Blogs, Linear Algebra

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

As always, I hope you learned something and had some fun along the way. If you did, make sure to share this on Social Media and subscribe to our YouTube channel.

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