Why Rank Is Needed
The number of rows and columns tells us the physical size of a matrix, but it does not always tell us how much independent information the matrix contains.
For example, a row may disappear during elimination because it is already a combination of earlier rows. A row that becomes
\[ 0=0 \]does not add a new constraint.
Strang therefore defines the rank of a matrix as
\[ \boxed{\operatorname{rank}(A)=\text{number of pivots}}. \]The rank is denoted by \(r\).
Rank and Redundant Columns
Consider the matrix
\[ A= \begin{bmatrix} 1&1&2&4\\ 1&2&2&5\\ 1&3&2&6 \end{bmatrix}. \]Its columns are
\[ a_1=\begin{bmatrix}1\\1\\1\end{bmatrix},\qquad a_2=\begin{bmatrix}1\\2\\3\end{bmatrix},\qquad a_3=\begin{bmatrix}2\\2\\2\end{bmatrix},\qquad a_4=\begin{bmatrix}4\\5\\6\end{bmatrix}. \]The third column is
\[ a_3=2a_1, \]so it adds no new direction.
The fourth column satisfies
\[ a_4=3a_1+a_2. \]Therefore columns 3 and 4 are redundant relative to the first two columns.
The reduced matrix has the form
\[ R= \begin{bmatrix} 1&0&2&3\\ 0&1&0&1\\ 0&0&0&0 \end{bmatrix}. \]There are two pivots, so
\[ \boxed{r=2}. \]The pivot columns identify the genuinely independent directions that survive elimination.
Free Columns and Special Solutions
The free variables are \(x_3\) and \(x_4\). Solving
\[ Rx=0 \]gives
\[ x_1+2x_3+3x_4=0, \] \[ x_2+x_4=0. \]Setting \(x_3=1\) and \(x_4=0\) gives the special solution
\[ s_1= \begin{bmatrix} -2\\0\\1\\0 \end{bmatrix}. \]Since \(As_1=0\), this means
\[ -2a_1+a_3=0, \]so
\[ \boxed{a_3=2a_1}. \]The second special solution is obtained from \(x_3=0\), \(x_4=1\):
\[ s_2= \begin{bmatrix} -3\\-1\\0\\1 \end{bmatrix}. \]Then
\[ -3a_1-a_2+a_4=0, \]which gives
\[ \boxed{a_4=3a_1+a_2}. \]This gives a concrete interpretation of special nullspace solutions:
\[ \boxed{\text{A special solution records how a free column is built from pivot columns.}} \]What Rank One Means
A rank-one matrix has exactly one pivot.
Strang’s example is
\[ A= \begin{bmatrix} 1&3&10\\ 2&6&20\\ 3&9&30 \end{bmatrix}. \]Its columns are
\[ a_1=\begin{bmatrix}1\\2\\3\end{bmatrix},\qquad a_2=3a_1,\qquad a_3=10a_1. \]All columns therefore lie on the same line. There is only one independent column direction.
After elimination,
\[ R= \begin{bmatrix} 1&3&10\\ 0&0&0\\ 0&0&0 \end{bmatrix}. \]There is only one pivot:
\[ \boxed{r=1}. \]Counting Free Variables
The rank-one example has
\[ n=3,\qquad r=1. \]Therefore the number of free variables is
\[ n-r=3-1=2. \]From
\[ x_1+3x_2+10x_3=0, \]the two special solutions are
\[ s_1=\begin{bmatrix}-3\\1\\0\end{bmatrix}, \qquad s_2=\begin{bmatrix}-10\\0\\1\end{bmatrix}. \]They encode the column relationships
\[ a_2=3a_1 \]and
\[ a_3=10a_1. \]The Rank-One Form A = uv^T
Because every column of a rank-one matrix is a multiple of one common vector, the matrix can be written as a column vector times a row vector.
For the example, let
\[ u= \begin{bmatrix} 1\\2\\3 \end{bmatrix} \]and
\[ v^T= \begin{bmatrix} 1&3&10 \end{bmatrix}. \]Then
\[ A=uv^T. \]Indeed,
\[ \begin{bmatrix} 1\\2\\3 \end{bmatrix} \begin{bmatrix} 1&3&10 \end{bmatrix} = \begin{bmatrix} 1&3&10\\ 2&6&20\\ 3&9&30 \end{bmatrix}. \]The meaning is:
\(u\) gives the common column direction, while \(v^T\) records how much of that direction appears in each column.
Rank One and the Nullspace
Since
\[ A=uv^T, \]the nullspace equation becomes
\[ Ax=uv^Tx=0. \]The product \(v^Tx\) is one scalar:
\[ v^Tx=x_1+3x_2+10x_3. \]Because \(u\neq0\), the equation
\[ u(v^Tx)=0 \]requires
\[ \boxed{v^Tx=0}. \]But \(v^Tx\) is the dot product \(v\cdot x\). Therefore
\[ v^Tx=0 \]means that every nullspace vector is perpendicular to \(v\).
The Geometry
For this rank-one example, the row space has only one independent direction, so it is a line.
The nullspace consists of all vectors perpendicular to that row-space direction.
In \(\mathbb R^3\), those perpendicular vectors form a plane through the origin:
\[ \boxed{\text{row space = line}} \]and
\[ \boxed{\text{nullspace = perpendicular plane}}. \]This agrees with the counting rule:
\[ n-r=3-1=2. \]The nullspace therefore has two independent directions, which geometrically form a plane.
Recognizing Rank One by Inspection
Strang’s Example 4 includes
\[ A= \begin{bmatrix} 1&3&4\\ 2&6&8 \end{bmatrix}. \]The second row is
\[ R_2=2R_1. \]So elimination gives
\[ \begin{bmatrix} 1&3&4\\ 0&0&0 \end{bmatrix}. \]There is only one pivot, so
\[ \boxed{\operatorname{rank}(A)=1}. \]The same conclusion appears from the columns:
\[ a_2=3a_1, \qquad a_3=4a_1. \]Therefore all columns also lie along one direction.
The Mental Picture
The main chain is:
\[ \text{number of pivots} \longrightarrow \text{rank }r \longrightarrow \text{number of genuinely independent directions}. \]For rank one:
\[ \boxed{ \begin{array}{c} 1\text{ pivot}\\ \Downarrow\\ r=1\\ \Downarrow\\ \text{all rows depend on one row direction}\\ \Downarrow\\ \text{all columns depend on one column direction}\\ \Downarrow\\ A=uv^T \end{array}} \]For an \(m\times n\) matrix:
\[ \boxed{\text{number of free variables}=n-r}. \]Each free variable produces a special solution, and those special solutions describe the nullspace.
Note metadata
- Note type: learning-note
- Subject: learning-notes
- Source: Introduction to Linear Algebra, 5th Edition — Chapter 3, Section 3.2