Question

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

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

Short answer

Characteristic polynomial:\({\lambda ^2} - 10\lambda + 16\)

Eigenvalues: \(\lambda = 8\) and \(\lambda = 2\).

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

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

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

So, the characteristic polynomial of is \({\left( {5 - \lambda } \right)^2} - 9\).

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\(\det \left( {A - \lambda I} \right) = 0\).

\[\begin{array}det\left[ {\begin{array}{*{20}{c}}{5 - \lambda }&3\\3&{5 - \lambda }\end{array}} \right] = 0\\{\left( {5 - \lambda } \right)^2} - 9 = 0\\{\lambda ^2} - 10\lambda + 25 - 9 = 0\\{\lambda ^2} - 10\lambda + 16 = 0\end{array}\]

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\({\lambda ^2} - 10\lambda + 16 = 0\)is obtained as follows:

\[\begin{array}{\lambda ^2} - 10\lambda + 16 = 0 = 0\\\lambda = \frac{{ - \left( { - 10} \right) \pm \sqrt {{{10}^2} - 4\left( { - 16} \right)} }}{2}\\ = \frac{{10 \pm 6}}{2}\\ = 8,\,2\end{array}\]

The eigenvalues of \(A\) are \(\lambda = 8\) and \(\lambda = 2\).