Consider a matrix $M \in \mathbb{R}^{n \times n}$ in the form:
$$ M = \begin{bmatrix}
\alpha & \beta & 0 & \cdots & 0 \\
\gamma & \alpha & \beta & \cdots & 0 \\
0 & \gamma & \alpha & \cdots & \vdots \\
\vdots & \vdots & \ddots & \ddots & \beta \\
0 & 0 & \cdots & \gamma & \operatorname{\alpha}
\end{bmatrix} $$
Then a closed form expression for the eigenvalues is
$$\mu_k = \alpha + 2 \beta \sqrt{\frac{\gamma}{\beta}} \cos \left( \frac{k \pi}{n+1} \right)$$
However, this also requires that $\beta \neq 0$, since the corresponding eigenvectors contain a division by $\beta$. Therefore, I am wondering if there exists a separate general formula for the case where $\beta = 0$.
This simplifies the above matrix $M$ to:
$$
\begin{bmatrix}
\alpha & 0 & 0 & \cdots & 0 \\
\gamma & \alpha & 0 & \cdots & 0 \\
0 & \gamma & \alpha & \cdots & \vdots \\
\vdots & \vdots & \ddots & \ddots & 0 \\
0 & 0 & \cdots & \gamma & \operatorname{\alpha}
\end{bmatrix}$$
Of course, it is simple enough to compute this manually for smaller matrices, but I would be interested in knowing what the general solution is in this case.
Note: the corresponding eigenvectors in the case where $\beta \neq 0$ are: $$v_k = \Big{(} \sqrt{ \frac{\gamma}{\beta}} \sin \big{(} \frac{ k \pi}{n+1} \big{)}, \big{(} \sqrt{ \frac{\gamma}{\beta}} \space \big{)}^2 \sin \big{(} \frac{ 2k \pi}{n+1} \big{)}, \cdots, \big{(} \sqrt{\frac{\gamma}{\beta}} \space \big{)}^n \sin \big{(} \frac{ nk \pi}{n+1} \big{)} \Big{)}^T$$
Best Answer
The matrix you've written is lower triangular, and the eigenvalues of triangular matrices are just their diagonal entries. In particular, your matrix has only one eigenvalue, $\alpha$, and it occurs with algebraic multiplicity $n$.