Since the null space is one dimensional, it follows that the column space has dimension 3. That means that the four columns cannot be independent. Of course, the test for independence is to look at
\(\displaystyle a<0,\, -2,\, 6,\, 2>\,+\, b<0,\, 2,\, 1,\, 0>\,+ \,c<0,\, -1,\, 0,\, 1>\,+\, d<0, \,4,\, 4, \,4>\)
. . .\(\displaystyle = \,<0,\, -2a\,+\, 2b\,-\, c\,+\, 4d,\, 6a\,+\, b\,+\, 4d,\, 2a\,+ \,c\,+ \,4d>\,= \,<0,\, 0,\, 0,\, 0>\)
Since we know the columns are not independent, there must exist some a, b, c, d, not all 0, such that this is true.
That is, -2a+ 2b- c+ 4d= 0, 6a+ b+ 4d= 0, and 2a+ c+ 4d= 0. If we subtract the first equation from the first, we eliminate d and have 8a- b+ c= 0. If we subtract the first equation from the third we also eliminate d and have 4a- 2b+ 2c= 0. If we multiply 8a- b- c= 0 by 2 and subtract that from 4a- 2b+ 2c= 0 we have -12a+ 4c= 0 or c= 3a So 4a- 2b+ 2c= 4a- 2b+ 6a= 10a- 2b= 0 or b= 5a. Also 2a+ c+ 4d= 2a+ 3a+ 4d= 5a+ 4d= 0 or d= (5/4)a.
So our
\(\displaystyle a<0,\, -2,\, 6,\, 2>\,+ \,b<0,\, 2,\, 1, \,0>\,+ \,c<0,\, -1,\, 0,\, 1>\,+ \,d<0,\, 4,\, 4, \,4>\)
. . .\(\displaystyle =\, <0,\, -2a\,+\, 2b\,- \,c\,+ \,4d,\, 6a\,+ \,b\,+\, 4d,\, 2a\,+\, c\,+\, 4d>\,=\, <0, \,0,\, 0,\, 0>\)
is
\(\displaystyle a<0,\, -2,\, 6,\, 2>\,+ \,5a<0,\, 2,\, 1,\, 0>\,+\, 3a<0, \,-1,\, 0,\, 1>\,+\, \left(\dfrac{5}{4}\right)a<0,\, 4,\, 4, \,4>\,=\, <0,\, 0,\, 0, \,0>\)
Multiplying by 4/a,
\(\displaystyle 4<0,\, -2,\, 6,\, 2>\,+\, 20<0,\, 2,\, 1,\, 0>\,+\, 12<0, \,-1, \,0, \,1>\,+\, 5<0, \,4, \,4,\, 4>\,=\, <0,\, 0,\, 0,\, 0>\)
We can solve that for any one of the four vectors as a linear combination of the other three. Since the original four spanned the column space, and we know the dimension of the columns space is three, we can, as above eliminate any one vector leaving the other three as a basis.
