How do you solve the eigenvalue equation f (x) y (x) = a y (x) where y (x) is the unknown function?

It was asked on quora how to answer the above. I need more space to answer this so I am doing it here and linking the two…

Their link is: https://www.quora.com/How-do-you-solve-the-eigenvalue-equation-f-x-y-x-a-y-x-where-y-x-is-the-unknown-function

First off, the question is somewhat vague so I am going to answer it based on what’s can gather it is asking. At first glance you might be tempted to divide both sides by y(x). You definitely do not want to do this. This is the issue with matrix math. The eigenvalue function is not a value but a matrix of answers. This is common in such problems. You need to multiply both sides by the inverse equation. f*y = a*y The following identities are important to note:

transposed(inv(Matrix))=inv(transposed(Matrix) and:

For matrices A, B and scalar c we have the following properties of transpose:

  1. {\displaystyle \left(\mathbf {A} ^{\operatorname {T} }\right)^{\operatorname {T} }=\mathbf {A} .}The operation of taking the transpose is an involution (self-inverse).
  2. {\displaystyle \left(\mathbf {A} +\mathbf {B} \right)^{\operatorname {T} }=\mathbf {A} ^{\operatorname {T} }+\mathbf {B} ^{\operatorname {T} }.}The transpose respects addition.
  3. {\displaystyle \left(\mathbf {AB} \right)^{\operatorname {T} }=\mathbf {B} ^{\operatorname {T} }\mathbf {A} ^{\operatorname {T} }.}Note that the order of the factors reverses. From this one can deduce that a square matrix A is invertible if and only if AT is invertible, and in this case we have (A−1)T = (AT)−1. By induction this result extends to the general case of multiple matrices, where we find that (A1A2…Ak−1Ak)T = AkTAk−1T…A2TA1T.
  4. {\displaystyle \left(c\mathbf {A} \right)^{\operatorname {T} }=c\mathbf {A} ^{\operatorname {T} }.}The transpose of a scalar is the same scalar. Together with (2), this states that the transpose is a linear map from the space of m × n matrices to the space of all n × m matrices.
  5. {\displaystyle \det \left(\mathbf {A} ^{\operatorname {T} }\right)=\det(\mathbf {A} ).}The determinant of a square matrix is the same as the determinant of its transpose.
  6. The dot product of two column vectors a and b can be computed as the single entry of the matrix product:{\displaystyle \left[\mathbf {a} \cdot \mathbf {b} \right]=\mathbf {a} ^{\operatorname {T} }\mathbf {b} ,}which is written as ai bi in Einstein summation convention.
  7. If A has only real entries, then ATA is a positive-semidefinite matrix.
  8. {\displaystyle \left(\mathbf {A} ^{\operatorname {T} }\right)^{-1}=\left(\mathbf {A} ^{-1}\right)^{\operatorname {T} }.}The transpose of an invertible matrix is also invertible, and its inverse is the transpose of the inverse of the original matrix. The notation A−T is sometimes used to represent either of these equivalent expressions.
  9. If A is a square matrix, then its eigenvalues are equal to the eigenvalues of its transpose, since they share the same characteristic polynomial.

Answer

The way the question was posed will work in MatLab because of how MathWorks designed the matrix math. However, it is more correct to say that v(A-lambda*I)=0 which is Av-v*lambda=0 or A*v = v*lambda which is exactly what is being asked… The way to solve it is to assume that the eigenvectors are not zero which would be uninteresting…

Thus, (A-lambda*I) = 0 and you can solve for it that way quite easily which will be held for another day. So Av=fy=v*lamda=ay.