# Set of vectors, linearly dependent or independent?

1. Feb 11, 2014

### Mutaja

1. The problem statement, all variables and given/known data

Check if the following set of vectors are linearly dependent or independent:

A) V1= $\stackrel{1}{1}$ V2= $\stackrel{1}{3}$

B) V1= $\stackrel{\stackrel{1}{2}}{3}$ V2= $\stackrel{\stackrel{2}{1}}{3}$

C) V1= $\stackrel{1}{3}$ V2= $\stackrel{2}{1}$ V3= $\stackrel{-1}{2}$

2. Relevant equations

Gauss-Jordan, equation with two unknowns.

3. The attempt at a solution

A) x1 * V1 + x2*v2 = 0

if x1 AND x2 = 0, then they're linearly independent. If not, they're linearly dependent.

$\stackrel{1}{1}$ $\stackrel{1}{3}$ * $\stackrel{X1}{X2}$ = 0

Using gauss jordan I get: $\stackrel{1}{1}$ $\stackrel{1}{3}$ $\stackrel{0}{0}$ -> $\stackrel{1}{0}$$\stackrel{1}{2}$$\stackrel{0}{0}$-> $\stackrel{1}{0}$$\stackrel{1}{1}$$\stackrel{0}{0}$-> $\stackrel{1}{0}$$\stackrel{0}{1}$$\stackrel{0}{0}$ which gives me x1 = 0, x2 = 0.

X1+ X2 = 0
X1 + 3X2 = 0

-2X2 = 0 -> X2 = 0
X1 = 0.

Both X1 and X2 has to be 0 -> linearly independent.

B) X1*V1 + X2*V2 = 0

Using the same method -> Gauss Jordan. I end up with the same, X1 an X2 has to be 0 -> linearly independent.

Please do let me know if I should write all my work here as well. Leaving it as it is for now as it takes 15 minutes, and I've done the exact same steps as above.

C) Here we can see that V1 - V2 = V3, therefore they are linearly dependent.

Is this sufficient to "check" if they're linearly dependent or independent, or do I have to do more work? If so, do you have any tips on how to proceed with three vectors?

Also, any tips regarding how to write vectors or matrices efficiently with latex or what it's called, please let me know. I'll be more than happy to include more of my work if I can do it somewhat efficiently.

Thanks for any input.

2. Feb 11, 2014

### dirk_mec1

3. Feb 11, 2014

### Mutaja

4. Feb 11, 2014

### Simon Bridge

To show the set is dependent you need only show a single case of dependence ... showing independence is the one that involves hard work.

Bottom of page:
http://www.math.oregonstate.edu/hom...ulusQuestStudyGuides/vcalc/lindep/lindep.html

you can use the "matrix" environment:$$v_1=\begin{pmatrix}1\\ 1 \end{pmatrix},\; v_2=\begin{pmatrix}1\\ 3 \end{pmatrix}$$
... hit the "quote" button to see what I did :)
http://www.math-linux.com/latex-26/faq/latex-faq/How-to-write-matrices-in-Latex [Broken]

Last edited by a moderator: May 6, 2017
5. Feb 11, 2014

### abitslow

Hmmm. I have no big problem with your answers. But they aren't 'optimum' (in some ways).
First V1 = ( 1,1), V2 = (1,3) ....these are NOT the same as (1,1,0) and (1,3,0)!!!!!
Second by simple inspection, there is no number x which will multiply (1,1) and produce (1,3). This is obvious....all you can get is (x,x). So, the set is independent.
Third the same thing is obviously true for (1,2,3) and (2,1,3)...there is exactly one number (scalar multiplication, of course) which will get you from the first element of vector 1 to the first element of vector 2. That number is 2... 2*(1,2,3) = (2,4,6) which is, again obviously, NOT (2,1,3). so also independent.
Fourth. It is IMPOSSIBLE to have MORE than 2 independent vectors in any (real) 2-vector. You didn't have to do any math to know that C could not be a set of three independent 2-vectors. Not possible. You can not have more than n independent n-vectors (over the real vector spaces).