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I'm trying to find the eigenvalues of a matrix $$A=\begin{bmatrix}2/3 & -1/4 & -1/4 \\ -1/4 & 2/3 & -1/4 \\ -1/4 & -1/4 & 2/3\end{bmatrix}$$

The eigenvalues of this matrix, are the roots $\lambda$ of the equation $det(A-\lambda I)=0$. Expanding this determinant with Sarrus's Rule gives a polynomial of a third degree, the solutions can apparently be estimated by iterative methods. Before I start exploring those avenues, however, I'd like to know if there is a more practical method to compute the eigenvalues of this matrix.

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It's equal to $- \frac{1}{4}J + \frac{11}{12} I$, where $I$ is the identity matrix and $J$ is the matrix of all $1$s. Note that $J$ has two eigenvalues: $0$ with multiplicity $2$ (since it has a rank of $1$) and $3$ with multiplicity $1$, with eigenvector $(1, 1, 1)$.

So, the eigenvalues of $- \frac{1}{4}J$ are $-3/4, 0, 0$, and hence $- \frac{1}{4}J + \frac{11}{12} I$ has eigenvalues $1/6, 11/12, 11/12$.

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. This is elegant; +1. Could you respond with a comment explaining why here the sums of the eigenvalues are the eigenvalues of the sums? $\endgroup$ – J. W. Tanner Mar 8 '20 at 14:43
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    $\begingroup$ Given the matrix $J$ admits a basis of three linearly independent eigenvectors $v_1, v_2, v_3$, corresponding respectively to eigenvalues $3,0,0$, we can observe that$$\left(-\frac{1}{4}J+\frac{11}{12}I\right)v_1=-\frac{1}{4}Jv_1+\frac{11}{12}Iv_1=-\frac{3}{4}v_1+\frac{11}{12}v_1=\frac{1}{6}v_1,$$making $v_1$ an eigenvector corresponding to eigenvalue $\frac{1}{6}$. We can do the same thing with the other eigenvectors:$$\left(-\frac{1}{4}J+\frac{11}{12}I\right)v_2=-\frac{1}{4}Jv_2+\frac{11}{12}Iv_2=0v_2+\frac{11}{12}v_2=\frac{11}{12}v_2,$$and similarly for $v_3$. $\endgroup$ – user757704 Mar 9 '20 at 8:39
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Check for the rational roots of the characteristic polynomial.

The eigenvalues are $$\frac {11}{12}, \frac {11}{12}, \frac {1}{6} $$

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