis is given by
$B=\left\{\begin{bmatrix}3\\6\\-9\\-3\end{bmatrix},\begin{bmatrix}6\\-2\\5\\1\end{bmatrix}\right\}$
Since the two vectors are linearly independent, they form the basis for the set and,
the dimension is 2.
