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If is an \(n \times n\)matrix, then \(det\left( {A - \lambda I} \right)\), iscalled the characteristic polynomial of matrix \(A\).

Let\(A = \left( {\begin{aligned}{{}}{\,3}&4\\4&9\end{aligned}} \right)\)and\(I = \left( {\begin{aligned}{{}}1&0\\0&1\end{aligned}} \right)\)is identity matrix. Find the matrix\(\left( {A - \lambda I} \right)\)as shown below:

\(\begin{aligned}{}A - \lambda I &= \left( {\begin{aligned}{{}}{\,3}&4\\4&9\end{aligned}} \right) - \lambda \left( {\begin{aligned}{{}}1&0\\0&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{3 - \lambda }&4\\4&{9 - \lambda }\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{}det\left( {A - \lambda I} \right) &= det\left( {\begin{aligned}{{}}{3 - \lambda }&4\\4&{9 - \lambda }\end{aligned}} \right)\\ &= \left( {3 - \lambda } \right)\left( {9 - \lambda } \right) + 11\\ &= {\lambda ^2} - 12\lambda + 11\end{aligned}\)

So, the characteristic polynomial of matrix \(A\) is \({\lambda ^2} - 12\lambda + 11\).

Thus, the eigenvalues of\(A\)are the solutions of the characteristic equation\(\det \left( {A - \lambda I} \right) = 0\). So, solve the characteristic equation\({\lambda ^2} - 12\lambda + 11 = 0\), as follows:

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\({\lambda ^2} - 12\lambda + 11 = 0\)is obtained as follows:

\(\begin{aligned}{}{\lambda ^2} - 12\lambda + 11 &= 0\\\lambda &= \frac{{ - \left( { - 12} \right) \pm \sqrt {{{\left( { - 12} \right)}^2} - 4\left( {11} \right)} }}{2}\\ &= \frac{{12 \pm \sqrt {100} }}{2}\\ &= 11,\,\,1\end{aligned}\)

The eigenvalues of \(A\)are \(\lambda = 11\) and \(\lambda = 1\) .

A matrix \(A\) is diagonalized as \(A = PD{P^{ - 1}}\), where \(P\) is orthogonal matrix of normalized Eigen vectors of matrix \(A\)and \(D\) is a diagonal matrix having Eigen values of matrix \(A\) on its principle diagonal.

For the Eigen value\(\lambda = 11\), the basis for Eigenspace is obtained by solving the system of equations\(\left( {A - 11I} \right){\bf{x}} = 0\). On solving it is obtained as\(\left( \begin{aligned}{}1\\2\end{aligned} \right)\).

Similarly, for the Eigen value\(\lambda = 1\), the basis for Eigenspace is obtained by solving the system of equations\(\left( {A - I} \right){\bf{x}} = 0\). On solving it is obtained as\(\left( \begin{aligned}{} - 2\\\,\,\,1\end{aligned} \right)\).

The normalized vectors are\({{\bf{u}}_1} = \left( \begin{aligned}{}1/\sqrt 5 \\\,2/\sqrt 5 \end{aligned} \right)\)and\({{\bf{u}}_2} = \left( \begin{aligned}{} - 2/\sqrt 5 \\\,\,\,1/\sqrt 5 \end{aligned} \right)\). So, the normalized matrix\(P\)is given as;

\(\begin{aligned}{}P &= \left( {{{\bf{u}}_1}\,\,{{\bf{u}}_2}} \right)\\ &= \left( {\begin{aligned}{{}}{1/\sqrt 5 }&{ - 2/\sqrt 5 }\\{2/\sqrt 5 }&{\,\,\,\,1/\sqrt 5 }\end{aligned}} \right)\end{aligned}\)

whereas the matrix \(D\) is obtained as \(D = \left( {\begin{aligned}{{}}{11}&0\\0&1\end{aligned}} \right)\).

The matrix\(A\)is diagonalized as\(A = PD{P^{ - 1}}\). Thus, the orthogonal diagonalization is as follows:

\(\begin{aligned}{}A &= PD{P^{ - 1}}\\ &= \left( {\begin{aligned}{{}}{1/\sqrt 5 }&{ - 2/\sqrt 5 }\\{2/\sqrt 5 }&{\,\,\,\,1/\sqrt 5 }\end{aligned}} \right)\left( {\begin{aligned}{{}}{11}&{\,\,0}\\0&1\end{aligned}} \right){\left( {\begin{aligned}{{}}{1/\sqrt 5 }&{ - 2/\sqrt 5 }\\{2/\sqrt 5 }&{\,\,\,\,1/\sqrt 5 }\end{aligned}} \right)^{ - 1}}\end{aligned}\)

The orthogonal diagonalization is \(A = \left( {\begin{aligned}{{}}{1/\sqrt 5 }&{ - 2/\sqrt 5 }\\{2/\sqrt 5 }&{\,\,\,\,1/\sqrt 5 }\end{aligned}} \right)\left( {\begin{aligned}{{}}{11}&{\,\,0}\\0&1\end{aligned}} \right)\left( {\begin{aligned}{{}}{1/\sqrt 5 }&{ - 2/\sqrt 5 }\\{2/\sqrt 5 }&{\,\,\,\,1/\sqrt 5 }\end{aligned}} \right)\).