Computation of Basis Computation of BasisWe illustrate another method for finding a basis of a column space through the same example.
| 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
This leads to the following method for finding a basis of the column space.
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: P2 → P2,
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.