Understanding Why System of Equations is Degenerate

[j-astron]

In my work (this has some practical application I won't get into), I ended up with a system of equations like this:

\(\displaystyle \displaystyle A_1 = x_1 + y_1 \)
\(\displaystyle \displaystyle A_2 = x_2 + y_2 \)
\(\displaystyle \displaystyle B_1 = x_1 + a_1y_1 \)
\(\displaystyle \displaystyle B_2 = x_2 + a_1y_2 \)
\(\displaystyle \displaystyle C_1 = x_1 + a_2y_1 \)
\(\displaystyle \displaystyle C_2 = x_2 + a_2y_2 \)

where \(\displaystyle \{A_1, A_2, B_1, B_2, C_1, C_2\}\) are known quantities, and \(\displaystyle \{x_1, y_1, x_2, y_2, a_1, a_2\} \) are unknown quantities that I would like to solve for. So, the system has 6 equations and 6 unknowns. If I go through and solve the system using substitution/elimination, I end up with equations for each of the variables in terms of the others (equations I have verified numerically):

\(\displaystyle \displaystyle y_2 = \frac{C_2 - B_2}{a_2 - a_1} \)

\(\displaystyle \displaystyle x_2 = C_2 - a_2\frac{C_2 - B_2}{a_2 - a_1} \)

\(\displaystyle \displaystyle y_1 = \frac{C_1 - B_1}{a_2 - a_1} \)

\(\displaystyle \displaystyle x_1 = C_1 - a_2\frac{C_1 - B_1}{a_2 - a_1} \)

and


\(\displaystyle \displaystyle a_1 = a_2\left(\frac{W+1}{W}\right) - \frac{1}{W} \)

where

\(\displaystyle \displaystyle W \equiv \frac{A_2 - C_2}{C_2-B_2} \)

But I've found out that you cannot solve for the last unknown, \(\displaystyle a_2\), because you end up with

\(\displaystyle \displaystyle a_2 = \frac{Va_1 + 1}{V+1}\)

where

\(\displaystyle \displaystyle V \equiv \frac{A_1 - C_1}{C_1-B_1} \)

Unfortunately, this leads to the equation

\(\displaystyle \displaystyle \left(1-\frac{V}{W}\right)a_2 = 1-\frac{V}{W} \)

You can show that \(\displaystyle V = W \) and hence we end up with

\(\displaystyle \displaystyle 0\cdot a_2 = 0 \)

I guess my question is, is there some way to understand, just by looking at the original system, that it will be degenerate, or to highlight what property of the original set of equations is leading to the degeneracy?
Thanks.

 

[tkhunny]

Two things stick out:

1) As a linear system, at first glance, one may construe a1 and a2 as arbitrary parameters, rather than "unknowns". At this point, albeit misunderstood, you've six equations and only four unknowns. One might expect this to be over-defined. This was my first warning that I should tread carefully as I go along.

2) Your demonstrated solution has difficult denominators as it goes along. Are we SURE, at each point, that these denominators are NOT zero (0)? I had no such confidence as I was looking at it. I wanted more constraints to raise this confidence before I thought I had a general solution.

Thoughts, really, rather than smoking guns.

 

[Dr.Peterson]

In my work (this has some practical application I won't get into), I ended up with a system of equations like this:

\(\displaystyle \displaystyle A_1 = x_1 + y_1 \)
\(\displaystyle \displaystyle A_2 = x_2 + y_2 \)
\(\displaystyle \displaystyle B_1 = x_1 + a_1y_1 \)
\(\displaystyle \displaystyle B_2 = x_2 + a_1y_2 \)
\(\displaystyle \displaystyle C_1 = x_1 + a_2y_1 \)
\(\displaystyle \displaystyle C_2 = x_2 + a_2y_2 \)

where \(\displaystyle \{A_1, A_2, B_1, B_2, C_1, C_2\}\) are known quantities, and \(\displaystyle \{x_1, y_1, x_2, y_2, a_1, a_2\} \) are unknown quantities that I would like to solve for. So, the system has 6 equations and 6 unknowns.
...
\(\displaystyle \displaystyle W \equiv \frac{A_2 - C_2}{C_2-B_2} \)
...
\(\displaystyle \displaystyle V \equiv \frac{A_1 - C_1}{C_1-B_1} \)
...
You can show that \(\displaystyle V = W \) and hence we end up with

\(\displaystyle \displaystyle 0\cdot a_2 = 0 \)

I guess my question is, is there some way to understand, just by looking at the original system, that it will be degenerate, or to highlight what property of the original set of equations is leading to the degeneracy?

Here are my two thoughts (without having spent much time working on the details):

1. I wouldn't be surprised if the symmetry of the equations were part of the issue. (That is, they come in pairs where subscripts 1 and 2 are interchanged.)

2. As I understand it, there is an unstated relationship among the constants, such that you can conclude that V = W, and this is also important. So the conclusion may depend on that additional knowledge (though likely that only determines whether there are no solutions or infinitely many). I wonder if there is a similar symmetry among the constants.

 

[JeffM]

In my work (this has some practical application I won't get into), I ended up with a system of equations like this:

\(\displaystyle \displaystyle A_1 = x_1 + y_1 \)
\(\displaystyle \displaystyle A_2 = x_2 + y_2 \)
\(\displaystyle \displaystyle B_1 = x_1 + a_1y_1 \)
\(\displaystyle \displaystyle B_2 = x_2 + a_1y_2 \)
\(\displaystyle \displaystyle C_1 = x_1 + a_2y_1 \)
\(\displaystyle \displaystyle C_2 = x_2 + a_2y_2 \)

where \(\displaystyle \{A_1, A_2, B_1, B_2, C_1, C_2\}\) are known quantities

I point out that this is a system of quadratic equations. I at least do not know a general method of determining whether quadratic equations are all consistent and independent. (I am not saying that there is no general method.)

Like Denis I find the notation ugly to work with so I am changing it.

\(\displaystyle a = w + y \implies w = a - y.\)

\(\displaystyle b = x + z \implies x = b - z.\)

\(\displaystyle c = w + py = a - y + py \implies y(p - 1) = c - a = g \implies ry = g.\)

\(\displaystyle d = x + pz = b - z + pz \implies z(p - 1) = d - b = h \implies rz = h.\)

\(\displaystyle e = w + qy = a - y + qy \implies y(q - 1) = e - a = j \implies sy = j.\)

\(\displaystyle f = x + qz = b - z + qz \implies z(q - 1) = f - b = k \implies sz = k. \)

So we have 4 equations with 4 unknowns, r, s, y, and z.

I shall ignore what happens if g, h, j, or k happens to be zero. If they are not zero,

\(\displaystyle \dfrac{y}{z} = \dfrac{g}{h}.\)

\(\displaystyle \dfrac{y}{z} = \dfrac{j}{k}.\)

If g/h equals j/k, the system does not have sufficient independent equations.

If g/h does not equal j/k, the system has inconsistent equations.