# Proving a set of vectors are linearly independent

1. Sep 17, 2010

### sakodo

Hi, I came across a question where I needed to prove that a set of vectors are linearly independent. The thing is, I am not sure how to reason the proof properly.

Say you have three vectors x1,x2,x3 E R3, and prove that they are linearly independent.

Put them into a 3x3 matrix A, row-reduce, if all the columns have leading terms, because there are no non-leading columns, the only solution of Ax=0 is when x=(0,0,0). Thus, x1,x2,x3 are linearly independent.

Is this reasoning good enough? I feel like I am missing something.

Any help would be appreciated.

2. Sep 17, 2010

### Char. Limit

I think you'll have to put some restrictions on your vectors. Right now, it seems to me that the vectors x1, x2, and x3 can be any vectors in R3. However, won't you need to put a bit more restriction on that? What if I give x1=<1,2,3> and x2=<2,4,6>... then they aren't linearly independent.

I hope you got some advice from all that babble.

3. Sep 17, 2010

### sakodo

Thanks for the reply Char.Limit.

Assuming that x1,x2,x3 are indeed linearly independent, is my reasoning good enough? I am not sure if my proof is sufficient.

Thanks.

4. Sep 17, 2010

### radou

You can't assume they are independent, if this is what you need to show.

If these are some given vectors, then your procedure is OK, since x = (0, 0, 0) represents the coefficients of your linear combination.

5. Sep 17, 2010

### sakodo

Yeah sorry I didn't put it clearly. What I meant was x1,x2,x3 are given vectors.

If the row-reduced matrix has no non-leading terms, you can deduce that the vectors are linearly independent already. Its just I don't know how to set up the proof properly lol.

6. Sep 17, 2010

### HallsofIvy

If the vectors are not independent, then row reduction will result in the last row being all "0"s and vice versa. That's essentially what you mean by "non-leading term" isn't it?

7. Sep 17, 2010

### sakodo

Yeah is that the right term? If a matrix has a row of zeros then it has no leading terms. It was either leading term or leading column. Sorry I forgot the exact name for it.

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