# Linear independence of 5 dimensional vectors in R^3

1. Nov 29, 2015

### quanta13

i am asked to determine whether 3 vectors which have 5 dimensions (x,y,z,w,u) are linearly dependent or independent in R^3.
it doesn't make any sense. should i ignore w and u dimensions and take x,y,z only? because if i dont, all answers would be same, doesnt matter in r^3 or R^4 etc.

the question is this:

determine whether the given vectors (2,0,1,-1,0) , (1,2,0,3,1) and (4,-4,3,-9,-2) are linearly dependent or independent in R^3?

Last edited: Nov 29, 2015
2. Nov 29, 2015

### phyzguy

Your question isn't even a complete sentence, so I don't know what you are asking. Please rewrite it in proper English. I've copied in the Physics Forums language guidelines below:

All posts must be in English. Posts in other languages will be deleted. Pay reasonable attention to written English communication standards. This includes the use of proper grammatical structure, punctuation, capitalization, spacing, and spelling. In particular, "I" is capitalized, there's a space after (but not before) a comma, a period, and other punctuation. Multiple exclamation marks are also discouraged. SMS messaging shorthand ("text-message-speak"), such as using "u" for "you", "plz" for "please", or "wanna" for "want to" is not acceptable.

3. Nov 29, 2015

### phyzguy

OK, that's better. You still should capitalize "I" and use punctuation, but it is now an intelligible question. Can you tell me what it means for three vectors to be linearly dependent?

4. Nov 29, 2015

### quanta13

I am so sorry for not capitalizing ''I''.
I wrote down the question:
''determine whether the given vectors (2,0,1,-1,0) , (1,2,0,3,1) and (4,-4,3,-9,-2) are linearly dependent or independent in R^3? ''
I think it is clear what is asking. Thanks.

5. Nov 29, 2015

### phyzguy

Yes, it is clear. My question is, do you know what it means to be linearly dependent (or linearly independent)? Can you give me the definition of linear dependence?

6. Nov 29, 2015

### WWGD

EDIT: OP: It may be a typo, and maybe it should be $\mathbb R^5$ instead of $\mathbb R^3$. You're right that it does not make sense otherwise.

7. Nov 29, 2015

### quanta13

No, I dont know the definition in English. I know that when we assume c1.V1 +c2.V2+c3.V3+...+cn.Vn=0 where c are coefficients and v are our vectors, if all c coefficients are zero, vectors are linearly independent, if not then they are dependent.

8. Nov 30, 2015

### phyzguy

OK. So are there constants c1, c2, c3, not equal to 0 such that c1 V1 + c2 V2 + c3 V3 = 0?

Ah, I see the confusion on ℝ3 vs ℝ5. WWGD is right. Perhaps you should ask your teacher to clarify.

9. Nov 30, 2015

### Staff: Mentor

Or correct it. The problem makes no sense whatever as written. The three given vectors don't belong to $\mathbb{R}^3$, so questions about whether they are linearly independent or dependent are meaningless.

10. Dec 10, 2015

### HallsofIvy

Staff Emeritus
Yes, this doesn't make sense! These vectors are in R^5, not R^3 so that last is probably a typo. Yes, these vectors are "independent" if and only if a(2, 0, 1, -1, 0)+ b(1, 2, 0, 3, 1)+ c(4, -4, 3, -9, -2)= (0, 0, 0, 0, 0) implies that a= b= c= 0. That is the same as (2a+ b+ 4c, 2b- 4c, a+ 3c, -a+ 3b- 9c, b- 2c)= (0, 0, 0, 0, 0) which implies 2a+ b+ 4c= 0, 2b- 4c= 0, a+ 3c= 0, -a+ 3b- 9c= 0, and b- 2c= 0. Solve those equations. a= b= c= 0 is an obvious solution. Are there others?