b Find b projCb Does b F b What does this tell you about the

b) Find b projC(b). Does b F b? What does this tell you about the number of solutions to Ax b

Solution

To determine C = Col(A), we will reduce A to its RREF as under:

Add 1 times the 1st row to the 2nd row; Add -1 times the 1st row to the 4th row

Add -1 times the 1st row to the 5th row; Interchange the 2nd row and the 3rd row

Multiply the 2nd row by ½; Add -2 times the 2nd row to the 4th row

Add -2 times the 2nd row to the 5th row; Multiply the 3rd row by 1/6

Add 6 times the 3rd row to the 4th row; Add -3/2 times the 3rd row to the 2nd row

Add -5 times the 3rd row to the 1st row; Add -3 times the 2nd row to the 1st row

Then the RREF of A is

1

0

0

0

1

0

0

0

1

0

0

0

0

0

0

Thus, a basis for Col(A) is {(1,0,0,0,0)^T,(0,1,0,0,0)^T,(0,0,1,0,0)^T} and C = span{ (1,0,0,0,0)^T,(0,1,0,0,0)^T,(0,0,1,0,0)^T} = span{v₁,v₂,v₃}(say).

Then ^b=proj_C(b)=proj_v1(b)+proj_v2(b)+proj_v3(b)= [(b.v₁)/(v₁.v₁)]v₁+[(b.v₂)/(v₂.v₂)]v₂+[(b.v₃)/(v₃.v₃)]v₃ = v₁+v₂+v₃= (1,0,0,0,0)^T+(0,1,0,0,0)^T+(0,0,1,0,0)^T= (1,1,1,0,0)^T . Thus ^b b.

Note: We could have taken the original columns of A also in C = Col(A), but computations with {v₁,v₂,v₃} are easier.

1	0	0
0	1	0
0	0	1
0	0	0
0	0	0