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Section3.7Basis and Dimension

  1. Understand the definition of a basis of a subspace.
  2. Recipes: basis for a column space, basis for a null space, basis of a span.
  3. Picture: basis of a subspace of R 2 or R 3 .
  4. Essential vocabulary words: basis, dimension.

Subsection3.7.1Basis of a Subspace

As we discussed in Section 3.6, a subspace is the same as a span, except we do not have a set of spanning vectors in mind. There are infinitely many choices of spanning sets for a nonzero subspace; to avoid reduncancy, usually it is most convenient to choose a spanning set with the minimal number of vectors in it. This is the idea behind the notion of a basis.


Let V be a subspace of R n . A basis of V is a set of vectors { v 1 , v 2 ,..., v m } in V such that:

  1. V = Span { v 1 , v 2 ,..., v m } , and
  2. the set { v 1 , v 2 ,..., v m } is linearly independent.

Recall that a set of vectors is linearly independent if and only if, when you remove any vector from the set, the span shrinks (Theorem 3.5.14). In other words, if { v 1 , v 2 ,..., v m } is a basis of a subspace V , then no proper subset of { v 1 , v 2 ,..., v m } will span V : it is a minimal spanning set.

A subspace generally has infinitely many different bases, but they all contain the same number of vectors.

We leave it as an exercise to prove that any two bases have the same number of vectors; one might want to wait until after learning the invertible matrix theorem in Section 4.5.


Let V be a subspace of R n . The number of vectors in any basis of V is called the dimension of V , and is written dim V .


The previous example implies that any basis for R n has n vectors in it. Let v 1 , v 2 ,..., v n be vectors in R n , and let A be the n × n matrix with columns v 1 , v 2 ,..., v n .

  1. To say that { v 1 , v 2 ,..., v n } spans R n means that A has a pivot in every row: see this theorem in Section 3.3.
  2. To say that { v 1 , v 2 ,..., v n } is linearly independent means that A has a pivot in every column: see this theorem in Section 3.5.

Since A is a square matrix, it has a pivot in every row if and only if it has a pivot in every column. We will see in Section 4.5 that the above two conditions are equivalent to the invertibility of the matrix A .

Subsection3.7.2Bases for Common Types of Subspaces

Now we show how to find bases for the types of subspaces we encountered in Section 3.6, namely: a span, the column space of a matrix, and the null space of a matrix.

A basis for the column space

First we show how to compute a basis for the column space of a matrix.


The above theorem is referring to the pivot columns in the original matrix, not its reduced row echelon form. Indeed, a matrix and its reduced row echelon form generally have different column spaces. For example, in the matrix A below:

A = 1 2 0 1 2 3 4 5 2 4 0 2 F G RREF −−→ 1 0 8 7 0 1 4 3 0 0 0 0 F G pivotcolumns = basis pivotcolumnsinRREF

the pivot columns are the first two columns, so a basis for Col ( A ) is

DB 1 22 C , B 2 34 CE .

The first two columns of the reduced row echelon form certainly span a different subspace, as

Span DB 100 C , B 010 CE = DB ab 0 CAAA a , b in R E ,

but Col ( A ) contains vectors whose last coordinate is nonzero.

A basis of a span

Computing a basis for a span is the same as computing a basis for a column space. Indeed, the span of finitely many vectors v 1 , v 2 ,..., v m is the column space of a matrix, namely, the matrix A whose columns are v 1 , v 2 ,..., v m :

A = B ||| v 1 v 2 ··· v m ||| C .
A basis for the null space

In order to compute a basis for the null space of a matrix, one has to find the parametric vector form of the solutions of the homogeneous equation Ax = 0.

In lieu of a proof, we illustrate the theorem with an example, and we leave it to the reader to generalize the argument.