Where This Fits in the Book
This note continues Gilbert Strang’s Introduction to Linear Algebra, Chapter 3, Section 3.2: The Nullspace of A: Solving Ax = 0 and Rx = 0.
The important lesson from this study session was that the reduced matrix should not appear from nowhere. To understand the nullspace properly, I need to see the complete chain:
\[\boxed{A \rightarrow Ax=0 \rightarrow \text{elimination} \rightarrow \text{free variables} \rightarrow \text{special solutions} \rightarrow N(A)}\]
The Problem We Are Trying to Solve
The nullspace is defined by
\[N(A)=\{x:Ax=0\}.\]
But knowing the definition is not enough. The practical question is:
Given an actual matrix A, how do I find every vector x that A sends to zero?
This is where Gaussian elimination becomes useful again.
Start With the Matrix A
Consider
\[A=\begin{bmatrix}1&2\\3&6\end{bmatrix}.\]
We want to solve
\[Ax=0.\]
Let
\[x=\begin{bmatrix}x_1\\x_2\end{bmatrix}.\]
Then
\[\begin{bmatrix}1&2\\3&6\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}.\]
Multiplying row by column gives the equations
\[x_1+2x_2=0\]
and
\[3x_1+6x_2=0.\]
Notice the Redundancy Before Eliminating
The second equation is exactly three times the first:
\[3(x_1+2x_2)=3x_1+6x_2.\]
Therefore the second equation gives no new information.
This connects directly with the mental picture developed earlier:
\[\boxed{\text{dependence means redundancy}.}\]
Although there are two written equations, there is really only one independent restriction on the two unknowns.
Elimination Exposes the Redundancy
Perform the row operation
\[R_2\leftarrow R_2-3R_1.\]
Then
\[\begin{bmatrix}1&2\\3&6\end{bmatrix}\longrightarrow\begin{bmatrix}1&2\\0&0\end{bmatrix}.\]
The zero row is elimination’s way of exposing the fact that the second equation contained no new information.
We are left with only
\[x_1+2x_2=0.\]
Why Does a Free Variable Appear?
We have one genuine equation but two unknowns.
The equation says
\[x_1=-2x_2.\]
It does not tell us one unique value for \(x_2\).
We may choose \(x_2\), and once we choose it, the equation determines \(x_1\).
This gives the useful mental distinction:
\[\boxed{\text{free variable = a choice we are still allowed to make}}\]
while a pivot variable is determined from those choices by the equations.
In this example, \(x_2\) is free and \(x_1\) is determined by
\[x_1=-2x_2.\]
Why Set the Free Variable Equal to 1?
We could choose any value for \(x_2\). For example, if \(x_2=5\), then \(x_1=-10\).
But choosing
\[x_2=1\]
is especially convenient because it exposes the basic direction of all the solutions.
Then
\[x_1=-2.\]
So we obtain
\[s=\begin{bmatrix}-2\\1\end{bmatrix}.\]
This is called a special solution.
The value 1 is not mathematically magical. It is simply the cleanest choice for isolating the direction associated with a free variable.
The Special Solution Is Not the Only Solution
If we choose
\[x_2=5,\]
then
\[x_1=-10\]
and
\[x=\begin{bmatrix}-10\\5\end{bmatrix}=5\begin{bmatrix}-2\\1\end{bmatrix}.\]
If instead
\[x_2=-3,\]
then
\[x_1=6\]
and
\[x=\begin{bmatrix}6\\-3\end{bmatrix}=-3\begin{bmatrix}-2\\1\end{bmatrix}.\]
Every possible solution is therefore a scalar multiple of the same special solution.
The Complete Nullspace
We can describe every solution at once by writing
\[x=c\begin{bmatrix}-2\\1\end{bmatrix},\qquad c\in\mathbb R.\]
Therefore
\[\boxed{N(A)=\left\{c\begin{bmatrix}-2\\1\end{bmatrix}:c\in\mathbb R\right\}.}\]
Geometrically, this nullspace is a line through the origin in the input space.
Connect This With the Earlier Mental Picture
Previously I learned to think of the nullspace as
\[\boxed{\text{input directions that A maps to zero}.}\]
Now we have actually calculated such a direction:
\[\begin{bmatrix}-2\\1\end{bmatrix}.\]
Check it directly:
\[A\begin{bmatrix}-2\\1\end{bmatrix}=\begin{bmatrix}1&2\\3&6\end{bmatrix}\begin{bmatrix}-2\\1\end{bmatrix}=\begin{bmatrix}-2+2\\-6+6\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}.\]
So this really is a direction that the transformation A completely erases.
Every multiple of this direction is also erased because
\[A(cs)=cAs=c0=0.\]
Why Elimination Matters
For this small matrix, the redundancy was easy to see by inspection. The second row was visibly three times the first.
For a large matrix, however, those relationships may be hidden.
Gaussian elimination systematically exposes them.
That is why elimination is appearing again in the study of nullspaces. We are not learning a completely unrelated procedure. We are using a procedure we already know to reveal the structure of the solution space.
The Important Chain to Remember
The conceptual flow is
\[A\]
\[\downarrow\]
\[Ax=0\]
\[\downarrow\]
\[\text{Gaussian elimination reveals the independent equations}\]
\[\downarrow\]
\[\text{identify pivot and free variables}\]
\[\downarrow\]
\[\text{set a free variable to 1}\]
\[\downarrow\]
\[\text{obtain a special solution}\]
\[\downarrow\]
\[\boxed{\text{linear combinations of special solutions give the nullspace}}\]
Current Mental Picture
I should not think of a special solution as something that appears mysteriously from a reduced matrix.
I start with the original matrix A and ask which inputs it sends to zero.
Elimination removes redundant information and exposes which variables are still free.
Choosing a simple value such as 1 for a free variable reveals one basic direction of freedom.
That direction is a special solution.
Its scalar multiples, and later combinations of multiple special solutions when there are multiple free variables, describe the complete nullspace.
The strongest memory chain from this lesson is:
\[\boxed{\text{redundant equations}\rightarrow\text{free variables}\rightarrow\text{special solutions}\rightarrow\text{nullspace}}.\]
Note metadata
- Note type: learning-note
- Subject: learning-notes
- Source: Introduction to Linear Algebra, Fifth Edition