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Consider the following system of equations with strictly positive unknowns $\lambda_1,\lambda_2,\lambda_3,\lambda_4$

$$ \begin{cases} \lambda_1 d=\lambda_4 a\\ \lambda_2 d=\lambda_4 b\\ \lambda_1 c=\lambda_3 a\\ \lambda_2 c=\lambda_3 b\\ \lambda_3 d=\lambda_4 c\\ \lambda_1+\lambda_2+\lambda_3+\lambda_4=1 \end{cases} $$

and parameters $a>0,b>0,c>0,d>0$ with $a+b+c+d=1$. Show that the unique solution of the system is

$$ \lambda_1=a\\ \lambda_2=b\\ \lambda_3=c\\ \lambda_4=d\\ $$

I tried by taking several routes but I couldn't find any clear pattern.

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3 Answers 3

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Using SymPy, we create the augmented matrix:

>>> from sympy import *
>>> a, b, c, d = symbols('a b c d', real=True, positive=True)
>>> 
>>> M = Matrix([[ d, 0, 0,-a, 0],
                [ 0, d, 0,-b, 0],
                [ c, 0,-a, 0, 0],
                [ 0, c,-b, 0, 0],
                [ 0, 0, d,-c, 0],
                [ 1, 1, 1, 1, 1]])

Imposing the constraint $a + b + c + d = 1$:

>>> M.subs(d,1-a-b-c)
Matrix([
[-a - b - c + 1,              0,              0, -a, 0],
[             0, -a - b - c + 1,              0, -b, 0],
[             c,              0,             -a,  0, 0],
[             0,              c,             -b,  0, 0],
[             0,              0, -a - b - c + 1, -c, 0],
[             1,              1,              1,  1, 1]])

Using Gaussian elimination to compute the RREF:

>>> _.rref()
(Matrix([
[1, 0, 0, 0, a/((-a - b - c + 1)*(a/(-a - b - c + 1) + b/(-a - b - c + 1) + c/(-a - b - c + 1) + 1))],
[0, 1, 0, 0, b/((-a - b - c + 1)*(a/(-a - b - c + 1) + b/(-a - b - c + 1) + c/(-a - b - c + 1) + 1))],
[0, 0, 1, 0, c/((-a - b - c + 1)*(a/(-a - b - c + 1) + b/(-a - b - c + 1) + c/(-a - b - c + 1) + 1))],
[0, 0, 0, 1,                    1/(a/(-a - b - c + 1) + b/(-a - b - c + 1) + c/(-a - b - c + 1) + 1)],
[0, 0, 0, 0,                                                                                       0],
[0, 0, 0, 0,                                                                                       0]]), [0, 1, 2, 3])

Simplifying:

>>> simplify(_)
(Matrix([
[1, 0, 0, 0,              a],
[0, 1, 0, 0,              b],
[0, 0, 1, 0,              c],
[0, 0, 0, 1, -a - b - c + 1],
[0, 0, 0, 0,              0],
[0, 0, 0, 0,              0]]), [0, 1, 2, 3])
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  • $\begingroup$ Thanks, but I'm not sure about the conclusion. Could you indicate in English what we can conclude from your lines? $\endgroup$
    – Star
    Mar 9, 2019 at 19:10
  • $\begingroup$ @STF We can conclude precisely what the question asked one to show. Note that the RREF'ed augmented matrix is $$\begin{bmatrix}1 & 0 & 0 & 0 & a\\0 & 1 & 0 & 0 & b\\0 & 0 & 1 & 0 & c\\0 & 0 & 0 & 1 & - a - b - c + 1\\0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0\end{bmatrix}$$ $\endgroup$ Mar 9, 2019 at 19:26
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Suppose there are two different solutions. Suppose also that these two solutions have common $\lambda$, for example $\lambda_1$. Then from the very first equation it is easily seen that $\lambda_4$ is also common, and so do others.That means that if there exist two solution they should contain different $\lambda$'s. Denoting $\lambda^{(1,2)}$ the ones from solution #1,2, we can write the following lines (substracting the equations for the first and second solution and dividing them by each other)

$$[\lambda_3^{(1)} - \lambda_3^{(2)}]=\frac{c}{a}[\lambda_2^{(1)} - \lambda_2^{(2)}]$$ $$[\lambda_1^{(1)} - \lambda_1^{(2)}]=\frac{b}{c}[\lambda_2^{(1)} - \lambda_2^{(2)}]$$ $$[\lambda_4^{(1)} - \lambda_4^{(2)}]=\frac{d}{a}[\lambda_2^{(1)} - \lambda_2^{(2)}]$$

Now let's substract the last equation for the first solution from the same equation for the second.

$$[\lambda_1^{(1)} - \lambda_1^{(2)}]+[\lambda_2^{(1)} - \lambda_2^{(2)}] +[\lambda_3^{(1)} - \lambda_3^{(2)}]+[\lambda_4^{(1)} - \lambda_4^{(2)}]=0$$

Putting it all together we write:

$$[\lambda_4^{(1)} - \lambda_4^{(2)}][\frac{b}{c}+\frac{a}{d}+\frac{c}{a}+1]=0$$

We have previously proven that the first term cannot be zero. It means that the second is zero. But that is impossible, because one of the parameters nust be negative in this case. So we conclude that there is only one solution.

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$$theorem:\\ \\ the \ system\ of\ equations\ \ \ y_{i}=\sum_{k=1}^{n}a_{ik}\ x_{k}+b_{i}\ \ \ \ \ , i=1,2,3....n\\ \\\ where\ y_{i},\ x_{i}\ and\ b_{i}\ are\ real\ numbers\ ,\ and\ (a_{ik})\ is\ a\ fixed\ real \ n\ *m\ matrix\ \\ \\ \\ with\ \sum_{k=1}^{n}\left | a_{ik} \right |\leqslant a\leqslant 1\ \ \ \ \ has\ \ a\ unique\ solution\ \ x=(x_{1},x_{2},.......)$$

now I will prove this theorem : first we choose the complete metric space $R^{n}$ with metric d defines by d(x,y)=$max_{1\leqslant i\leqslant n}\left | x_{i}-y_{i} \right |$

and we define a mapping T by the above system of linear equations as $y_{i}=Tx_{i}\sum_{k=1}^{n}a_{ik}\ x_{k}+b_{i}$

now , we prove that the mapping T is contraction , let Wi=Tzi defined by :

$$so\ we \ have\ \ d(T_{X_{i}}, T_{X_{i}})=d(y_{i}, w_{i})=max_{1\leqslant i\leqslant n}\ \left | y_{i-w_{i}} \right |\\ \\ \\$$ $$=max_{1\leqslant i\leqslant n}\ \left | \sum_{k=1}^{n} a_{ik}(x_{k}-Z_{k})\right |\leqslant max_{i}\ \sum_{k=1}^{n}\left | a_{ik} \right |\left | x_{k-Z_{k}} \right |\leq max_{i}\sum_{k=1}^{n}|a_{ik}|max|x_{k}-Z_{k}|\ =max_{i}\sum_{k=1}^{n}|a_{ik}|d(x_{i},Z_{i})\leqslant \alpha d(x_{i}\ , Z_{i})\\ \\$$ \ $$Hence\ T\ is\ contraction\ mapping\ , therfore\ there\ exists\ at\ least\ one\ point\ x \in \ R^{n}\ \ such\ that\ \ \ \\ \\$$

$$T_{x}=x\ \ \ \ , \ \ x_{i}\ =\sum_{k=1}^{n}a_{ik}\ x_{k}\ +b_{i}\\ \\$$

and this finishes proof

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