Question
Find the characteristic polynomial of the matrix
$$ A = \left( \begin{matrix}0&\frac{ 1 }{ 2 }&\frac{ 3 }{ 5 }&\frac{ 1 }{ 2 }\\\frac{ 1 }{ 2 }&0&\frac{ 3 }{ 5 }&\frac{ 2 }{ 5 }\\\frac{ 3 }{ 5 }&\frac{ 3 }{ 5 }&0&\frac{ 3 }{ 5 }\\\frac{ 1 }{ 2 }&\frac{ 2 }{ 5 }&\frac{ 3 }{ 5 }&0\end{matrix} \right) $$Answer
The characteristic polynomial for matrix A is:
$$ p(\lambda) = $$Explanation
The characteristic polynomial is given by
$$ p(\lambda) = det(A - \lambda I) $$In this example we have:
$$ \begin{aligned} p(\lambda) &= det(A - \lambda I) = det \left( \left[ \begin{matrix}0&\frac{ 1 }{ 2 }&\frac{ 3 }{ 5 }&\frac{ 1 }{ 2 }\\\frac{ 1 }{ 2 }&0&\frac{ 3 }{ 5 }&\frac{ 2 }{ 5 }\\\frac{ 3 }{ 5 }&\frac{ 3 }{ 5 }&0&\frac{ 3 }{ 5 }\\\frac{ 1 }{ 2 }&\frac{ 2 }{ 5 }&\frac{ 3 }{ 5 }&0\end{matrix} \right] - \left[ \begin{matrix} \lambda & 0 & 0 & 0 \\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda \end{matrix} \right] \right) = \\ &= \left| \begin{matrix}0 - \lambda &\frac{ 1 }{ 2 }&\frac{ 3 }{ 5 }&\frac{ 1 }{ 2 }\\\frac{ 1 }{ 2 }&0 - \lambda &\frac{ 3 }{ 5 }&\frac{ 2 }{ 5 }\\\frac{ 3 }{ 5 }&\frac{ 3 }{ 5 }&0 - \lambda &\frac{ 3 }{ 5 }\\\frac{ 1 }{ 2 }&\frac{ 2 }{ 5 }&\frac{ 3 }{ 5 }&0 - \lambda \end{matrix} \right| = \\ & = \end{aligned} $$This page was created using
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