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Question: Decide on the maximum number of linearly independent vectors among a given a set of vectors. $$ A = (1,-1,0,0) \quad B = (1,0,-1,0) \quad C = (1,0,0,-1) \\ D = (0,1,-1,0) \quad E = (0,1,0,-1) \quad F = (0,0,1,-1) $$

I don't understand that question. how can I solve this question?

edit: Actually in that question, I though like that, how can write that vectors with another two vectors and so how many vectors there are like that?? is this true. can anyone explain

I mean,

A-B = (0,1,-1,0)

so, A-B = C I can write C instead of A and B.. It is my understanding.

mvw
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Njx
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  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos May 29 '18 at 11:05
  • Okay. I edited. my thinking is this. – Njx May 29 '18 at 11:09
  • Create a matrix with rows your vectors. Do row reduction to put it in echelon form. Any rows which are not all 0's are linearly independent. You probably did a few examples in class. – Paul May 29 '18 at 11:11
  • I mention that in my edited. but I am not sure. yes we did it in classroom like in my edited shape. Is it true? I mentioned just one of part of that question. – Njx May 29 '18 at 11:13

2 Answers2

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If your set consists of vectors $\{u_1,u_2,\dots u_k\}$, it may not be a linearly independent set but you can always take a subset that is linearly independent (usually we want the maximum number of these). For example if my set consists of $u_1=(1,1), u_2=(0,1),u_3=(2,3)$, then a set with maximum number of linearly independent vectors, that I can take from this set is $\{(1,1),(0,1)\}$ since $u_3$ is a linear combination of the first two. Equally it would be fine to take any two here, since any pair out of this set will be linearly independent.

As pointed out in the comments, you probably were given a procedure to follow. If unsure of why and how that works, you can ask.

EDIT:

Putting your vectors into a matrix, as columns, and row reducing will give a rank 3 matrix, which means we have three linearly independent vectors. The first three columns consist of elementary vectors and the remaining columns contain coefficients of linear dependence for the corresponding vectors. Columns 4 is $(-1,1,0,0)$ which tells me that $D=-A+B$

AnyAD
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  • oooo. my opinion was true. thank you so much. I am clear now. – Njx May 29 '18 at 11:14
  • I will accepted your answer. but now there is minute limit. – Njx May 29 '18 at 11:15
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    I am confused about one thing. for example, if we write A-B = C, and C is not independent. normally we don't come in answer. but if I write A = C+B. so this time A is being dependent. so We don't come "A" in answer. So answer can be changeable? Is it important. or we must explain how we solve the question – Njx May 29 '18 at 11:38
  • You're right, there can be multiple correct answers. Usually we apply something called the 'column method' (described in the edit section) and take the column vectors in the original matrix (original bectors) which have 'transformed' (via row reduction) to elementary vectors. – AnyAD May 29 '18 at 11:44
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    thank you again. I see :) – Njx May 29 '18 at 11:50
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E.g. if $\{ A, B, C \}$ are linear independent, then the equation $$ c_1 A + c_2 B + c_3 C = 0 $$ has only the trivial solution $c = (c_1, c_2, c_3) = (0, 0, 0) = 0$.

There are different methods to determine this. E.g. you can formulate this as system of linear equations $$ (A^\top B^\top C^\top) c = M c = 0 $$ and use Gauss elimination to find out the rank of $M$, where $\top$ means matrix transposition (here turning row vectors into column vectors).

You can think in advance about how many vectors can be linear independent at most.

Choosing the right subset is bit trial and error.

mvw
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