### Computation of Basis

##### 2. Basis of column space, first method

In the next example, we find a basis of the column space of a matrix.

Example The column space of the matrix

 A = [ 1 3 2 0 1 ] -1 -1 -1 1 0 0 4 2 4 3 1 3 2 -2 0

is the span of the five column vectors. By the row operations in this example, we get a row echelon form.

 [ 1 3 2 0 1 ] 0 2 1 -1 0 0 0 0 2 1 0 0 0 0 0

Since [col 3] and [col 5] are not pivot, [col 5] is a linear combination of [col 1] through [col 4], and [col 3] is a linear combination of [col 1] and [col 2]. Then by this property, we have

colA
= span{[col 1], [col 2], [col 3], [col 4], [col 5]}
= span{[col 1], [col 2], [col 3], [col 4]}
= span{[col 1], [col 2], [col 4]}.

On the other hand, if we pick [col 1], [col 2], [col 4] only, then

 [ 1 3 0 ] -1 -1 1 0 4 4 1 3 -2

has the following row echelon form

 [ 1 3 0 ], 0 2 -1 0 0 2 0 0 0

in which all columns are pivot. Thus the three columns are linearly independent. Since they also span colA, we conclude that

[col 1] of A = (1, -1, 0, 1)
[col 2] of A = (3, -1, 4, 3)
[col 4] of A = (0, 1, 4, -2)

form a basis of colA.

The argument in the example can be applied to any matrix A. We have

1. The nonpivot columns of A are linear combinations of the preceeding columns. Therefore they can be deleted without affecting the columns space.
2. Tautologically, the matrix formed by the pivot columns of A has all columns pivot. Therefore the pivot columns are linearly independent.

This leads to the following method for finding a basis of the column space.

For a matrix A, the pivot columns of A form a basis of colA.

We emphasis that it is the columns of A, not the columns of the row echelon form, that form the basis.

Example We would like to find a basis of

 H = span{ [ 1 -1 ], [ 2 1 ], [ 4 2 ], [ 3 -3 ] }. 1 1 -1 3 -2 1 3 2

By the translation

 A = [ a b ] ↔ [A] = (a, b, c, d), c d

the problem becomes the basis of the column space of

 [ 1 2 4 3 ]. -1 1 2 -3 1 -1 -2 3 1 3 1 2

Row operations give us the row echelon form.

 [ 1 2 4 3 ] 0 1 2 0 0 0 5 1 0 0 0 0

Thus the first three columns of the matrix form a basis of the column space. Correspondingly, the first three 2 by 2 matrices

 [ 1 -1 ], [ 2 1 ], [ 4 2 ] 1 1 -1 3 -2 1

form a basis of the span H.

Example We would like to find a basis for the range of the linear transformation

T(p(t)) = p(0) + p(1)t + (p'(0) + p'(1))t2: P2P2.

The range is spanned by

T(1) = 1 + t, T(t) = t + 2t2, T(t2) = t + 2t2.

By the translation [a + bt + ct2] = (a, b, c), the problem becomes the basis of the column space of

 [ [T(1)] [T(t)] [T(t2)] ] = [ 1 0 0 ]. 1 1 1 0 2 2

By row operations, we get

 [ 1 0 0 ]. 0 1 1 0 0 0

Thus the first two columns form a basis of the column space. Correspondingly, T(1) = 1 + t and T(t) = t + 2t2 form a basis of rangeT.