Math, Linear Algebra question?

Question

Consider the matrix B =
1; -2; 1
3; 4; 1
6; -2; 4
(a) Find out if the columns of B are independent.
(b) Are the rows of B dependent?
(c) Suppose that Bu = Bv. Does it follow that u = v? Why or why not?

Answer 1

a) The columns of B are independent only if the following system has a unique solution if a1 = a2 = a3 = 0

a1(1, 3, 6) + a2(-2,4,-2) + a3(1,1,4) = (0,0,0)

b) The rows of B are independent only if the following system has a unique solution for a1 = a2 = a3 = 0

a1(1,-2, 1) + a2(3,4,1) + a3(6,-2,4) = (0,0,0)

c) if Bu = Bv, this means that

1u ; -2u ; 1u...................1v ; -2v ; 1v
3u ; 4u ; 1u.........=.........3v ; 4v ; 1
6u ; -2u , 4u...................6v ; -2v , 4v

the only way that this is true if is v = u

Answer 2

(a)You can check if they are independent by seeing if one of the columns can be obtained from any other by multiplying it with a number. (b) Same goes for rows - if any row can be obtained from the other by multiplying.

If you take row #1 and multiply it by 3 and then add it to row #2 then you'll obtain row #3. Row #3 is not independent.

(c) I'm not sure. It's been too long since my formal training is complete :-) I'll post if I remember. I think it follows... I think the easiest way to prove/check it is to assume that it doesn't follow and see if you can prove the opposite.

Answer 3

a) They aren't. Consider that 3/5 v_1 - 1/5 v_2 - v_3 = 0
b) Yes. 3 r_1 + r_2 - r_3 = 0
c) No. Consider that the vectors [3/5, -1/5, -1] and [0, 0, 0] (those are column vectors - just can't be bothered to write them as such) both become zero when multiplied by B, but are not equal.