eigenvalue problem for $n$ dimensional matrix Lets say I have $n$-dimensional matrix
$$
\hat T = 
\begin{pmatrix}
e^{Y} & 1 & \cdots & 1 \\
1 & e^{Y} & \cdots & 1 \\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 & \cdots & e^{Y} \\
\end{pmatrix}
$$
and I need to solve eigenvalue problem
$$\hat T |x\rangle = \lambda |x\rangle$$
For precice finite dimensional matrix  I can solve this
by finding determinant 
$$
\det
\begin{pmatrix}
e^{Y}-\lambda & 1 & \cdots & 1 \\
1 & e^{Y}-\lambda & \cdots & 1 \\
\vdots & \vdots & \ddots & \vdots\\
1 & 1 & \cdots & e^{Y}-\lambda \\
\end{pmatrix} = 0.
$$
However I am stuck on this $n$-dimensional case.
I have tried to solve this problem in index notation:
$$
e^Yx_i + \sum^n_{j\neq i}x_j = \lambda x_i
$$
After rearranging:
$$
\lambda = e^Y - 1 + \frac{1}{x_i}\sum^n_{j\neq i}x_j
$$
It seams that term $\frac{1}{x_i}\sum^n_{j\neq i}x_j$ is same for any $i$. Maybe 
someone can advice me, how to solve this and similar problems.

The characteristic polynomial of the $n\times n$ matrix
$$
U_n=\begin{bmatrix}
1 & 1 & \dots & 1 \\
1 & 1 & \dots & 1 \\
\vdots & \vdots & \ddots & \vdots \\
1 & 1 & \dots & 1
\end{bmatrix}
$$
is $p(\lambda)=\det(U_n-\lambda I_n)=(0-\lambda)^{n-1}(n-\lambda)$, because the rank is $1$ and there's a nonzero eigenvalue $n$.
Since $T=U_n-(1-e^Y)I_n$, you have
\begin{align}
\det(T-\lambda I_n)&=\det(U_n-(1-e^Y)I_n-\lambda I_n)\\
&=\det(U_n-(\lambda+1-e^Y)I_n)\\
&=p(\lambda+1-e^Y)\\
&=(0-(\lambda+1-e^Y))^{n-1}(n-(\lambda+1-e^Y))\\
&=(e^Y-1-\lambda)^{n-1}(n+e^Y-1-\lambda)
\end{align}
You can use a similar idea to find the eigenvectors.
