Show vectors are linearly independent and finding a basis Show that the following vectors are linearly independent -
$v1 = (1, 2, 0, 2)$
$v2 = (1,1,1,0)$
$v3 = (2,0,1,3)$
Find a fourth vector v4 so that the set { v1, v2, v3, v4 } is a basis fpr $\mathbb{R}^4$?
Answer:
I put the 3 vectors into the columns of a matrix and row reduced and got
$$
\begin{bmatrix}
1 & 1 & 2\\
0 & -1 & -4\\
0 & 0 & -3\\
0 & 0 & 7\\
\end{bmatrix}
$$
But something doesnt look right about that to me. And linear independence means there are the same number of pivots as columns, but I have 4 pivots...
I have no idea how to find a fourth vector that makes the set a basis for $\mathbb{R}^4$.
 A: Your reduced matrix is not in echelon form; you need to do one more stage with your row reduction. Then you'll see the fourth row will consist of  all zeroes. So, then, the three columns of the row reduced form, and thus the three columns of the original matrix, are linearly independent.
To find the one vector $(a,b,c,d)$ that you need to complete a basis (the space has dimension four), you could start with the matrix
$$\tag{1} 
\left[\matrix {1&1&2&a\cr2&1&0&b\cr 0&1&1&c\cr 2&0&3&d}\right].
$$
Row reduce this and in the end select $a,b,c,d$ so that the reduced form has independent columns (four pivots).

I did the row-reduction suggested above and obtained
$$\tag{2}
\left[\matrix {1&1&2&a\cr0&-1&-4&b-2a\cr 0&0&21&7b-14a+7c\cr 0&0&0&\color{maroon}{-20a+13b+7c-3d}}\right]
$$
Assuming I made no errors, 
taking $a=1$ and $b=c=d=0$ in $(2)$ will give us a matrix with independent columns (we want
$\color{maroon}{-20a+13b+7c-3d}\ne 0$).  So we can take $(1,0,0,0)$ to be the forth vector needed to complete a basis. 
A: Showing that a set of vectors is linearly independent can be done by putting them together in a matrix and verifying that it has full rank. Completing a basis can be done by adding a basis for the left nullspace of the matrix with the given basis vectors as columns.
Calculating the rank of a matrix can be done in many ways, see Wikipedia, “Rank (linear algebra)” section “Computation”. Summarized: bringing the matrix into echelon form is a pedagogical approach, but may not be the most numerically dependable.
The left nullspace is the same as the (right) nullspace of the transpose of the matrix. Calculating the (right) nullspace can also be done by bringing the matrix into echelon form, see Wikipedia, “Kernel (matrix)” section “Basis”. Also here, another approach may be numerically preferable, see Wikipedia, “Kernel (matrix)” section “Computation on a computer”.
For your example, starting from
\begin{equation}
  A = 
  \begin{pmatrix}
     1 & 1 & 2\\
     2 & 1 & 0\\
     0 & 1 & 1\\
     2 & 0 & 3
  \end{pmatrix}
\end{equation}
the (row reduced) echelon form is
\begin{equation}
  A_{\text{rref}} =
  \begin{pmatrix}
     1 & 0 & 0\\
     0 & 1 & 0\\
     0 & 0 & 1\\
     0 & 0 & 0
  \end{pmatrix},
\end{equation}
which shows that the matrix has full rank (i.e., $3$).
A basis for the left nullspace is
\begin{equation}
  A_{\text{lnull}} =
  \begin{pmatrix}
     -8\\
     1\\
     7\\
     3
  \end{pmatrix};
\end{equation}
you can verify that 
\begin{equation}
  A_{\text{lnull}}^t A =
  \begin{pmatrix}
     0 &
     0 &
     0
  \end{pmatrix},
\end{equation}
which shows that the columns of $A_{\text{lnull}}$ (here a single one) are indeed orthogonal to the columns of $A$.
