Understanding Ranks and Nullity through Systems of Equations
Historical Background
Historically, the invention of matrices was primarily aimed at simplifying and conveniently notating systems of linear equations. For instance, observing the system of linear equations
$$ \begin{cases} 2x_{1} & + & x_{2} & + & x_{3} =& 0 \\ & x_{2} & =& 0 \end{cases} $$
reveals the inconvenience of having to use the same variables multiple times. Representing this as a matrix leads to
$$ \begin{bmatrix} 2 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$
where it can be neatly expressed in an organized form. The importance of simplifying equations and manipulating them easily goes without saying. As systems of linear equations become larger and more complex, such techniques become increasingly useful, leading to the development of matrix algebra. Furthermore, the exploration of linear structures and the generalization of spaces culminated in the creation of what is known today as linear algebra.
Reasons for Difficulty in Understanding
However, from the learning perspective, it can be challenging to keep up with linear algebra as it becomes more abstract with the introduction of concepts like column spaces and null spaces. Initially, vectors might have looked like $(1,0, \cdots , 0)$, but suddenly matrices disappear and new concepts start pouring in.
For example, expressions like $\dim \mathcal{C} (A) = \text{rank}(A)$ or $\dim \mathcal{N} (A) = \text{nullity} (A)$ might be neat to write, but difficult to comprehend. Returning to the concept of systems of linear equations, one can easily understand $\text{rank}$ and $\text{nullity}$.
Example
Looking at the earlier example,
$$ \begin{cases} 2x_{1} & + & x_{2} & + & x_{3} =& 0 \\ & x_{2} & =& 0 \end{cases} $$
there are 2 equations but 3 unknowns. As it is known, a system of linear equations will not have a unique and obvious solution unless the number of equations matches the number of unknowns. Therefore, if $x_2 = 0$ and consequently $2 x_1 = - x_3$, one can express the solution using only 2 variables. In this sense, think of $\text{rank}$ as the ’number of variables actually used’, and $\text{nullity}$ as the ’number of variables not used'.
For $A \in \mathbb{R}^{ m \times n }$,
$$ \begin{align*} \text{rank} (A) + \text{nullity} (A) =& \dim \mathbb{R}^{n} = n \\ \text{rank} (A^{T}) + \text{nullity} (A^{T}) =& \dim \mathbb{R}^{m} = m \end{align*} $$
Let’s verify if the explanation obtained using the theorem aligns.
$$ A = \begin{bmatrix} 2 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} $$ then $A \in \mathbb{R}^{ 2 \times 3 }$.
$$ \dim \mathcal{C} (A) = \dim \text{span} \left\{ \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \right\} = 2 $$
and
$$ \dim \mathcal{N} (A) = \dim \text{span} \left\{ \begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix} \right\} = 1 $$
thus
$$ \text{rank} (A) + \text{nullity}(A) = 2 + 1 = 3 = \dim (\mathbb{R}^{3}) $$