Question

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

15. \(\left[ {\begin{array}{*{20}{c}}4&- 7&0&2\\0&3&- 4&6\\0&0&3&{ - 8}\\0&0&0&1\end{array}} \right]\)

Short answer

The eigen values of the matrix are 4, 3, 3, and 1.

Step-by-step solution

Definition of the characteristic polynomial

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

When\(A\)is an\(n \times n\)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.

\[\begin{array}{c}\det \left( {A - \lambda I} \right) = \det \left[ {\begin{array}{*{20}{c}}{4 - \lambda }&{ - 7}&0&2\\0&{3 - \lambda }&{ - 4}&6\\0&0&{3 - \lambda }&{ - 8}\\0&0&0&{1 - \lambda }\end{array}} \right]\\ = \left( {4 - \lambda } \right){\left( {3 - \lambda } \right)^2}\left( {1 - \lambda } \right)\end{array}\]

Thus, the eigenvalues of the matrix are \(4\left( {multiplicity\,\,1} \right),\) \(3\left( {multiplicity\,\,2} \right),\) and \(1\left( {multiplicity\,\,1} \right)\).

Thus, the eigenvalues of the matrix are 4, 3, 3, and 1.