# Basis for the column space of a matrix: Consider the matrix A = [[-2,-2,0,-8,-5,-6,27],[-9,-9,-2,-28,-9,4,-22],[4,4,-9,52,2,2,-9],[-2,-2,4,-24,-5,-4,11],[-4,-4,-1,-12,-9,0,-14]] and assume that its...

Basis for the column space of a matrix:

Consider the matrix

A = [[-2,-2,0,-8,-5,-6,27],[-9,-9,-2,-28,-9,4,-22],[4,4,-9,52,2,2,-9],[-2,-2,4,-24,-5,-4,11],[-4,-4,-1,-12,-9,0,-14]]

and assume that its reduction is

B = [[1,1,0,4,0,0,-3],[0,0,1,-4,0,0,-1],[0,0,0,0,1,0,3],[0,0,0,0,0,1,-6],[0,0,0,0,0,0,0]]

We know that the vectors

[[-2],[-9],[4],[-2],[-4]] , [[0],[-2],[-9],[4],[-1]] , [[-5],[-9],[2],[-5],[-9]] , [[-6],[4],[2],[-4],[0]]

form a basis for the column space of A. Write each of the following columns of A in terms of this basis.

a) [[-2],[-9],[4],[-2],[-4]]

b) [[-8],[-28,],[52],[-24],[-12]]

c) [[27],[-22],[-9],[11],[-14]]

*print*Print*list*Cite

### 1 Answer

Let the rows of `A` be `c_1,c_2,c_3,c_4,c_5,c_6,c_7`.

The reduction of `A` is ` `

`B = ([1,1,0,4,0,0,-3],[0,0,1,-4,0,0,-1],[0,0,0,0,1,0,3],[0,0,0,0,0,1,-6],[0,0,0,0,0,0,0]) `

The linearly independent columns are 1, 3, 5 and 6, so columns 1, 3, 5 and 6 of `A` (`c_1,c_3,c_5,c_6`) form a basis for the column space of `A`.

Then, examining column 4 of `B`, `c_4 = 4c_1 - 4c_3` ` `

(check: (-8,-28,52,-24,-12) = 4(-2,-9,4,-2,-4) - 4(0,-2,-9,4,-1) )

and, examining column 7 of `B`, `c_7 = -3c_1 - c_3 + 3c_5 -6c_6 `

(check: (27,-22,-9,11,-14) = -3(-2,-9,4,-2,-4) - (0,-2,-9,4,-1)

+ 3(-5,-9,2,-5,-9) - 6(-6,4,2,-4,0) )

**a) c1 = 1.c1 + 0.c3 + 0.c5 + 0.c6**

**b) c4 = 4.c1 - 4.c3**

**c) c7 = -3.c1 - 1.c3 + 3.c5 - 6.c6**