Question

Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

4. \(\left[ {\begin{array}{*{20}{c}}5&-3\\-4&3\end{array}} \right]\)

Short answer

Characteristic polynomial:\({\lambda ^2} - 8\lambda + 3\).

Eigenvalues: \(\lambda = 4 + \sqrt {13} \) and \(\lambda = 4 - \sqrt {13} \).

Step-by-step solution

Find the characteristic polynomial

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

It is given that \(A = \left[ {\begin{array}{*{20}{c}}5& - 3\\ - 4&3\end{array}} \right]\) and \(I = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\) is identity matrix. Find the matrix\(\left( {A - \lambda I} \right)\) as shown below:

\[\begin{array}A - \lambda I = \left[ {\begin{array}{*{20}{c}}5& - 3\\ - 4&3\end{array}} \right] - \lambda \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{5 - \lambda }&{ - 3}\\{ - 4}&{3 - \lambda }\end{array}} \right]\;\end{array}\]

Now, calculate the determinant of the matrix\(\left( {A - \lambda I} \right)\)as shown below:

\[\begin{array}det\left( {A - \lambda I} \right) = det\left[ {\begin{array}{*{20}{c}}{5 - \lambda }&{ - 3}\\{ - 4}&{3 - \lambda }\end{array}} \right]\\ = \left( {5 - \lambda } \right)\left( {3 - {\rm{\lambda }}} \right) - 12\\ = {\lambda ^2} - 8\lambda + 15 - 12\\ = {\lambda ^2} - 8\lambda + 3\end{array}\]

So, the characteristic polynomial of is \({\lambda ^2} - 8\lambda + 3\).

Describe the characteristic equation

To find the eigenvalues of the matrix, we must calculate all the scalarssuch that\(\left( {A - \lambda I} \right)x = 0\) has a non-trivial solution which is equivalent to finding allsuch that the matrix\(\left( {A - \lambda I} \right)\)is not invertible, that is, when determinant of\(\left( {A - \lambda I} \right)\)is zero.

Thus, the eigenvalues of \(A\) are the solutions of the characteristic equation\(\det \left( {A - \lambda I} \right) = 0\). So, find the characteristic equation \(\det \left( {A - \lambda I} \right) = 0\).

\[\begin{array}det\left[ {\begin{array}{*{20}{c}}{5 - \lambda }&{ - 3}\\{ - 4}&{3 - \lambda }\end{array}} \right] = 0\\\left( {5 - \lambda } \right)\left( {3 - {\rm{\lambda }}} \right) - 12 = 0\\{\lambda ^2} - 8\lambda + 15 - 12 = 0\\{\lambda ^2} - 8\lambda + 3 = 0\end{array}\]

Find roots of characteristic equation

For the quadratic equation,\(a{x^2} + bx + c = 0\), the general solution is given as\(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} \;\;}}{{2a}}\).

Thus, the solution of the characteristic equation \[{\lambda ^2} - 8\lambda + 3 = 0\] is obtained as follows:

\[\begin{array}{c}{\lambda ^2} - 8\lambda + 3 = 0\\\lambda = \frac{{ - \left( { - 8} \right) \pm \sqrt {{{\left( { - 8} \right)}^2} - 4\left( 3 \right)} }}{2}\\ = \frac{{8 \pm 2\sqrt {13} }}{2}\\ = 4 \pm \sqrt {13} \end{array}\]

The eigenvalues of \(A\) are \(\lambda = 4 + \sqrt {13} \) and \(\lambda = 4 - \sqrt {13} \).