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.

Hints:Note that the sum of each column(row) is equal. So you can do the summation first and disjunct common factor,which is $(n-1)+e^Y-\lambda$ then you'll get a column(row) of $1$. 
And things become easily.
