24 1. Random Walk and Discrete Heat Equation
This nice form leads us to try to find eigenvalues and eigenfunctions
of Q, i.e., to find λ, φ such that
(1.16) Qφ(x) = λ φ(x),
with φ 0 on ∂A.

The “algorithmic” way to find the eigenvalues and eigenvectors for
a matrix Q is first to find the eigenvalues as the roots of the characteristic
polynomial and then to find the corresponding eigenvector for each eigenvalue.
Sometimes we can avoid this if we can make good guesses for the eigenvectors.
This is what we will do here.
The sum rule for sine,
sin((x ± 1)θ) = sin(xθ) cos(θ) ± cos(xθ) sin(θ),
tells us that
Q{sin(θx)} = λθ {sin(θx)}, λθ = cos θ,
where {sin(θx)} denotes the vector whose component associated to
x A is sin(θx). If we choose θj = πj/N, then φj(x) = sin(πjx/N),
which satisfies the boundary condition φj(0) = φj(N) = 0. Since
these are eigenvectors with different eigenvalues for a symmetric ma-
trix Q, we know that they are orthogonal, and hence linearly inde-
pendent. Hence every function f on A can be written in a unique
way as
(1.17) f(x) =
N−1
j=1
cj sin
πjx
N
.
This sum in terms of trigonometric functions is called a finite Fourier
series. The solution to the heat equation with initial condition f is
pn(y) =
N−1
j=1
cj cos

N
n
φj(y).
Orthogonality of eigenvectors tells us that
N−1
x=1
sin
πjx
N
sin
πkx
N
= 0 if j = k.
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