The Toeplitz matrix is
\begin{align*} T_r(a,b,c)= \begin{bmatrix} b & c \\ a & b & c\\ &a&b&c\\ &&\ddots&\ddots&\ddots\\ &&&\ddots&\ddots&\ddots\\ &&&&\ddots&\ddots&\ddots\\ &&&&&\ddots&\ddots&c\\ 0&&&&&&a&b \end{bmatrix} \end{align*}
I found that the eigenvalue $\lambda_j$ of this matrix = $b+2c\sqrt{\frac{a}{c}}\cos(\frac{j\pi}{N+1}), \quad j=1,\dots,N$.
Thus, I'll only check that the eigenvectors $\vec{v_j}$ according to the eigenvalues $\lambda_j$ be $(v_1,v_2,\dots,v_N)^{T}$, where $v_k=2(\sqrt{\frac{a}{c}})^k\sin(\frac{kj\pi}{N+1}), \quad k=1,\dots,N$.
Since I knew the eigenvalue, I tried $T_r(a,b,c)\vec{v_j}=\lambda_j\vec{v_j}$. At first row, we could get
\begin{align*}
bv_1+cv_2&=\lambda_jv_1\\
&=bv_1+2c\sqrt{\frac{a}{c}}\cos(\frac{j\pi}{N+1})v_1
\end{align*}
This implies that $v_2=2\sqrt{\frac{a}{c}}\cos(\frac{j\pi}{N+1})v_1$.
How to get explicit the eigenvector according to the eigenvalue although we don't know $v_1$??
Any help is appreciated...
Thank you!