Vector Dependancy: Show Linear Dependence & Independence

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Here is the question:

Let w, x, y, z be vectors in a vector space. Show that
(i)the vectors w + x, x + y, y + z and z +w are linearly dependant
(ii) if w, x, y, z are linearly independent the so are the vectors
w + x + y, x + y + z and w + 2z.

Anyone know what that is asking, i have been putting this into simulataneous equations and solving to find the constants.

Can I do that with this?

Cheers
 
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You are given a problem to show that vectors are "linearly dependent" or "linearly independent" and no one has ever told you what those words mean? How awful of them! I recommend you look up the words in your textbook!
 
Yes I know what linear dependancy and independcy are, the whole book is on that.

Simplified I know it as:
Vectors in V are linearly dependant if there are scalars that do not equal zero.

Otherwise the vectors are linearly independant, if all the scalars have to equal zero.

What I am asking is how I work it out.

I am used to finding out if
(1,0,1,2)^T and (0,1,1,2)^T and (1,1,1,3)^T are linearly independant.

and solving for

1(landa1) + + 1(landa3) = 0
1(landa2) + 1(landa 3) = 0
1(landa1) + 1(landa2) + 1(landa 3) = 0
2(landa1) + 2(landa2) + 3(landa 3) = 0

Not x + y and such like, I am asking what they mean?

Cheers
 
My point was that you asked "What does that mean?" in reference to a problem about independence! I could only interpret that as asking what "independence" meant.

Simplified I know it as:
Vectors in V are linearly dependant if there are scalars that do not equal zero.

Otherwise the vectors are linearly independant, if all the scalars have to equal zero.
That's way over simplified. There are always scalars that are not equal to zero! You mean "If a linear combination of the vectors (sum of scalars times the vectors) is equal to 0, with not all the scalars being 0, then the vectors are dependent" and the opposite for independence. It is important to mention that the scalars are in the linear combination and that the linear combination is equal to 0.

(i)Let w, x, y, z be vectors in a vector space. Show that
(i)the vectors w + x, x + y, y + z and z +w are linearly dependant
Okay look a a(w+x)+ b(x+ y)+ c(y+ z)+ d(z+ w)= 0. Multiply out and combine like vectors. Is it possible to choose a, b, c, d not all 0?
Notice that nothing is assumbed about x, y, z, w.

(ii) if w, x, y, z are linearly independent the so are the vectors
w + x + y, x + y + z and w + 2z.
Again look at a(w+ x)+ b(x+y+ z)+ c(w+ 2z)= 0. Is it possible to choose a, b, c, not all 0 so that this is true? Here, you are given that w, x, y, z are linearly independent.
 
ok yes that's all i needed thank,

will have a go at that, I did not know what was meant by w, x, y and z.

I did not know whether each letter meant a vector in in direction, or a vector in all directions, if you get what I mean.

i did not know whether it meant w = 3 or w = (1,2,1,4)^T
 
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