## Another basis and dimension question

Consider the set $T=\{\mathbf{v_1, v_2, v_3, v_4, x}\}$ where

$\mathbf{v_1}=\begin{pmatrix}1\\ 2\\ -1\\0\end{pmatrix}, \mathbf{v_2}=\begin{pmatrix}3\\ 6\\ -3\\0\end{pmatrix}, \mathbf{v_3}=\begin{pmatrix}2\\ 1\\ -1\\4\end{pmatrix}, \mathbf{v_4}=\begin{pmatrix}-1\\ -5\\ 2\\4\end{pmatrix}, \mathbf{v_4}=\begin{pmatrix}6\\ -3\\ -1\\20\end{pmatrix}$

a) Find a basis B for span $({\mathbf{v_1, v_2, v_3, v_4, x})$

For this question, would you just write the columns out in a matrix, then reduce into reduced row-echelon form.
The leading (pivot) columns would then form a basis B

b) Explain why $\mathbf{x}$ belongs to span span $({\mathbf{v_1, v_2, v_3, v_4})$. Write $\mathbf{x}$ as a linear combination of B.

Since the vectors in part (a) form a basis, does that mean that any vector in $\mathbb{R}^4$ can be written as a linear combination of the vectors in B

c) Suppose the matrix A has columns $\mathbf{v_1, v_2, v_3, v_4}$. what is the dimension of the column space of A?

I don't really understand what column space is?