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A standard application of Linear Algebra is circuits and Kirchhoff's Laws. Does anyone know of a proof of uniqueness of a solution of a system given by these laws? There are many, many examples, but little theory regarding why there is always a unique solution.

For reference (Wikipedia)

  • At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node

  • The algebraic sum of the products of the resistances of the conductors and the currents in them in a closed loop is equal to the total emf available in that loop.

My thoughts are as follows: Initially these laws setup two systems $Ax = 0$ and $Bx= b$, respectively. If there are $n$ nodes and m currents, then $A$ is a $n \times m$ matrix. If there are l loops in the circuit, then $B$ is a $l \times m$ matrix. I tried to work with the augmented matrix $$ \left[\begin{matrix} A \\B\end{matrix} \right|\left.\begin{matrix}0\\b\end{matrix}\right]$$ But I see no reason why this always has a unique solution.

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    $\begingroup$ @NoChance that's not true; the equations need to be independent and consistent. $\endgroup$
    – Ben Grossmann
    Apr 14, 2016 at 18:04
  • $\begingroup$ I would recommend looking at Gilbert Strang's Introduction to Linear Algebra if you have it on hand; he does look at Kirchhoff's laws in this way. I don't remember if he proves uniqueness, though. $\endgroup$
    – Ben Grossmann
    Apr 14, 2016 at 18:08
  • $\begingroup$ If it's not in there, you might need some results about "incidence matrices" from graph theory (at least, that would save you some work). $\endgroup$
    – Ben Grossmann
    Apr 14, 2016 at 18:09
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    $\begingroup$ You can't even be sure there is a solution -- suppose, for example, you short the two terminals of an EMF; then the resulting system of equations has no solution. $\endgroup$
    – hmakholm left over Monica
    Apr 14, 2016 at 18:15
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    $\begingroup$ And if you have negative resistances, then solutions are not necessarily unique either -- so a proof of uniqueness will need to depend on properties of order, not just on $\mathbb R$ being a field. $\endgroup$
    – hmakholm left over Monica
    Apr 14, 2016 at 18:17

1 Answer 1

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There is a statement that works for your purposes (proposition 9.4) presented in Markov Chains and Mixing Times (Levin, Peres, Wilmer) that you can follow by reading pages 115-118.

Perhaps there are other, better references for you purpose, but this is the first that comes to mind.

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  • $\begingroup$ Yes, this is what I'm looking for. Thank you. $\endgroup$
    – CarlOlimb
    Apr 14, 2016 at 18:35

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