Finding a basis for the null space of a matrix The problem is to find the null space of the matrix
$$
A = \begin{bmatrix}1&2&1&-3\\2&4&4&-1\\3&6&7&1\end{bmatrix}.
$$
Does this have something to do with linear independence or dependence? A hint or solution would be appreciated!
 A: The nullspace (also called the kernel) of a matrix is the set of all vectors $x$ that satisfy $Ax = 0$. It so happens that the nullspace is a linear subspace of the domain of the map, so it is enough to find a basis for the nullspace (as so much else in linear algebra, this is related to linear independence since a basis for the nullspace is a maximal set of linearly independent vectors in the nullspace).
The easiest way to find the nullspace is to do row reudction on $A$. Wikipedia has an example: Wikipedia on nullspace.
A: First of all, if someone asks you to find a vector space satisfying some conditions, then you have to provide the basis of that space only. Because using that basis he or she can generate the whole space.
Now, the row-echelon form of $A$ is $\begin{pmatrix}1 & 2 & 1 & -3\\ 0 & 0 & 2 & 5\\ 0 & 0 & 0 & 0\end{pmatrix}$. So $rank(A) = 2$.
As we know, $rank(A) + dim(nullspace(A)) = \text{No. of columns of A}$, it gives $dim(nullspace(A)) = 4 - 2 = 2$, which means a particular basis of $nullspace(A)$ contains two linearly independent vectors. 
Again, $nullspace(A) = \left\{\underline{x} : A\underline{x} = \underline{0}\right\}$. Take any arbitrary $\underline{y} = (y_1, y_2, y_3, y_4)'$, where $\underline{y}$ is not necessarily a null vector. This $\underline{y}$ will belong to the $nullspace(A)$ iff it satisfies $A\underline{y} = \underline{0}$. Since rank of $A$ is 2, we will be able to have solution of only two variables freely, rest of the variables will be expressed in terms of other two variables.
Let's find the $y_i$'s. Here it is enough to solve the system,
$$
\begin{pmatrix}1 & 2 & 1 & -3\\
 0 & 0 & 2 & 5\\
 0 & 0 & 0 & 0\end{pmatrix}\underline{y} = \underline{0}
$$
which gives, $y_1 + 2y_2 + y_3 - 3y_4 = 0$ and $2y_3 + 5y_4 = 0$, which I have obtained as,
$$
y_3 = -\frac{1}{2}y_1 - y_2
$$ and 
$$
y_4 = \frac{1}{5}y_1 + \frac{2}{5}y_2
$$
Hence the general solution is,
$$
\underline{y} = \begin{pmatrix}y_1\\ y_2\\ -\frac{1}{2}y_1 - y_2\\ \frac{1}{5}y_1 + \frac{2}{5}y_2\end{pmatrix} = y_1\begin{pmatrix}1\\ 0\\ -\frac{1}{2}\\ \frac{1}{5}\end{pmatrix} + y_2\begin{pmatrix}0\\ 1\\ -1\\ \frac{2}{5}\end{pmatrix}\qquad(*)
$$, where $y_1$ and $y_2$ are arbitrary. Now it is clear that every vector belonging to nullspace will be of the form $(*)$, which is nothing but the linear combination of $(1, 0, -\frac{1}{2}, \frac{1}{5})'$ and $(0, 1, -1, \frac{2}{5})'$. Hence these two vectors form the basis of $nullspace(A)$.
Please check that these two vectors are linearly independent.
