Why is this linearly independent? (linear algebra)

[warwick]

Determine if the columns of the matrix form a linearly independent set. I put this in echelon form and found that the bottom row is all zeroes. There are three variables, and each variable is a basic variable. I just now realized that this is an overdetermined set. So, it is linear independent because there are no free variables.

0 -8 5
3 -7 4
-1 5 -4
1 -3 2

 

[pka]

Determine if the columns of the matrix form a linearly independent set.

It appears that you are not working with the rows and not the columns.
You want row reduce the transpose of the matrix. I found the columns to be independent.

 

[warwick]

Determine if the columns of the matrix form a linearly independent set.

It appears that you are not working with the rows and not the columns.
You want row reduce the transpose of the matrix. I found the columns to be independent.

I did work with the rows. I found the matrix to be linearly independent because it is an overdetermined system with no free variables. Thus, it has no non-trivial solution, which makes it independent.

 

[pka]

You do not understand the question do you?
If that is true then how do you expect us to help you?
Please reread my post.
If it makes no sense to you then seek some sit down help with a live instructor.

 

[warwick]

You do not understand the question do you?
If that is true then how do you expect us to help you?
Please reread my post.
If it makes no sense to you then seek some sit down help with a live instructor.

What the crap are you talking about? I damn well understand the question. We have not gone over the transpose of anything, by the way.

Your post does not make sense either.

It appears that you are not working with the rows and not the columns.
You want row reduce the transpose of the matrix. I found the columns to be independent.

You said I'm not working with the rows and columns.

At any rate, I reworked the problem. I put the matrix in reduced echelon form and found there are only basic variables that each equal zero. Thus, the only solution contains zeroes and it is linearly independent, as well as being overdetermined. Correct?

 

[mmm4444bot]

You do not understand the question do you? ...

What the crap are you talking about? I damn well understand the question ...

At any rate, I reworked the problem ... [the three columns comprise a] linearly independent [set] ...Correct?

Yes, the columns (vectors) form a linearly-independent set of columns (vectors).

You've discovered that the only way to combine these columns (vectors) linearly -- in order to result in a zero column (zero vector) -- is by setting all of the variables to zero. This is sufficient to show linear independence among a set of columns (vectors). In other words, none of the three columns (vectors) can be formed by a linear combination of the other two.

I believe the intent of PKA's question is to clarify. (Maybe you're working with a new system where the rows come from the columns of an old system, for example). If you wish to clarify something in the future, then I would like to request that you adopt a more civilized tone.

It is not always easy to communicate mathematical concepts through a bulletin board, so we need to realize that the entire process often takes longer to work out than if we were speaking in person.

Cheers,

~ Mark :|