Rank-Nullity Theorem: Dimensions of Subspaces

1. Defining Rank and Nullity

First, let's give formal names to the dimensions of the subspaces we studied.

• The rank of a matrix $A$, denoted $\text{rank}(A)$, is the dimension of its column space.
• The nullity of a matrix $A$, denoted $\text{nullity}(A)$, is the dimension of its null space.

From our last lesson, we know how to find these dimensions by row reducing a matrix to its reduced row echelon form (RREF):

• $\text{rank}(A)=\dim (\text{Col}(A))=\text{number of pivot columns}$
• $\text{nullity}(A)=\dim (\text{Nul}(A))=\text{number of free variables}$

Since every column in the RREF either contains a pivot or corresponds to a free variable, the total number of columns must be the sum of these two counts. This simple observation gives us our theorem.

So, the theorem is surprisingly straightforward to state:

The Rank-Nullity Theorem
For any $m\times n$ matrix $A$,

where $n$ is the number of columns of $A$.

Thinking in terms of a linear transformation $T:{\mathbb{R}}^n\to {\mathbb{R}}^m$ where $T(\mathbf{x})=A\mathbf{x}$, this translates to:

This is a profound statement. It says the dimension of the input space is split between the dimensions of the part that gets collapsed to zero (the kernel) and the part that survives as the output (the image).

2. Building Intuition: A Conservation of Dimension

To truly understand why this theorem holds, it's helpful to visualize what's happening to the basis vectors of the domain.

Let's say our domain $V$ has dimension $n$. We can construct a basis for $V$ in a clever way. First, find a basis for the null space (kernel) of our transformation $T$. Let's say the nullity is $k$, so this basis has $k$ vectors: $\{{\mathbf{v}}_1,\dots ,{\mathbf{v}}_k\}$.

These $k$ vectors all get sent to $0$ by the transformation. We can then extend this set to a full basis for the domain $V$ by adding $n-k$ more vectors: $\{{\mathbf{v}}_1,\dots ,{\mathbf{v}}_k,{\mathbf{v}}_{k+1},\dots ,{\mathbf{v}}_n\}$.

What happens when we apply the transformation $T$ to these basis vectors?

• $T({\mathbf{v}}_1),\dots ,T({\mathbf{v}}_k)$ are all $0$. They contribute nothing to the image.
• What about $T({\mathbf{v}}_{k+1}),\dots ,T({\mathbf{v}}_n)$? It turns out these $n-k$ vectors form a basis for the image of $T$. Therefore, the dimension of the image (the rank) is exactly $n-k$.

So, we have:

• Nullity = $k$
• Rank = $n-k$
• Rank + Nullity = $(n-k)+k=n$, which is the dimension of the domain.

This intuitive proof shows that the dimensions must add up. The dimensions of the domain are perfectly partitioned—some dimensions collapse into the null space, and the remaining dimensions are preserved to form the image.

3. Applying the Theorem: Computation

The Rank-Nullity theorem is extremely practical. If you know any two of the three quantities (rank, nullity, number of columns), you can immediately find the third. Often, it's easier to compute the rank and then use the theorem to find the nullity, rather than solving the entire homogeneous system $A\mathbf{x}=0$.

The standard workflow is:

1. Take your matrix $A$ and reduce it to row echelon form.
2. Count the number of pivots. This is the rank.
3. The nullity is then simply the total number of columns minus the rank.

4. Application Beyond Computation

The real power of the Rank-Nullity Theorem is in making qualitative statements about systems and transformations without solving them. It provides constraints on what's possible.

Let's consider a practical example.

Let's break down the logic from that example:

• The Problem: We have a system of 3 equations in 5 variables, represented by a $3\times 5$ matrix $A$. We are told the system $A\mathbf{x}=\mathbf{b}$ is consistent for all possible right-hand sides $\mathbf{b}$ in ${\mathbb{R}}^3$. We need to find the number of linearly independent solutions to the homogeneous system $A\mathbf{x}=0$.
• Translation:
  • "5 variables" means the domain is ${\mathbb{R}}^5$, so $n=5$.
  • "Consistent for all $\mathbf{b}$ in ${\mathbb{R}}^3$" means the column space of $A$ must be all of ${\mathbb{R}}^3$. Therefore, $\dim (\text{Col}(A))=\text{rank}(A)=3$.
  • "Number of linearly independent solutions to $A\mathbf{x}=0$" is just another way of asking for the dimension of the null space, $\text{nullity}(A)$.
• Applying the Theorem:
  $\text{rank}(A)+\text{nullity}(A)=n$
  $3+\text{nullity}(A)=5$
  $\text{nullity}(A)=2$
• Conclusion: There are exactly 2 linearly independent solutions to the homogeneous system.

This shows how the theorem connects the properties of the input space, output space, and the transformation itself into a single, powerful equation.

Conclusion

In this lesson, we formalized the relationship between the dimensions of the kernel and image of a linear transformation.

Key Takeaways:

• Rank is the dimension of the column space (image), and it equals the number of pivot columns in the matrix.
• Nullity is the dimension of the null space (kernel), and it equals the number of free variables.
• The Rank-Nullity Theorem states that for an $m\times n$ matrix $A$, $\text{rank}(A)+\text{nullity}(A)=n$.
• This theorem expresses a fundamental trade-off: for a given input dimension, a transformation that collapses more dimensions (higher nullity) must produce a lower-dimensional output (lower rank), and vice-versa.

Preview of the Next Lesson:
We've seen that a transformation with a nullity greater than zero maps multiple inputs to the same output (it's not one-to-one). This has a direct consequence for whether the transformation can be "reversed." In our next lesson, we will investigate the concept of invertibility. We will see how the Rank-Nullity Theorem provides one of several powerful criteria for determining if a linear transformation is invertible.