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Problem:

Consider the vectors $u = (1,2,3)$ and $v = (2,3,1)$ in $R^3$. Write $w = (1,3,8)$ as a linear combination of $u$ and $v$.

Answer:

$$ c_1( 1,2,3) + c_2( 2,3,1 ) = (1,3,8) $$ Now to find $c_1$ and $c_2$ we setup a matrix. $$ \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 3 \\ 3 & 1 & 8 \end{bmatrix} \sim \begin{bmatrix} 1 & 2 & 1 \\ 0 & -1 & 1 \\ 0 & -5 & 5 \end{bmatrix} $$ This gives us $c_2 = -1$. \begin{align*} c_1 + 2c_2 &= 1 \\ c_1 + 2(-1) &= 1 \\ c_1 &= 3 \\ \end{align*} Hence the solution is $w = 3u - v$. This answer matches the back of the book; however it seems to me that there should be more than one solution. Is this solution unique?

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    $\begingroup$ You found the unique solution of the defining equations so yes, it must be unique. Out of interest, why do you think it shouldn't be? $\endgroup$
    – Paul
    Feb 9, 2023 at 19:36
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    $\begingroup$ If one was a scalar multiple of the other, it would not be unique. But here, that isn't happening. $\endgroup$
    – Randall
    Feb 9, 2023 at 19:38
  • $\begingroup$ @Paul I guess I feel it is not unique because the associated matrix is not square. $\endgroup$
    – Bob
    Feb 9, 2023 at 19:41
  • $\begingroup$ As @Randall says, if $u$ and $v$ were multiples of each other the solution would not be unique. Your overdetermined system might have three possibilities, a unique consistent solution, an infinite set of solutions (the case just mentioned) or no solution so that not such combination is possible $\endgroup$
    – Paul
    Feb 9, 2023 at 19:55
  • $\begingroup$ @Bob If you were you thinking that there are 3 linear equations and 2 unknowns, think that we are basically solving the problem in 2d, placing all three vectors on the same plane. We can do it, provided that the third vector is not orthogonal to the other two. $\endgroup$
    – Vasili
    Feb 9, 2023 at 19:57

2 Answers 2

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If $\{ v_1, v_2 \}$ is linearly independent then the linear combinations are unique. This is also true in the case $\{v_1, \ldots, v_n \}.$

For see this suppose we have $w=c_1 v_1 + c_2 v_2 = d_1v_1 + d_2v_2$. Then

$$ (c_1 - d_1) v_1 + (c_2-d_2)v_2=0. $$

As $\{v_1, v_2\}$ is linearly independent we know that $c_1-d_1=0$ and $c_2 - d_2=0$, i.e. $c_1=d_1$ and $c_2=d_2$.

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To check for uniqueness, the basic approach is to assume there exists two different solutions, and see what happens. If the problem is truly unique, then you will run into a contradiction.

Suppose there were two combinations, $c_1, c_2$ and $d_1, d_2$. Then, $$(1,3,8) = c_1(1,2,3) + c_2(2,3,1) = d_1(1,2,3) + d_2(2,3,1)$$

This gives, \begin{align} c_1 + 2c_2 &= d_1 + 2d_2 \\ 2c_1 + 3c_2 &= 2d_1 + 3d_2 \\ 3c_1 + c_2 &= 3d_1 + d_2 \\ \end{align}

Rearranging, we get $$ c_1 - d_1 = 2(d_2 - c_2) = \frac{3}{2} (d_2 - c_2) = \frac{1}{3} (d_2 - c_2) $$

If the two combinations are different (i.e. $c_1 \neq d_1, c_2 \neq d_2$), then the above yields a contradiction that $2 = \frac{3}{2} = \frac{1}{3}$. To avoid this contradiction, the two combinations must be the same. Therefore, the problem has at most one unique solution.

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