Question
Find the characteristic polynomial of the matrix
$$ A = \left( \begin{matrix}\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 }\\\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 }\\\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 }\end{matrix} \right) $$Answer
The characteristic polynomial for matrix A is:
$$ p(\lambda) = -\lambda^3+\lambda^2 $$Explanation
The characteristic polynomial is given by
$$ p(\lambda) = det(A - \lambda I) $$In this example we have:
$$ \begin{aligned} p(x) &= det(A - x I) = det \left( \left[ \begin{matrix}\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 }\\\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 }\\\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 }\end{matrix} \right] - \left[ \begin{matrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{matrix} \right] \right) = \\ &= \left| \begin{matrix}\frac{ 1 }{ 3 } - \lambda &\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 }\\\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 } - \lambda &\frac{ 1 }{ 3 }\\\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 }&\frac{ 1 }{ 3 } - \lambda \end{matrix} \right| = \\ &= (-\lambda+\frac{ 1 }{ 3 }) (-\lambda+\frac{ 1 }{ 3 }) (-\lambda+\frac{ 1 }{ 3 }) + \frac{ 1 }{ 3 } \cdot \frac{ 1 }{ 3 } \cdot \frac{ 1 }{ 3 } + \frac{ 1 }{ 3 } \cdot \frac{ 1 }{ 3 } \cdot \frac{ 1 }{ 3 } - (-\lambda+\frac{ 1 }{ 3 }) \cdot \frac{ 1 }{ 3 } \cdot \frac{ 1 }{ 3 } - \frac{ 1 }{ 3 } \cdot \frac{ 1 }{ 3 } \cdot (-\lambda+\frac{ 1 }{ 3 }) - \frac{ 1 }{ 3 } \cdot (-\lambda+\frac{ 1 }{ 3 }) \cdot \frac{ 1 }{ 3 } = \\ & = -\lambda^3+\lambda^2-\frac{ 1 }{ 3 }\lambda+\frac{ 1 }{ 27 } + \frac{ 1 }{ 27 } + \frac{ 1 }{ 27 } - (-\frac{ 1 }{ 9 }\lambda+\frac{ 1 }{ 27 }) - (-\frac{ 1 }{ 9 }\lambda+\frac{ 1 }{ 27 }) - (-\frac{ 1 }{ 9 }\lambda+\frac{ 1 }{ 27 }) = \\ & = -\lambda^3+\lambda^2 \end{aligned} $$This page was created using
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