Linear combination issue

Question

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?

Answer 1

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!

Answer 2

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 $v$.

Answer 3

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 $u_1,\cdots u_4$

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