Question

Given that {u,v,w} is a linearly independent set in V , determine which of the following are also linearly independent in V . Explain

(a) {u + v − w,v + w}
(b) {u,v − w,v + w}
(c) {u + v,v + w,u − w}

Answer 1

Given that \(\large\{u,v,w\}\) is a linearly independent set in \(\large V\), does it follow that

(a) \(\large a (u+v-w)+b(v+w) = 0\implies a=b=0\)? If so, then the set is L.I.

(b) \(\large au+b(v-w)+c(v+w) = 0\implies a=b=c=0\)? If so, then the set is L.I.

(c) \(\large a(u+v)+b(v+w)+c(u-w) = 0\implies a=b=c=0\)? If so, then the set is L.I.

Do you know how to proceed from here?

Answer 2

let me try the first one, if I am wrong, please correct me
we have to prove that (u + v -w) and (v+w) is linearly independent
\(\left[\begin{matrix}1&1&-1\\0&1&1\end{matrix}\right]\)\(\left[\begin{matrix}u\\v\\w\end{matrix}\right]\)= \(\left[\begin{matrix}u+v-w\\v+w\end{matrix}\right]\)
since the left matrix is linearly independent --> the right one is linearly independent (However, I feel how WEAK this logic is, hihihi...)
so a(u+v-w) + b(v+w) =0 iff a =0, b=0

Answer 3

Since I'm allergic to matrixes, I'm with @ChristopherToni on this one :>

Answer 4

If you want to go down the matrix route, here's a more succinct/right-to-the-point version: the matrix \(\large \begin{bmatrix} 1 & 1 & -1\\ 0 & 1 & 1\end{bmatrix}\) has rank 2, and that is the dimension of the spanning set \(\{u+v-w,v+w\}\). Hence, that set is linearly independent.

Answer 5

Thank you, I learned it from you. XD

Answer 6

@MarcLeclair
still there?

Answer 7

Yeah but.... I mean I know what linear independence is... in the sense that they have to "add up" to 0? I mean it sounds wrong but I mean it in the sense

a(u1)+ a(u2)+...+a(Un)=0 That is how I learned it so I'm all thrown off by this question without any... numerical vectors (if you can say that)

Answer 8

Then first, let me make one thing about linear independence clear...ok? ^_^

u, v, and w, are linearly independent (vectors!) right?
That means if there are constants (aka, scalars) a, b, c for which THIS equation holds true:
\[\Large a\vec u + b\vec v + c\vec w = \vec 0\]

Then a, b, and c must ALL be zero, without question.

Catch me so far?

Answer 9

Yeah That I do understand.

Answer 10

Okay, let me demonstrate how to do it the way Chris Toni said it.
But to avoid confusion, I will not use a, b, and c for my scalars, all right.

Let's try the first choice, your two vectors are u+v-w and v+w
right?
Suppose there are scalars p and q such that this
\[\Large p(\vec u + \vec v - \vec w) +q(\vec v + \vec w) = \vec 0\]
holds... catch me so far?

Answer 11

yeah your 2 scalars have to be equal to 0 for the equation to hold true.

Answer 12

Well, that's what we want to prove, namely, that, since the equation holds, p and q have to both be zero...
Can you do that?

Answer 13

hummm thats where I get stuck, so if i have 3 vecots u ( u1 + u2 + u3) + (v1+v2+v3)-(w1+w2+w3) , do I just take random vectors ? In the sense that I can say u = (1,1,1) and v (1,0,1) and so on?

If I can't do that I have no idea how to prove it.

Answer 14

You want me to demonstrate? ^_^
I'll do the first, and hopefully, you'll already see the pattern, and be able to do the second and third with ease, you ready?

Answer 15

yeah ahahha I just want to understand the process of it. doesn't seem to be clicking in my head

Answer 16

@MarcLeclair
Well, anyway, pay attention:
Suppose this\[\Large p(\vec u + \vec v - \vec w) +q(\vec v + \vec w) = \vec 0\]hold...
Let's distribute p and q...
\[\Large p\vec u + p\vec v - p\vec w +q\vec v + q\vec w = \vec 0\]

Answer 17

Now, let's regroup, regroup the terms with the same vector, and then 'factor out' that vector...
\[\Large \Large p\vec u + p\vec v +q\vec v + q\vec w- p \vec w = \vec 0\]\[\Large \Large p\vec u + (p +q)\vec v + (q- p) \vec w = \vec 0\]
You follow so far?

Answer 18

yeah simple algebra in this case, I'm okay with that

Answer 19

Well, remember that u, v, and w ARE linearly independent, so, as I said earlier, *without question*
p, p+q and q-p must ALL be zero...


p = 0
p+q = 0
q-p = 0

...a simple system of linear equations, one that easily leads to the conclusion that both p and q must be zero, yes? ^_^

Answer 20

wait I don't understand how you took the assumption they must all be zero from the fact that our vectors are initially linearl independent. I mean these vectors combination give new vectors after all no?

Answer 21

I know you said the au + bv + cw = 0 then a,b,c must all be = 0 which I agree completely

Answer 22

ohh...you're getting confused...well, that at least means you're paying attention... or I'm poor at explaining things (T.T)

...anyway, let me bring you through it again, a bit more quickly this time...
--YES, u, v, and w are linearly independent, that is given...
--YES we have two *new* vectors, u+v-w and v+w

So... to check if they are linearly independent, we assume that there are constant scalars p and q such that
p(u+v-w) + q(v+w) = 0

To show linear independence, we must prove that p=q=0

NOW...
from here:
p(u+v-w) + q(v+w) = 0
Algebraic manipulation gives us
pu + (p+q)v + (q-p)w = 0

Get it?
We manipulated our equation so that it's now a linear combination of u,v,w, the vectors that we ACTUALLY know something about....
and, again, since u, v, and w are independent, if ever there are scalars that lead to
au + bv + cw = 0
then a = b = c = 0

It just so happened that a in this case is p... b in this case is p+q...and c in this case is q-p... and they all have to be zero, sinve u, v, and w are linearly independent...

p = 0
p+q = 0
q-p = 0

Thus, p = q = 0.

Answer 23

OOOHHH we came back to our initial equation which was equal to 0 and (p+q) = b and so on. makes sense ahahha thanks!

Answer 24

Hey, this isn't done yet, perhaps you could try the second one?
You do a very similar process...

Answer 25

yeah sure let me see

Answer 26

You have three vectors, so... I suggest p, q, and r? lol :>

Answer 27

yeah so you'd do the same process ( I won't use the whole notation system it takes too long)

So I'll havev p(u)+q(v-w)+c(v+w)=0
which simplifies to pa+qv-qw+cv+cw=0
Group similar vectors and factor out the scalar so...
p(u)+(q+c)v+(-q+c)w = 0
Therefore it is like our initial equation
a(u)+b(v)+c(w) = 0 so it is Linear independent

Answer 28

whoa now, you skipped a step...
it means that all the constants, p, q+c and -q+c have to all be zero...
p = 0
q + c = 0
-q + c = 0
and you solve that system...

Answer 29

I thought it was implied because I referred to my "theorem" ( well for R3 that is) but you're right it does seem kind of misleading without mentioning that the scalars = 0

Answer 30

Okay, do the third, without missing a step... (pretty please? :3 )
^_^

Answer 31

however the last one isn't suppose to be linear independent , it is suppose to be dependent. so I'm not too sure ( I just did it on paper but it seems kinda odd)

Answer 32

How do you know that it's supposed to linearly dependent?

Answer 33

I have the answers without the work ( review sheet) so I just know the the answer is "no" haha

Answer 34

Well then.... what's the opposite of linearly independent?

Answer 35

...wait, don't answer that, ^that above there is just asking for trouble :>

To show that a set of vectors is linearly dependent, either, you provide specific scalars, a, b, and c such that a linear combinations involving those result in the zero-vector (or course, excluding a = b = c = 0)

OR you could express one of the vectors as a linear combination of the other two...

Answer 36

Well I would say it is dependent when one vector is the result of the two other vectors, but I can't really see it in this case... I end up with (p+r)u + (p+q)v+(q-r)w

Answer 37

All I can't think of is that one of the scalars =/= 0 because they are all combination of one another. but I mean it is a weird thought to say.

Answer 38

I'm not too good at teaching how to prove linear dependence... I usually just take a good look at them and see which one is a linear combination of the others...

uhh, did you notice that the first vector is just the sum of the second and third? :P

Answer 39

how do you notice that.. is it because the q is like an intermediate variable ( I'm thinking partial derivative... haha)

Answer 40

Mercy...
I don't know how I notice it... I just do T.T

(Linear Algebra is definitely not my strong point, sorry)

But yeah, first vector is the sum of the second and third... so, linearly dependent :)

Answer 41

alright thanks a lot!! that was really helpful of you thanks so much. Not a lot of people here help with linear... haha

Answer 42

That's because Linear Algebra is the work of the devil >:)

Answer 43

it is a mind numbing class :)

Answer 44

College maths?

Answer 45

Yeah I'm in something called cegep, it's in quebec, its mandatory...we have uni classes / grade 12 classes. So I'm taking cal 3, differential equations and Linear algebra as science courses :/

Answer 46

Canada :D
Then how old are you?
:?

Answer 47

I'm 19! turning 20 soon and yes Canada, the land of the snow :)

Answer 48

haha... vieux :P