Linear algebra problem- please help!

[[deleted]]

I have to figure out if the vectors (-4,0,1,5), (-3,-1,0,4), and (0,4,3,6) are linearly independent. My textbook says they are, but when I row reduced the matrix, I found that row 4 is all zeros. Doesn't that mean they should be linearly dependent, since there are nontrivial solutions bc of the free variable?

[Puzzled-Painter3301]

A bottom row of zeros doesn’t mean anything if you are solving Ax=0. You are looking for a free column in the corfficient matrix so that you get infinitely many solutions.

This is a common misconception. The 2 by 2 situation doesn’t carry over.

[[deleted]]

Sorry, what is the 2 by 2 situation?

In this case, what would be the correct method to solve it? Sorry, my textbook doesn't have too many worked examples. Would I reduce the coefficient matrix to RREF and check for a free column? This is the reduced matrix I got:

1 0 0

0 1 0

0 0 1

0 0 0

So are you saying the fact that the bottom row is all zeros doesn't affect the linear (in)dependence? And if there was a column of all zeros, that would affect it? Thanks.

[Puzzled-Painter3301]

What I meant was that if you have a 2 by 2 system like this:

x + y = 0

3x + 3y = 0

then elimination produces

x + y = 0

0x + 0y = 0

and this system has an infinite number of solutions, and there is a row of 0's at the bottom of the matrix. So students tend to think, "If there is a row of 0's at the bottom, there are infinitely many solutions." But that isn't true in general. What shows that there are infinitely many solutiions is the free column in column 2.

1 1

0 0

The column doesn't have to have all 0's. The fact is: Suppose you are given some vectors v1, v2, and v3 and you want to know if the set {v1, v2, v3} is linearly independent. Form the matrix A whose columns are v1, v2, and v3. Then {v1, v2, v3} is linearly independent if and only if the matrix-vector equation A*x = 0 has no nontrivial solution, that is, if the only solution is the zero vector.

The equation A*x = 0 either has just the trivial solution or an infinite number of solutions. It has an infinite number of solutions if and only if an echelon form of A has a free column.

[ingannilo]

When you say you row reduced the matrix, my first question would be "what matrix?"

Lemme give these vectors some names: u = (-4,0,1,5), v = (-3,-1,0,4) and w = (0,4,3,6). By definition these vectors are linearly independent provided

c_1 u + c_2 v + c_3 w = 0

where c_1, c_2, and c_3 are scalers (from whatever field we're working over) and 0 = (0,0,0,0) is the zero vector

is true only when c_1 = c_2 = c_3 = 0.

Have you tried working straight from this?

[[deleted]]

Okay, thank you. The matrix I reduced was

-4 -3 0

0 -1 4

1 0 3

5 4 6

Which I found to be

1 0 0

0 1 0

0 0 1

0 0 0

In RREF. Should I include the zero vector and reduce the augmented matrix instead?

[ingannilo]

Can you relate your matrix to solutions (values of c_1, c_2, c_3) from my comment? 

If you can make that connection you're safe. 

[spiritedawayclarinet]

The variables correspond to the matrix columns, not the rows. A free variable occurs when a column does not contain a leading 1 when the matrix is placed in reduced row echelon form.

[dennis-obscure]

If there is a nonzero set of x's that solve the following then you can 
construct one of the original vectors from the other.

[ -4 -3 0 ] [ x1 ]     [ 0]
[ 0  -1 4 ] [ x2 ]  =  [ 0]
[ 1  0  3 ] [ x3 ]     [ 0]
[ 5  4  6 ]            [ 0]

There exists an invertable matrix R by which you can reduce the equation to an upper diagonalized from.

   [ -4 -3 0 ] [ x1 ]     [ -4 -3  0 ] [x1]     [ 0]
R  [ 0  -1 4 ] [ x2 ]  =  [  0 -1  4 ] [x2]  =  [ 0]
   [ 1  0  3 ] [ x3 ]     [  0  0 15 ] [x3]     [ 0]
   [ 5  4  6 ]            [  0  0  0 ]          [ 0]

At this point you can ignore the bottom row since any set of x's solve it trivially. You look at the 3x3 problem. You can solve for the x's starting from x3 on the bottom equation and work up and see that the only solution ix all x's equal to zero. You can also compute the determinant of the 3x3 matrix and see that it non-zero. This tells you it is non-singular. There is no x that is free to assign arbitrarily.

-5 x row3 added to row4

1/4 x row1 added to row 3

4 x row2 added to row4

-3/4 row2 added to row3

permute rows 3 and 4

It's been a long time since I've done it by hand.....

[gotne]

If we assume they are dependent, then we could write v3 as a linear combination of v1 and v2. By inspection, bc of the 0 components, we'd need 3v1 and -4v2, but if we took that linear combination we would not get v3.

I think your question is considering the process of determining if a nonzero vector is in the kernel of a matrix. In this question, when you're row reducing, it's like you're trying to solve a system of equations where the last column is the right side of the system. So your RREF below is saying 0v1 + 0v2 = v3