The Question We Were Trying to Answer
After learning that row elimination changes a matrix \(A\) into a simpler matrix \(R\) without changing the solutions of the homogeneous system, the next question was:
How do we actually calculate \(N(R)\)?
The answer is:
\[ N(R)=\{x:Rx=0\}. \]So finding the nullspace of \(R\) means solving \(Rx=0\).
Why We Can Work With R Instead of A
Gaussian elimination performs reversible row operations. If elimination changes \(A\) into \(R\), the systems
\[ Ax=0 \]and
\[ Rx=0 \]have exactly the same solution vectors \(x\). Therefore:
\[ N(A)=N(R). \]This can also be seen using an elimination matrix. If \(EA=R\) and \(Ax=0\), then
\[ EAx=E0=0, \]which gives
\[ Rx=0. \]Because the elimination operation is reversible, the implication works in both directions. Row reduction changes the appearance of the equations but not their solution vectors.
This is useful because \(R\) exposes the pivot variables and free variables much more clearly than the original matrix.
Calculating N(R): A Small Example
Suppose elimination produces
\[ R=\begin{bmatrix}1&2\\0&0\end{bmatrix}. \]To calculate \(N(R)\), solve
\[ R\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}. \]This gives
\[ x_1+2x_2=0. \]Column 1 contains a pivot, so \(x_1\) is the pivot variable. Column 2 has no pivot, so \(x_2\) is free.
Set
\[ x_2=t. \]Then
\[ x_1=-2t, \]and therefore
\[ x=t\begin{bmatrix}-2\\1\end{bmatrix}. \]Thus
\[ N(R)=\left\{t\begin{bmatrix}-2\\1\end{bmatrix}:t\in\mathbb R\right\}. \]There is no separate mysterious formula for calculating \(N(R)\). We simply solve \(Rx=0\).
Moving to More Than One Free Variable
To see why special solutions are useful, consider a reduced system such as
\[ R=\begin{bmatrix}1&0&2&3\\0&1&4&5\end{bmatrix}. \]There are four variables because there are four columns:
\[ x=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}. \]Solving \(Rx=0\) gives
\[ x_1+2x_3+3x_4=0, \] \[ x_2+4x_3+5x_4=0. \]The pivots are in columns 1 and 2. Therefore \(x_1\) and \(x_2\) are pivot variables. Columns 3 and 4 contain no pivots, so \(x_3\) and \(x_4\) are free variables.
The practical meaning is:
We are free to choose \(x_3\) and \(x_4\). Once those choices are made, the equations determine \(x_1\) and \(x_2\).
Rearranging the equations gives
\[ x_1=-2x_3-3x_4, \] \[ x_2=-4x_3-5x_4. \]Therefore every vector in the nullspace has the form
\[ x=\begin{bmatrix}-2x_3-3x_4\\-4x_3-5x_4\\x_3\\x_4\end{bmatrix}. \]Checking a Particular Choice
For example, choose
\[ x_3=2,\qquad x_4=1. \]Then
\[ x_1=-7,\qquad x_2=-13, \]so
\[ x=\begin{bmatrix}-7\\-13\\2\\1\end{bmatrix}. \]Checking directly:
\[ R\begin{bmatrix}-7\\-13\\2\\1\end{bmatrix} =\begin{bmatrix}-7+4+3\\-13+8+5\end{bmatrix} =\begin{bmatrix}0\\0\end{bmatrix}. \]Therefore this vector belongs to \(N(R)\).
Why Special Solutions Are Introduced
The general nullspace vector contains two independent choices, \(x_3\) and \(x_4\). We can isolate the effect of each free variable.
First choose
\[ x_3=1,\qquad x_4=0. \]This gives the first special solution:
\[ s_1=\begin{bmatrix}-2\\-4\\1\\0\end{bmatrix}. \]Then choose
\[ x_3=0,\qquad x_4=1. \]This gives the second special solution:
\[ s_2=\begin{bmatrix}-3\\-5\\0\\1\end{bmatrix}. \]The choices 0 and 1 are not restrictions on the free variables. They are simply a convenient way to isolate one independent freedom at a time.
Reconstructing Every Solution
Return to the general solution:
\[ x=\begin{bmatrix}-2x_3-3x_4\\-4x_3-5x_4\\x_3\\x_4\end{bmatrix}. \]It can be separated into
\[ x=x_3\begin{bmatrix}-2\\-4\\1\\0\end{bmatrix}+x_4\begin{bmatrix}-3\\-5\\0\\1\end{bmatrix}. \]Therefore
\[ x=x_3s_1+x_4s_2. \]This shows that every vector in the nullspace can be constructed from the two special solutions.
Nullspace as a Span
The word span means all possible linear combinations. Therefore:
\[ N(R)=\operatorname{span}\{s_1,s_2\}. \]If \(R\) came from \(A\) by row elimination, then
\[ N(A)=N(R), \]so the same special solutions describe the nullspace of \(A\):
\[ N(A)=\operatorname{span}\{s_1,s_2\}. \]The Mental Picture
The full reasoning chain is:
\[ A\xrightarrow{\text{elimination}}R \xrightarrow{\text{solve }Rx=0} \text{free variables} \xrightarrow{\text{set one free variable to 1 at a time}} \text{special solutions} \xrightarrow{\text{all linear combinations}} N(A). \]The key intuition is:
Each free variable gives an independent freedom in the solution. A special solution isolates that freedom. Combining all the special solutions produces the entire nullspace.
Note metadata
- Note type: learning-note
- Subject: learning-notes
- Source: Introduction to Linear Algebra, 5th Edition — Chapter 3, Section 3.2