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I have 4 vectors:

$u_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix} $, $\; u_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \\ \end{pmatrix} $, $\; u_3 = \begin{pmatrix} 1 \\ 1 \\ 0 \\ \end{pmatrix} $, $\; u_4 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ \end{pmatrix} $

and I wanna express the following vector in terms of them:

$\; v = \begin{pmatrix} 2 \\ 3 \\ 4 \\ \end{pmatrix} $

I'm working this way:

First put vectors in a matrix and then put in rref:

$ \left[ \begin{array}{cccc|c} 1&1&1&0&2\\ 1&1&1&1&3\\ 1&0&0&0&4 \end{array} \right] $ => $ \left[ \begin{array}{cccc|c} 1&1&1&0&2\\ 0&0&0&1&1\\ 0&-1&-1&0&2 \end{array} \right] $ => $ \left[ \begin{array}{cccc|c} 1&1&1&0&2\\ 0&-1&-1&0&2\\ 0&0&0&1&1 \end{array} \right] $ =>

$ \left[ \begin{array}{cccc|c} 1&1&1&0&2\\ 0&1&1&0&-2\\ 0&0&0&1&1 \end{array} \right] $ => $ \left[ \begin{array}{cccc|c} 1&1&1&0&2\\ 0&1&1&0&-2\\ 0&0&0&1&1 \end{array} \right] $ => $ \left[ \begin{array}{cccc|c} 1&0&0&0&4\\ 0&1&1&0&-2\\ 0&0&0&1&1 \end{array} \right] $

but this result (4 -2 1) is meaningless to me (because $4u_1 -2u_2 + u_3 \ne v$)... does it make any sense? Can it represent some sort of coeficients of the linear combination?

If I swap $u_3$ for $u_4$ in the matrix, the (4 -2 1) is the same but it DOES make sense, because $4u_1 -2u_2 + u_4 = v$

How can I write v as combination of u's ? If I did the right way, how does this make sense?

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  • $\begingroup$ Are $u_2$ and $u_3$ supposed to be identical? $\endgroup$ Commented Jul 15, 2016 at 2:53
  • $\begingroup$ yes, just for purpose of testing, and anyway i think this shouldnt matter, once I have enough vectors to write v as combination of u's $\endgroup$ Commented Jul 15, 2016 at 2:55
  • $\begingroup$ @Daniel Check your RREF. That $-2$ in the last column does NOT signify that you have -2 $u2$'s $\endgroup$ Commented Jul 15, 2016 at 3:02

3 Answers 3

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If I rewrite your augmented matrix as follows:

$\begin{bmatrix} 1&0&0&0&\\ 0&1&1&0&\\ 0&0&0&1&\\ \end{bmatrix}$ $\begin{bmatrix} a_1\\ a_2\\ a_3\\ a_4\\ \end{bmatrix}$ $= \begin{bmatrix} 4\\ -2\\ 1\\ \end{bmatrix}$

Where $a_1, \cdots a_4$ are the coefficients of $u_1,\cdots u_4$

$\begin{bmatrix} a_1\\ a_2+a_3\\ a_4\\ \end{bmatrix}$ $= \begin{bmatrix} 4\\ -2\\ 1\\ \end{bmatrix}$

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Note that $(4,-2,1)$ by itself cannot be the right answer since you need four coefficients to give a linear combination of your four vectors.

Since your reduced matrix has a non-leading (non-pivot) column, the system will have infinitely many solutions. You could take the third coefficient to be zero, giving the solution $(4,-2,0,1)$ which is correct since $$4u_1-2u_2+0u_3+u_4=v\ .$$ Or you could take the second to be zero giving $(4,0,-2,1)$, which is also correct since $$4u_1+0u_2-2u_3+u_4=v\ .$$ The general solution of your system is $$(4,-2-\alpha,\alpha,1)$$ where $\alpha$ is a scalar, and this is also correct for any $\alpha$ since $$4u_1+(-2-\alpha)u_2+\alpha u_3+u_4=v\ .$$ Hope this helps!

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  • $\begingroup$ how can i find v in terms of u's? Is that method wrong? $\endgroup$ Commented Jul 15, 2016 at 2:59
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    $\begingroup$ No it's fine, but I think you are misinterpreting the final reduced matrix. Remember that the matrix represents a system of linear equations, and you don't have any answer to your problem until you have solved the system. $\endgroup$ Commented Jul 15, 2016 at 3:01
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You need three linearly independent vectors to form a basis for three dimensions.

As $u_2$ and $u_3$ are in fact identical and the same, the redundant and duplicated one is unnecessary and also not needed.

Reapply your process with only three column vectors: $u_1, u_2, u_4$, and $v$.

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  • $\begingroup$ no, i dont need three. for example, i can write (1,1,0) as combination of (1,0,0) and (0,1,0) only.. $\endgroup$ Commented Jul 15, 2016 at 3:02
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    $\begingroup$ @Daniel you don't need three to write some examples. However to be able to write any example, you will need three. Read the sentence again, "you need three linearly independent vectors to form a basis for three dimensions." As it so happens, $(1,1,0)=1(1,0,0) + 1(0,1,0) + 0(0,0,1)$ happens to use a zero multiple of the third standard basis element. The point is that without needing to change those three, we could also express $(0,1,1)$ or $(5,2,-1)$ or $(\pi,e,-\pi)$ or any other as well. Sometimes zeroes will be used, that's fine and expected. $\endgroup$ Commented Jul 15, 2016 at 3:25

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