## On Eigen(patch(ix))values - (RWLA,MCT,GT,AM Part VIII)

So in the continuation of this series, I have been thinking long and hard about the curious property of the existence of eigen(patch(ix))values that I have talked about in a previous post.  I began to question whether such eigen(patch(ix))values are limited to a finite set (much as in finite matrices) or whether there was some other fundamental insight, like, if 1 is an eigen(patch(ix))value, then all elements of $\mathbb{R}$ are too (or all of $\mathbb{R}$ minus a finite set).  In my latest attempt to understand this, the question comes down to, using the "star" operator, whether

has discrete values of $\lambda$ or, "what values can lambda take for the equation to be true," in direct analogy with eigenvalues when we're dealing with discrete matrices.  I am not using yet "integral transform notation" because this development seemed more intuitive to me, and thus I'm also limiting the treatment to "surfaces" that are smooth and defined on $[0,1] \times [0,1]$, like I first thought of them. Thus, the above equation translates to:

and, if we recall our construction of the patch (or patchix if we relax the assumption that integrating with respect to x is 1) $p(x,y) = f_1(x) g_1(y) + f_2(x) g_2(y)$:

where $B_1, B_2$ are constants.  It is very tempting to divide $\lambda$ as

must hold provided $\lambda \neq 0$.  So we have excluded an eigen(patch(ix))value right from the start, which is interesting.

We can systematically write the derivatives of $a(x)$, as we're going to need them if we follow the algorithm I delineated in one of my previous posts (NB: we assume a finite number of derivatives or periodic ones, or infinite derivatives such that the subsequent sums we'll write are convergent):

provided, as before, $\lambda \neq 0$.  We want to calculate the constants $B_1, B_2$, to see if they are restricted in some way by a formula, and we do this by integrating by parts as we did in a previous post to obtain the cool "pasquali series." Thus, we have that if $B_1 = \int_0^1 a(1-y) g_1(y) dy$, the tabular method gives:

and so,

if we remember the alternating sign of the multiplications, and we are allowed some leeway in notation.  Ultimately, this last bit means: $\sum_{i=0}^\infty a^i(0) G_1^{i+1}(1) - \sum_{i=0}^\infty a^i(1) G_1^{i+1}(0)$.

Since we have already explicitly written the derivatives of $a(x)$, the $a^i(0), a^i(1)$ derivatives can be written as $\frac{B_1}{\lambda} f_1^i(0) + \frac{B_2}{\lambda} f_2^i(0)$ and $\frac{B_1}{\lambda} f_1^i(1) + \frac{B_2}{\lambda} f_2^i(1)$ respectively.

We have then:

Since we aim to solve for $B_1$, multiplying by $\lambda$ makes things easier, and also we must rearrange all elements with $B_1$ in them, so we get:

Subtracting both sides the common term and factoring the constant we endeavor to solve for, we get:

or

A similar argument for $B_2$ suggests

where the new constants introduced emphasizes the expectation that the sums converge.  Plugging in the one into the other we get:

and now we seem to have additional restrictions on lambda: $\lambda \neq F$ and $\lambda \neq C$.  Furthermore, the constant $B_1$ drops out of the equation, suggesting these constants can be anything we can imagine (all of $\mathbb{R}$ without restriction), but then we have the constraint:

which is extraordinarily similar to its analogue in finite matrix or linear algebra contexts.  Expanding suggests:

which we can solve by the quadratic equation of course, as:

So not only is $\lambda$ not equal to a few values, it is incredibly restricted to two of them.

So here's a sort of conjecture, and a plan for the proof.  The allowable values of $\lambda$ is equal to the number of x terms $a(x)$ (or $p(x,y)$) carries.  We have already shown the base case, we need only show the induction step, that it works for $k$ and $k+1$ terms.