Question

Question: For the matrices in Exercises 15-17, list the eigenvalues, repeated according to their multiplicities.

17. \(\left[ {\begin{array}{*{20}{c}}3&0&0&0&0\\- 5&1&0&0&0\\3&8&0&0&0\\0&- 7&2&1&0\\- 4&1&9&- 2&3\end{array}} \right]\)

Short answer

The eigenvalues of the matrix are \(3\), \(3\), \(1\), \(1\), and \(0\).

Step-by-step solution

Definition of the characteristic polynomial

The eigenvalue of an \(n \times n\) matrix \(A\) is a scalar \(\lambda \) such that if \(\lambda \) 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 thecharacteristic 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.

Determine the eigenvalues repeated according to their multiplicities

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.

\[\begin{array}\det \left( {A - \lambda I} \right) = \det \left[ {\begin{array}{*{20}{c}}{3 - \lambda }&0&0&0&0\\{ - 5}&{1 - \lambda }&0&0&0\\3&8&{0 - \lambda }&0&0\\0&{ - 7}&2&{1 - \lambda }&0\\{ - 4}&1&9&{ - 2}&{3 - \lambda }\end{array}} \right]\\ = \left( {3 - \lambda } \right)\left( {1 - \lambda } \right)\left( {0 - \lambda } \right)\left( {1 - \lambda } \right)\left( {3 - \lambda } \right)\\ = {\left( {3 - \lambda } \right)^2}{\left( {1 - \lambda } \right)^2}\left( { - \lambda } \right)\end{array}\]

Therefore, the eigenvalues of the matrix are \(3\left( {multiplicity\,\,2} \right),\) \(1\left( {multiplicity\,\,2} \right),\) and \(0\left( {multiplicity\,\,1} \right)\).

Thus, the eigenvalues of the matrix are \(3\), \(3\), \(1\), \(1\), and \(0\).