Skip to main content

Section3.9The Rank Theorem and the Basis Theorem

Objectives
  1. Learn to use the rank theorem and the basis theorem.
  2. Picture: the rank theorem.
  3. Theorems: rank theorem, basis theorem.
  4. Vocabulary words: rank, nullity.

In this section we present two important general facts about dimensions and bases.

Subsection3.9.1The Rank Theorem

With the rank theorem, we can finally relate the dimension of the solution set of a matrix equation with the dimension of the column space.

Definition

The rank of a matrix A , written rank ( A ) , is the dimension of the column space Col ( A ) .

The nullity of a matrix A , written nullity ( A ) , is the dimension of the null space Nul ( A ) .

According to this theorem in Section 3.7, the rank of A is equal to the number of columns with pivots. On the other hand, this theorem in Section 3.7 implies that nullity ( A ) equals the number of free variables, which is the number of columns without pivots. To summarize:

rank ( A )= dimCol ( A )= thenumberofcolumnswithpivotsnullity ( A )= dimNul ( A )= thenumberoffreevariables = thenumberofcolumnswithoutpivots

Clearly (the number of columns with pivots) plus (the number of columns without pivots) equals (the number of columns of A ), so we have proved the following theorem.

In other words, for any consistent system of linear equations,

(dimofcolumnspan) + (dimofsolutionset) = (numberofvariables).

Subsection3.9.2The Basis Theorem

Recall from this example in Section 3.7 that { v 1 , v 2 ,..., v n } forms a basis for R n if and only if the matrix A with columns v 1 , v 2 ,..., v n has a pivot in every row and column. Since A is an n × n matrix, these two conditions are equivalent: the vectors span if and only if they are linearly independent. The basis theorem is an abstract version of the preceding statement, that applies to any subspace.

Proof

In other words, if you already know that dim V = m , and if you have a set of m vectors B = { v 1 , v 2 ,..., v m } in V , then you only have to check one of:

  1. B is linearly independent, or
  2. B spans V ,

in order for B to be a basis of V . If you did not already know that dim V = m , then you would have to check both properties.

For example, if V is a plane, then any two noncollinear vectors in V form a basis.