Understanding Rank and Nullity Through Systems of Equations
Historical Background
Historically, the very reason matrices were devised was to represent systems of equations more easily and conveniently. For example, if we look carefully at the system of equations
$$ \begin{cases} 2x_{1} & + & x_{2} & + & x_{3} =& 0 \\ & x_{2} & =& 0 \end{cases} $$
we notice the inconvenience of having to write the same variable multiple times. Expressing this as a matrix,
$$ \begin{bmatrix} 2 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$
it can be represented neatly in a well-organized form. The importance of techniques for expressing formulas simply and manipulating them easily needs no further mention. The larger and more complex a system of equations became, the more useful such techniques proved to be, and matrix algebra developed accordingly. Then, as the exploration of linear structures and the generalization of spaces took place, it was completed under the name of linear algebra.
Why It Is Hard to Understand
However, from the learner’s standpoint, it can be difficult to keep up with linear algebra as it becomes increasingly abstract along with concepts such as column space and null space. What clearly started out as vectors that looked like $(1,0, \cdots , 0)$ suddenly have their matrices nowhere to be found, and a flood of new concepts pours in.
For example, expressions like $\dim \mathcal{C} (A) = \text{rank}(A)$ or $\dim \mathcal{N} (A) = \text{nullity} (A)$ are neat to write but hard to grasp in meaning. If we return here to the concept of systems of equations, we can easily understand this $\text{rank}$ and $\text{nullity}$.
Example
Looking at the
$$ \begin{cases} 2x_{1} & + & x_{2} & + & x_{3} =& 0 \\ & x_{2} & =& 0 \end{cases} $$
we used as an example above, there are 2 equations but 3 unknowns. As you know, a system of equations has a nontrivial yet unique solution only when the number of equations equals the number of unknowns. Therefore $x_2 = 0$ and $2 x_1 = - x_3$, so the solution can be expressed using only 2 variables. In this sense, when expressing the solution, let us 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 us check whether this matches the explanation we obtained using the theorem.
$$ A = \begin{bmatrix} 2 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} $$ If we let this be the case, 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
$$ \dim \mathcal{N} (A) = \dim \text{span} \left\{ \begin{bmatrix} 1 \\ 0 \\ -2 \end{bmatrix} \right\} = 1 $$
so
$$ \text{rank} (A) + \text{nullity}(A) = 2 + 1 = 3 = \dim (\mathbb{R}^{3}) $$
results.