Question

Orthogonally diagonalize the matrices in Exercises 13–22, giving an orthogonal matrix\(P\)and a diagonal matrix\(D\). To save you time, the eigenvalues in Exercises 17–22 are: (17)\( - {\bf{4}}\), 4, 7; (18)\( - {\bf{3}}\),\( - {\bf{6}}\), 9; (19)\( - {\bf{2}}\), 7; (20)\( - {\bf{3}}\), 15; (21) 1, 5, 9; (22) 3, 5.

13. \(\left( {\begin{aligned}{{}}3&1\\1&{\,\,3}\end{aligned}} \right)\)

Short answer

The orthogonal diagonalization is \(A = \left( {\begin{aligned}{{}}{1/\sqrt 2 }&{ - 1/\sqrt 2 }\\{1/\sqrt 2 }&{\,\,\,\,1/\sqrt 2 }\end{aligned}} \right)\left( {\begin{aligned}{{}}2&0\\0&4\end{aligned}} \right){\left( {\begin{aligned}{{}}{1/\sqrt 2 }&{ - 1/\sqrt 2 }\\{1/\sqrt 2 }&{\,\,\,\,1/\sqrt 2 }\end{aligned}} \right)^{ - 1}}\).

Step-by-step solution

Find the characteristic polynomial

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

Let\(A = \left( {\begin{aligned}{{}}3&1\\1&{\,\,3}\end{aligned}} \right)\)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:

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

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

\(\begin{aligned}{}det\left( {A - \lambda I} \right) &= det\left( {\begin{aligned}{{}}{3 - \lambda }&1\\1&{3 - \lambda }\end{aligned}} \right)\\ &= \left( {3 - \lambda } \right)\left( {3 - \lambda } \right) - 1\\ &= {\lambda ^2} - 6\lambda + 8\end{aligned}\)

So, the characteristic polynomial of matrix \(A\) is \({\lambda ^2} - 6\lambda + 8\).

Find the Eigen values

Thus, the eigenvalues of\(A\)are the solutions of the characteristic equation\(\det \left( {A - \lambda I} \right) = 0\). So, solve the characteristic equation\({\lambda ^2} - 6\lambda + 8 = 0\), 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\({\lambda ^2} - 4\lambda + 5 = 0\)is obtained as follows:

\(\begin{aligned}{}{\lambda ^2} - 6\lambda + 8 &= 0\\\lambda &= \frac{{ - \left( { - 6} \right) \pm \sqrt {{{\left( { - 6} \right)}^2} - 4\left( 8 \right)} }}{2}\\ &= \frac{{6 \pm \sqrt 4 }}{2}\\ &= 2,\,4\end{aligned}\)

The eigenvalues of \(A\)are \(\lambda = 2\) and \(\lambda = 4\).

Find the matrix \(P\) and \(D\)

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

Similarly, for the Eigen value \(\lambda = 4\), the basis for Eigenspace is obtained by solving the system of equations \(A - 4I = 0\). On solving it is obtained as \(\left( \begin{aligned}{l} - 1\\\,\,\,1\end{aligned} \right)\).

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

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

Here, the matrix \(D\) is obtained as \(D = \left( {\begin{aligned}{{}}2&0\\0&4\end{aligned}} \right)\).

Diagonalize matrix \(A\)

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

\(\begin{aligned}{}A &= PD{P^{ - 1}}\\ &= \left( {\begin{aligned}{{}}{1/\sqrt 2 }&{ - 1/\sqrt 2 }\\{1/\sqrt 2 }&{\,\,\,\,1/\sqrt 2 }\end{aligned}} \right)\left( {\begin{aligned}{{}}2&0\\0&4\end{aligned}} \right){\left( {\begin{aligned}{{}}{1/\sqrt 2 }&{ - 1/\sqrt 2 }\\{1/\sqrt 2 }&{\,\,\,\,1/\sqrt 2 }\end{aligned}} \right)^{ - 1}}\end{aligned}\)

The orthogonal diagonalization is \(A = \left( {\begin{aligned}{{}}{1/\sqrt 2 }&{ - 1/\sqrt 2 }\\{1/\sqrt 2 }&{\,\,\,\,1/\sqrt 2 }\end{aligned}} \right)\left( {\begin{aligned}{{}}2&0\\0&4\end{aligned}} \right){\left( {\begin{aligned}{{}}{1/\sqrt 2 }&{ - 1/\sqrt 2 }\\{1/\sqrt 2 }&{\,\,\,\,1/\sqrt 2 }\end{aligned}} \right)^{ - 1}}\).