Did you know there’s an easy way to describe the fundamental relations between the dimensions of the column space, row space, and null space?
Jenn, Founder Calcworkshop®, 15+ Years Experience (Licensed & Certified Teacher)
Understanding the Row Space, Column Space, and Null Space
As we know, the row space is the set of all linear combinations of the row vectors and has the same dimension as the column space.
Defining Rank and Nullity
The rank of \(A\) is the dimension of the columns space of \(A\) and represents the number of pivot columns.
The nullity of \(\mathrm{A}\) is the dimension of the null space of \(\mathrm{A}\) and represents the number of free variables in the equation \(\mathrm{Ax}=0\).
The Rank Theorem
And together, the rank theorem states that the dimensions of the column space and the row space of an \(\mathrm{m} \times \mathrm{n}\) matrix \(\mathrm{A}\) are equal to the number of columns in \(\mathrm{A}\)! This common dimension, known as the rank of \(\mathrm{A}\), also equals the number of pivot positions in A and satisfies the following equations
\begin{align*}
\left\{\begin{array}{c}
\text { number of } \\
\text { pivot columns }
\end{array}\right\}+\left\{\begin{array}{c}
\text { number of } \\
\text { non-pivot columns }
\end{array}\right\}=\left\{\begin{array}{c}
\text { number of } \\
\text { columns }
\end{array}\right\}
\end{align*}
In other words, if \(\mathrm{A}\) is a matrix with \(\mathrm{n}\) columns, then
\begin{align*}
\underbrace{\operatorname{dim} \operatorname{Col}(A)}_{\operatorname{rank}(A)}+\underbrace{\operatorname{dim} \operatorname{Nul}(A)}_{\text {nullity }(A)}=\underbrace{n}_{\text {#of columns }}
\end{align*}
Additionally, if \(\mathrm{A}\) is an \(\mathrm{m} \times \mathrm{n}\) matrix, then the \(\operatorname{dim} \operatorname{Col}(\mathrm{A})\) and \(\operatorname{dim} \operatorname{Row}(\mathrm{A})\), known as the column rank and the row rank of \(\mathrm{A}\) respectively, are equal because \(\mathrm{A}\) and \(A^{T}\) have the same number of pivots.
Practical Application: Finding the Bases of Different Spaces
For example, assume that the matrix \(\mathrm{A}\) is row equivalent to \(\mathrm{B}\) and \(\mathrm{C}\).
Without calculations, find the bases for \(\operatorname{Col}(\mathrm{A})\), \(\operatorname{Row}(\mathrm{A})\), and \(\operatorname{Nul}(\mathrm{A})\) as well as the rank and nullity of \(A\).
\begin{align*}
\underbrace{\left[\begin{array}{cccc}
1 & -4 & 9 & -7 \\
-1 & 2 & -4 & 1 \\
5 & -6 & 10 & 7
\end{array}\right]}_{A} \sim \underbrace{\left[\begin{array}{cccc}
1 & 0 & -1 & 5 \\
0 & -2 & 5 & -6 \\
0 & 0 & 0 & 0
\end{array}\right]}_{B} \sim \underbrace{\left[\begin{array}{cccc}
1 & 0 & -1 & 5 \\
0 & 1 & -5 / 2 & 3 \\
0 & 0 & 0 & 0
\end{array}\right]}_{C}
\end{align*}
Calculation of Column Space, Row Space, and Null Space
First, as we learned from our Null, Column, Row Space video lesson, the column space is the original pivot columns from \(\mathrm{A}\), and the row space are the pivot rows from \(\mathrm{B}\).
\begin{aligned}
\operatorname{col} A & =\left\{\left[\begin{array}{c}
1 \\
-1 \\
5
\end{array}\right],\left[\begin{array}{c}
-4 \\
2 \\
-6
\end{array}\right]\right\} \\
\operatorname{row} A & =\left\{\left[\begin{array}{cccc}
1 & 0 & -1 & 5
\end{array}\right],\left[\begin{array}{cccc}
0 & -2 & 5 & -6
\end{array}\right]\right\}
\end{aligned}
And the null space is found from the non-pivot (free) columns in parametric vector form.
\begin{equation}
\left[\begin{array}{cccc}
1 & 0 & -1 & 5 \\
0 & 1 & -5 / 2 & 3 \\
0 & 0 & 0 & 0
\end{array}\right]
\end{equation}
\begin{aligned}
x_1 & =x_3-5 x_4 \\
x_2 & =\frac{5}{2} x_3-3 x_4 \\
x_3 & =x_3 \\
x_4 & =x_4
\end{aligned}
\begin{aligned}
\Rightarrow \vec{x} & =x_3\left[\begin{array}{c}
1 \\
5 / 2 \\
1 \\
0
\end{array}\right]+x_4\left[\begin{array}{c}
-5 \\
-3 \\
0 \\
1
\end{array}\right]
\end{aligned}
\begin{aligned}
n u l A & =\left\{\left[\begin{array}{c}
1 \\
5 / 2 \\
1 \\
0
\end{array}\right],\left[\begin{array}{c}
-5 \\
-3 \\
0 \\
1
\end{array}\right]\right\}
\end{aligned}
And the dimension for the column space and the row space (i.e., the rank of A) are both two, as there are two vectors in each basis.
The dimension of the null space (i.e., the nullity of \(\mathrm{A}\) ) is also 2, as there are two vectors in the basis.
And we can check the Rank Theorem to verify its authenticity, seeing as there are four columns in our matrix, \(2+2=4\), showing rank + Nullity \(=\mathrm{n}\)
Nice!
Importance of Rank and Nullity
But why is it important to know the rank and the nullity?
Because they help us to determine the number of solutions for a given system of linear equations, and in control theory, the rank of a matrix can be used to determine whether a linear system is controllable or observable.
This is to say, it helps us to determine the ability to move a system around in its configuration space using only specific manipulations.
Expansion of the Invertible Matrix Theorem (IMT)
And armed with this new information, we can expand our Invertible Matrix Theorem (IMT).
Letting \(\mathrm{A}\) be an \(\mathrm{n} \times \mathrm{n}\) matrix, the following statements are each equivalent to the statement that \(\mathrm{A}\) is an invertible matrix (Invertible Matrix Theorem continued):
- The columns of \(\mathrm{A}\) form a basis of \(\mathbb{R}^{n}\)
- \(\operatorname{Col}(A)=\mathbb{R}^{n}\)
- \(\operatorname{dim} \operatorname{Col}(A)=n\)
- \(\operatorname{rank}(A)=n\)
- \(\operatorname{Nul}(A)=\{\overrightarrow{0}\}\)
- \(\operatorname{dim} \operatorname{Nul}(A)=0\) (no free variables)
Next Steps
In this lesson, you will:
- Work through finding bases for the column space, row space, and null space
- Find the dimensions of rank \(A\) and the nullity
- Use this knowledge along with the invertible matrix theorem to answer facts about homogeneous and nonhomogeneous matrices
Jump right in and get started!
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