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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.

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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.

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  • $\begingroup$ Thanks, that was nice and easy :). $\endgroup$ – Baranas Aug 5 '15 at 17:08
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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.

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