How to show this set is a basis? I want to show that a basis for $\mathbb{R}^4$ is: $ \begin{pmatrix} 1 \\ 2 \\ 0 \\ 0 \end{pmatrix},\begin{pmatrix} 2 \\ 2 \\ 1 \\ -2 \end{pmatrix}, \begin{pmatrix} 1 \\ 4 \\ 0 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$. 
I know that I need to show linear independence and that these span $\mathbb{R}^4$. However, this was a past exam question and I am not really sure that row reduction/solving systems (for only 4 marks) is what they want. Is there an easier/alternative way to show that they span the space and are LI? 
Thank you :)
 A: For this matrix, it is easy to compute the determinant:
$$
\begin{vmatrix}
1 & 2 & 1 & 0 \\
2 & 2 & 4 & 0 \\
0 & 1 & 0 & 0 \\
0 & -2 & 0 & \color{red} 1 \\
\end{vmatrix} 
=
\begin{vmatrix}
1 & 2 & 1 \\
2 & 2 & 4 \\
0 & \color{red} 1 & 0 \\
\end{vmatrix} 
=
-\begin{vmatrix}
1 & 1 \\
2 & 4 \\
\end{vmatrix}
\ne 0
$$
A: While I suggest to show this using the determinant, here is one easy and quick way to do it without determinants:
As we have 4 vectors, we only need to show that they are linearly independent. Let $a,b,c,d\in \mathbb R$ with
$$a\cdot\begin{pmatrix} 1 \\ 2 \\ 0 \\ 0 \end{pmatrix}+b\cdot\begin{pmatrix} 2 \\ 2 \\ 1 \\ -2 \end{pmatrix} + c\cdot \begin{pmatrix} 1 \\ 4 \\ 0 \\ 0 \end{pmatrix}+d\cdot\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}.$$
It follows immediately that $b=0$ by looking at the third row of each vector, so this leaves us with
$$a\cdot\begin{pmatrix} 1 \\ 2 \\ 0 \\ 0 \end{pmatrix} + c\cdot \begin{pmatrix} 1 \\ 4 \\ 0 \\ 0 \end{pmatrix}+d\cdot\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}.$$
The same argument yields $d=0$ by looking at the fourth row and from
$$a\cdot\begin{pmatrix} 1 \\ 2 \\ 0 \\ 0 \end{pmatrix} + c\cdot \begin{pmatrix} 1 \\ 4 \\ 0 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$
we can easily conclude $a=c=0$ (otherwise we would get a contradiction with $a=-c$ and $a=-2c$). 
A: If calculating determinate of the matrix sounds difficult to you, you may show that A^(-1) exists.
Showing that determinate is non zero is one way. Another way that finding A^(-1) by Gauss- Jordan method.
Start  using elementary matrices to change A to U (upper triangle). Then continue to change U into a diagonal matrix. In the last step, factor any numbers which are not equal to 1. You have identity matrix . You can claim that A was invertible because after multiplying to a matrix, it has changed into identity and you can find A^(-1) by writing elementary matrices and multiplying them respectively and also because “For a square matrix, left and right inverses are equal” so can claim that it is A^(-1) and because its existence, for rows(and columns) of A is independent. 
Because this set of vectors is maximal independent (4 vectors is maximum number of vectors in R^4 to be independent) So, it is a basis for R^4. There are so many other ways, another one is showing that it is minimal generator
