91Ó°ÊÓ

The eigenvalue of a \(n \times n\) matrix \(A\) is a scalar \(\lambda \) such that \(\lambda \) satisfies the characteristic equation \(\det \left( {A - \lambda I} \right) = 0\).

When \(A\) is a \(n \times n\) matrix, \(\det \left( {A - \lambda I} \right)\) is the characteristic polynomial of \(A\), which is the polynomial of degree \(n\).

In particular, the multiplicity of an eigenvalue \(\lambda \) represents its multiplication as a root of the characteristic equation.

The product of the diagonal entries of \(A\) becomes the determinant of a triangular matrix \(A\).

Use the above fact to obtain the eigenvalue of the matrix, as shown below.

\[\begin{array}\det \left( {A - \lambda I} \right) = \det \left[ {\begin{array}{*{20}{c}}{5 - \lambda }&0&0&0\\8&{ - 4 - \lambda }&0&0\\0&7&{1 - \lambda }&0\\1&{ - 5}&2&{1 - \lambda }\end{array}} \right]\\ = \left( {5 - \lambda } \right)\left( { - 4 - \lambda } \right){\left( {1 - \lambda } \right)^2}\end{array}\]

Therefore, the eigenvalues of the matrix are \(5\left( {multiplicity\,\,1} \right),\) \( - 4\left( {multiplicity\,\,1} \right),\) and \(1\left( {multiplicity\,\,2} \right)\).

Thus, the eigenvalues of the matrix are \(5,1,1,\) and \( - 4\).