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Let S={(1,k,k+1,k-1)|k is real} and W is subspace of vector space R4. Prove that a=(0,1,1,1) is in W

Linear Algebra
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What does S have to do with this though?
sorry :D W is generated by S :)
meaning... S spans W?

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Well, then, all you need to show, for the meantime, is that (0,1,1,1) can be written as a linear combination of elements in S.
I know, but I don't know how
Unfortunately, I know no other way than to figure out exactly what linear combination of S could possibly be equal to (0,1,1,1). Stand by...
should I take vectors from S for different k and make linear combination... for example (0,1,1,1)=a*(1,2,3,1)+b*(1,1,2,0)+... how many vectors should I take? 4?
It can be as many vectors as you like, but there may well be a minimum. Yes, you form vectors from S using different values for k.
ok, thanks :) that's what was confusing me :)
You can do it from here?
you mean to write it here?
I'm not even finished thinking about it XD
// too comfy to stand up and look for a pencil and paper XD
I'll write it :)
Never mind, it's actually a lot simpler than I thought... just consider the elements of S when k=0 and k=1.
but I think I can't take less than 4 vectors from S since dimension of R4 is 4, so I have to write (0,1,1,1) as linear combination of 4 vectors
No... the dimension of R4 is 4, so that means any vector can be written as a linear combination of 4 vectors, it doesn't mean that it can't be smaller than that. For instance, in R4 itself, the vector (1,0,0,0) can certainly be written as a linear combination of four vectors: (1,-1,0,0) + (0,1,-1,0)+(0,0,1,-1)+(0,0,0,1) But it doesn't mean it can't be written as a linear combination of two: (1,-1,0,0) + (0,1,0,0) Or even just one (1,0,0,0) Get the drift? XD
yes, but... it's safer to write with 4 vectors and get some 0 coefficients :D never mind... I go now, have exam at 5 :) wish me luck :) and thanks again :)
No problem :)

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