Following the hint, we know that det ( P − λ I ) = | 1 / 2 − λ 0 1 / 2 0 1 / 2 − λ 1 / 2 1 / 2 1 / 2 − λ | = ( 1 / 2 − λ ) | 1 / 2 − λ 1 / 2 1 / 2 − λ | + 1 / 2 | 0 1 / 2 − λ 1 / 2 1 / 2 | = ( 1 / 2 − λ ) ( − λ ( 1 / 2 − λ ) − 1 / 4 ) + 1 / 2 ( − 1 / 2 ( 1 / 2 − λ ) ) = ( 1 / 2 − λ ) ( λ + 1 / 2 ) ( λ − 1 ) {\displaystyle {\begin{aligned}\det(P-\lambda I)&={\begin{vmatrix}1/2-\lambda &0&1/2\\0&1/2-\lambda &1/2\\1/2&1/2&-\lambda \\\end{vmatrix}}\\&=(1/2-\lambda ){\begin{vmatrix}1/2-\lambda &1/2\\1/2&-\lambda \end{vmatrix}}+1/2{\begin{vmatrix}0&1/2-\lambda \\1/2&1/2\\\end{vmatrix}}\\&=(1/2-\lambda )(-\lambda (1/2-\lambda )-1/4)+1/2(-1/2(1/2-\lambda ))\\&=(1/2-\lambda )(\lambda +1/2)(\lambda -1)\end{aligned}}} .
Thus, the eigenvalues are λ = 1 / 2 , − 1 / 2 , 1 . {\displaystyle \color {blue}{\lambda =1/2,-1/2,1}.}
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