linearly independent/linearly dependent sets

[wtfbbq]

To find if a set is linearly dependent you can either put the vectors in rows OR columns and see if there is a non trivial solution, right?

 

[BrownTown]

wouldn't you just check if the rank of the matrix was equal to the minimum of the numbers of rows or columns? If it is then they are independant?

 

[CycloWizard]

Originally posted by: BrownTown
wouldn't you just check if the rank of the matrix was equal to the minimum of the numbers of rows or columns? If it is then they are independant?

:thumbsup:

 

[wtfbbq]

ah... looking back in my notes, I see that you can "decide if the equations have a solution by comparing the rank of the augmented matrix to the rank of the coefficient matrix," is this what you guys are talking about? if not could you please elaborate? i'm pretty bad at this stuff. And I'm still not sure which is preferred: sticking them in rows or columns. I know that if you have a problem like: A = {(4,1,1), (1,2,2), (1,1,1)} find if (2,4,3) is in the span of the set, you'd put them all in columns and augment the 2,4,3 on and rref/solve to see if there's a non triv solution. What i was wonderif was if the question is simply "L.I. or L.D.?" if it was okay to put them in rows and augment on a 0 column and see if there is a sol?

 

[BrownTown]

OK, i really should know this since I just had an exam on it like 2 days ago, but alass I have already forgotten. Pluss, we were trying to find solutions to overdetermined problems, so this wasnt so much the issue. Anyways, let me take a stab at it. If the rank of the augmented matrix is greater than the rank of the origional matrix then there is no exact solutions and you have to find a least squares solution. If the rank of the augmented matrix is equal to the rank of the origional matrix AND the origional matrix is full rank than there is a single exact solution. And If the rank of the augmented matrix is the same as the rank of the origional matrix, but the origional matrix is NOT full rank than there are an infinite number of exact solutions.

That sound right to anyone? Probably not, but its a start...

 

[blahblah99]

Originally posted by: BrownTown
wouldn't you just check if the rank of the matrix was equal to the minimum of the numbers of rows or columns? If it is then they are independant?

I think what the OP was asking was how do you prove linear independence mathematically.

You have to find a1, a2, a3, ..., an , not all equal to zero so that the following is true:

a1*Col1 + a2*Col2 + a3*Col3 + .. + an*Col n = 0 (zero vector).

If a1, a2, a3, ... , an, exists and they are not all equal to zero, then the vectors are linearly dependent. If no solution exists other than a1, a2, a3, ... an being all zeros, then the vectors are linearly independent.

 

[wtfbbq]

well my main question was about the difference in putting the vectors from the set into columns vs. putting them into rows, and what the difference is

 

[flyboy84]

I remember something about taking the Wronskian of the matrix being nonzero? Or am I thinking of something else?

 

[Unitary]

Originally posted by: wtfbbq
well my main question was about the difference in putting the vectors from the set into columns vs. putting them into rows, and what the difference is

You can do it both ways. Put your vectors into the columns or rows of a matrix then row reduce. If you end up with less "leading ones" than vectors, your set is linearly dependent.

Of course there is a more indepth answer explaning why this works, but this is the quick explanation.

Originally posted by: flyboy84
I remember something about taking the Wronskian of the matrix being nonzero? Or am I thinking of something else?

This is a method to show functions are linearly independent/dependent.