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If $\Bbb{C}$ is the field of complex numbers, which vectors in $\Bbb{C}^3$ are linear combinations of $(1, 0, -1)$, $(0, 1, 1)$, and $(1, 1,1)$?

I didn't understand the question actually. It's an exercise in Hoffman's linear algebra book (page 41, exercise 3).

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Do you understand what a linear combination is? Where are you stuck? –  Ben Millwood Aug 31 '12 at 14:01
    
Yes, I know what is a linear combination. What do mean "spanned by (1,0,−1), (0,1,1) and (1,1,1)"? –  TalBin Aug 31 '12 at 14:12
    
The word "spanned" doesn't appear anywhere in your question. Can you tell me any linear combination of the given vectors? –  Ben Millwood Aug 31 '12 at 14:16
    
@BenMillwood span refers to an earlier comment by me. I removed it because obviously OP (from his/her comments down) doesn't know what a span is. –  user2468 Aug 31 '12 at 14:17
    
@JenniferDylan: oh, okay. I still think this is an incomplete question. –  Ben Millwood Aug 31 '12 at 14:18

3 Answers 3

up vote 4 down vote accepted

Well, you can make $(1, 0, 0) = (1, 1, 1) - (0, 1, 1)$.

You can make $(0, 0, 1) = (1, 1, 1) - (0, 1, 1) - (1, 0, -1)$

Therefore, we can get $(0, 1, 0) = (0, 1, 1) - (0, 0, 1) = 2 * (0, 1, 1) - (1, 1, 1) + (1, 0, -1)$.

So, we can write $(1, 0, 0), (0, 1, 0), (0, 0, 1)$ as linear combinations of these. What does that tell us?

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That (a,b,c), when a,b and c in C, are linear combinations of (1,0,−1), (0,1,1), and (1,1,1)? –  TalBin Aug 31 '12 at 15:11
    
@TalBin I'm not sure what you just said. The conclusion you can draw, which is perhaps what you said, is that every vector in $\mathbb{C}^3$ can be written as a linear combination of the three you started with. Do you understand why? –  Graphth Aug 31 '12 at 15:15
    
Yes, I need to find vectors in C3 which equivalant to the linear combination c1(1,0,1)+c2(0,1,1)+c3(−1,1,1).You used elementary row operations to get the last linear combination. So c1(1,0,0)+c2(0,1,0)+c3(0,0,1)=(a,b,d). c1=a, c2=b, c3=d. correct? –  TalBin Aug 31 '12 at 15:38
    
@TalBin Yes, this means that for any vector, $(x, y, z)$ in $\mathbb{C}^3$, there exist $c_1, c_2, c_3$ such that $c_1 (1, 0, 1) + c_2 (0, 1, 1) + c_3 (1, 1, 1) = (x, y, z)$. And, to show this, instead of figuring out $c_1, c_2, c_3$ directly for every possible vector, I showed that you can make the standard basis vectors, $(1, 0, 0), (0, 1, 0), (0, 0, 1)$, because then it is obvious that every other vector can be formed as a linear combination of these. –  Graphth Sep 4 '12 at 12:54

You need to figure which vectors $(a,b,c)$ can be written as

$$c_1(1,0,1)+c_2(0,1,1)+c_3(-1,1,1)=(a,b,c)$$

This is s system of equations in $c_1,c_2,c_3$, and the question asks you for which $(a,b,c)$ does this system have solution.

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@TalBin: this is precisely the definition of a linear combination. If someone were to give you a vector $(a,b,c)$ you could try to determine what the values of $c_1, c_2, c_3$ are. When you do, you will find out what vectors it works for. It is a set of simultaneous equations that has been made fairly easy to solve. –  Ross Millikan Aug 31 '12 at 14:24

HINT: $\det\pmatrix{1 &0 &1\\0&1&1\\-1&1&1}\neq 0$

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I didn't learn determinants yet. –  TalBin Aug 31 '12 at 14:13
    
I can't imagine this answer being useful to anyone who didn't already know everything necessary to answer the question. –  Ben Millwood Aug 31 '12 at 14:15
    
Why not starting to learn something about them: ...the determinant of a matrix (with real or complex entries, say) is zero if and only if the column vectors of the matrix are linearly dependent. (Determinant/Applications/Linear Dependance) –  draks ... Aug 31 '12 at 14:18

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