Section 4.6: The Rank of a Matrix
We will use the concept of “rank” to develop and study the four fundamental subspaces associated to a matrix
A.
1. Definition: The row space of A, Row(A) is the span of the rows of A (which is Col(AT )).
2. Theorem: If two matrices A and B are row equivalent, then there row spaces are equivalent. If B is
in Echelon Form, then the nonzeros rows of B form a basis for both Row(A) and Row(B). We can use
either the RREF of A or the Echelon Form of A.
3. Remark: Row operations do not preserve the linear dependence relations among the rows (although
they do for the columns). This means, for example, that while the first k rows of B may be linearly
independent, while the first k rows of A may not be. (In the row reduction process, all the rows of zeros
are shuffled to the bottom).
4. Definition: The rank of matrix A is the dimension of the column space (i.e., the number of independent
columns).
5. The Rank Theorem:
• Rank of A = # of pivots in A
• dim(Col(A)) = dim(Row(A))
or Rank of A = Rank of AT .
• rank(A) + dim(Null(A)) = n
or the number of pivots + number of free vars = n
6. Let A be an m × n matrix with rank k. Let B be the RREF of A, and D be the RREF of AT . The four
subspaces and their dimensions are:
Symbol
Col(A)
Basis found by
Look for pivots in B, use cols of A
Space
IR
k
n
n−k
Null(A)
Use B to solve Ax = 0
IR
Row(A)
Use the nonzero rows of B
IRn
T
Null(A )
T
Use D to solve A x = 0
7. We have the following additions to the Invertible Matrix Theorem:
(m.) The cols of A form a basis for IRn
(n.) Col(A) = IRn
(o.) dim(Col(A)) = n
(p.) Null(A) = {0}
(q.) dim(Null(A)) = 0
8. If u, v ∈ IRn , then show that the n × n matrix uv T has rank 0 or 1.
1
IR
Dim
m
m
k
m−k