### Computation of Basis

##### 3. Basis of column space, second method

We illustrate another method for finding a basis of a column space through the same example.

Example The column space of

 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. The five spanning vectors can be simplified by column operations. The computation in this example gives us a column echelon form.

 B = [ 1 0 0 0 0 ] -1 1 0 0 0 0 2 1 0 0 1 0 -1 0 0

Since zero columns do not contribute to the span, we have

colA = colB = span{first three columns of B}.

On the other hand, the first three columns of B are linearly independent.

c1[col 1] + c2[col 2] + c3[col 3] = 0
c1 = 0 (look at the 1st coordinate of c1[col 1] + c2[col 2] + c3[col 3] = 0)
c2 = 0 (look at the 2nd coordinate of c2[col 2] + c3[col 3] = 0)
c3 = 0. (look at the 3rd coordinate of c3[col 3] = 0)

Thus the nonzero columns of B

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

form a basis of colB = colA.

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

1. If B is the column echelon form of A, then colB = colA.
2. The nonzero columns in B are linearly independent and span colB.

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

For a matrix A, the nonzero columns of a column echelon form is a basis of colA.

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

Example To find a basis of

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

we use the translation

 A = [ a b ] ↔ [A] = (a, b, c, d) c d

to convert the problem to the basis of the column space of

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

Column operations give us a column echelon form.

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

The nonzero columns (1, -1, 1, 1), (0, 3, -3, 1), (0, 0, 0, 1) form a basis of the column space. The corresponding 2 by 2 matrices

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

form a basis of the span H.

Example 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,

we use the translation [a + bt + ct2] = (a, b, c) to convert the problem to the basis of the column space of

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

By column operations, we get a column echelon form

 [ 1 0 0 ], 1 1 0 0 2 0

and the nonzero columns (1, 1, 0), (0, 1, 2) of the column echelon form is a basis of the column space. Correspondingly, 1 + t and t + 2t2 form a basis of rangeT.