I have been taking the Numerical Analysis this year. And the lecturer asked us to show the eigenvalue of following matrix satisfies certain equation and he gave pretty good hint.
$A = \left[\begin{array}{ccccc}
a & b & & & \\
b & a & b & & \\
& \ddots & \ddots & \ddots & \\
& & b & a & b\\
& & & b & a
\end{array}\right]_{N \times N} B = \left[\begin{array}{ccccc}
a & c & & & \\
b & a & c & & \\
& \ddots & \ddots & \ddots & \\
& & b & a & c\\
& & & b & a
\end{array}\right]_{N \times N}$
For $A$ I have already developed a quite good solution as follows.
First we consider the continuous eigenvalue problem.
\begin{equation}
\left\{\begin{array}{l}
u'' = \lambda u\\
u (0) = u (1) = 0
\end{array}\right. \label{eq:1}
\end{equation}
The general solution of the above problem is $u = c_1 e^{\sqrt{\lambda} x} + c_2
e^{- \sqrt{\lambda} x}$.
Substitute the solution into boundary condition, we get
\begin{equation}
\left\{\begin{array}{l}
c_1 + c_2 = 0\\
c_1 e^{\sqrt{\lambda}} + c_2 e^{- \sqrt{\lambda}} = 0
\end{array}\right. \label{eq:2}
\end{equation}
Since (2) admits a non-zero solution $u (x) \neq 0$.
\begin{equation}
\left|\begin{array}{cc}
1 & 1\\
e^{\sqrt{\lambda}} & e^{- \sqrt{\lambda}}
\end{array}\right| = e^{- \sqrt{\lambda}} – e^{\sqrt{\lambda}} = 0
\label{eq:3}
\end{equation}
From above we can know that
\begin{equation}
\lambda = – k^2 \pi^2 \text{ where } k = 1, 2, \ldots
\end{equation}
Substitute $\lambda_k$ into the equation we have $u_k = \sin (k \pi x)$.
Now for the matrix version, we can guess the eigenvector has a similar form
and verify it.
Denote $k_{th}$ eigenvector as $v_k$.
\begin{equation}
v_k = \left[\begin{array}{c}
\sin (k \pi x_1)\\
\ldots .\\
\sin (k \pi x_n)
\end{array}\right] = \left[\begin{array}{c}
\sin (k \pi h)\\
\ldots\\
\sin (Nk \pi h)
\end{array}\right] = \left[\begin{array}{c}
\sin (\theta_k)\\
\ldots\\
\sin (N \theta_k)
\end{array}\right]
\end{equation}
Now plug our $v_k$ back to the matrix.
First consider $Av_k = \lambda_k v_k$.
\begin{equation}
\left\{\begin{array}{l}
a \sin (\theta_k) + b \sin (2 \theta_k) = \lambda_k \sin (\theta_k)\\
b \sin ((j – 1) \theta_k) + a \sin (j \theta_k) + b \sin ((j + 1)
\theta_k) = \lambda_k \sin (j \theta_k)\\
b \sin ((N – 1) \theta_k) + a \sin (N \theta_k) = \lambda_k \sin (N
\theta_k)
\end{array}\right. \label{eq:6}
\end{equation}
Above can be simplified into one equation as $\sin (0) = \sin (k \pi) =
0$.
\begin{equation}
b \sin ((j – 1) \theta_k) + a \sin (j \theta_k) + b \sin ((j + 1) \theta_k)
= \lambda_k \sin (j \theta_k) \quad j = 1 \ldots N \label{eq:7}
\end{equation}
By sum-to-product identity of trignometric functions. We have
\begin{eqnarray*}
2 b \sin (j \theta_k) \cos (\theta_k) + a \sin (j \theta_k) & = & \lambda_k
\sin (j \theta_k)\\
2 b \sin (j \theta_k) \cos \left( \frac{k \pi}{N + 1} \right) + a \sin (j
\theta_k) & = & \lambda_k \sin (j \theta_k)\\
\lambda_k & = & a + 2 b \cos \left( \frac{k \pi}{N + 1} \right) \quad k = 1
\ldots N
\end{eqnarray*}
We have solved $\lambda_k$ for $A$.
Here is my problem. Will the same procedure work on $B$? I've tried a bit.
but it always got stuck at the trignometric part.
Best Answer
By a standard procedure well described here : https://en.wikipedia.org/wiki/Tridiagonal_matrix#Similarity_to_symmetric_tridiagonal_matrix with an explicit solution you will find here : https://math.stackexchange.com/q/1148874