# Row, Column and Null Space

We have now seen what a vector space is and how to determine if a given set forms a basis of that vector space. We now turn our attention to finding a basis a vector space spanned by a given set of vectors. One such way to do this is to turn the vector into either the rows or columns of a matrix, then find the row or column space of that matrix. That is what we will focus on in this post.

## Matrix

We let
\begin{align*}
\begin{bmatrix}
1 & 2 & 4 & 66 \\
3 & 6 & 1 & 11 \\
7 & 14 & 6 & 88
\end{bmatrix}
\end{align*}
with this matrix we will rank of the matrix, and the bases of the row, column and null space.

### Row Reduce

In order to find all of these, we will need to begin by row reducing the matrix. If you would like to review row reducing, I would suggest seeing the posts Solving a System of Equations, Inverse Matrix or Finding a Determinant (Part 2). Since we have seen how to do this, we will instead using a calculator to find that this matrix rew reduces to the matrix
\begin{align*}
\begin{bmatrix}
1 & 2 & 0 & -2 \\
0 & 0 & 1 & 7 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}

### Rank, row space, column space

Now that we have the row reduced echelon form of the matrix, we are ready to provide our solution. In particular, we find the rank of $$A$$ by counting the left most 1s in the matrix. That is, if we look at the first row, the left most nonzero number is in the first column and is a 1. In the second row, the left most nonzero number is in the third column and is a 1. However, there are no nonzero numbers in the third row.

Because we have two rows with leading 1s, the rank$$(A)=2$$.

Furthermore, the resulting rows with left most 1s will form a basis of the row space of $$A$$. That is, the basis of the row space of $$A$$ is the set $$\left\{(1,2,0,-2),(0,0,1,7)\right\}$$.

Also, the columns that have left most ones tell us which vectors to use as the basis for the column space. That is, we should use the original column vectors of $$A$$ that are in the columns with left most 1s. Therefore, the basis of the column space of $$A$$ is the set $$\left\{(1,3,7),(4,1,6)\right\}$$.

### Null space

In order to find the null space of $$A$$, we have to find solutions to $$A\mathbf{x}=\mathbf{0}$$. While doing this, we will notice that columns that do not correspond to left most 1s will be free variable. Therefore, we will let $$x_{2}=s$$ and $$x_{4}=t$$ be the free variables.

Using the first row, we see that $$x_{1}+2x_{2}-2x_{4}=0$$ in $$A\mathbf{x}=\mathbf{0}$$, so $$x_{1}=-2x_{2}+2x_{4}$$, since $$x_{2}=s$$ and $$x_{4}=t$$, we then get $$x_{1}=-2s+2t$$. Then, doing the same thing for the second row, we find that $$x_{3}-7x_{4}=0$$, so $$x_{3}=7x_{4}$$. Hence $$x_{3}=7t$$.

This gives us that the vector $$\mathbf{x}$$ is in the null space of $$A$$ if
\begin{align*}
\mathbf{x}=\begin{bmatrix} -2s+2t \\ s \\7t \\ t\end{bmatrix}
\end{align*}
for some $$s,t \in \mathbb{R}$$. Therefore, if we want to find a basis, we find the vectors that correspond to $$s=1,t=0$$ and $$s=0,t=1$$. This gives us that the basis of the null space is the set $$\left\{(-2,1,0,0),(2,0,1,1)\right\}$$.

## Conclusion

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