Question

In Exercises 27 and 28, find a factorization of the given matrix A in the form \(A = PC{P^{ - 1}}\),where \(C\) is a block-diagonal matrix with \(2 \times 2\) blocks of the form shown in Example 6. (For each conjugate pair of eigenvalues, use the real and imaginary parts of one eigenvector in \({\mathbb{}^4}\) to create two columns of P.)

27.\(\left( {\begin{aligned}{}{.7}&{}&{1.1}&{}&{2.0}&{}&{1.7}\\{ - 2.0}&{}&{ - 4.0}&{}&{ - 8.6}&{}&{ - 7.4}\\0&{}&{ - .5}&{}&{ - 1.0}&{}&{ - 1.0}\\{1.0}&{}&{2.8}&{}&{6.0}&{}&{5.3}\end{aligned}} \right)\)

Short answer

The factorization is \(P = \left( {\begin{aligned}{}{.5}&{}&{ - .5}&{}&{ - .5}&{}&0\\{ - 2}&{}&0&{}&0&{}&{.5}\\0&{}&0&{}&{ - .75}&{}&{ - .25}\\1&{}&0&{}&1&{}&0\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{}{.2}&{}&{ - .5}&{}&0&{}&0\\{.5}&{}&{.2}&{}&0&{}&0\\0&{}&0&{}&{.3}&{}&{.1}\\0&{}&0&{}&{.1}&{}&{.3}\end{aligned}} \right)\).

Step-by-step solution

Input the matrix

\({\rm{ > > A = }}\left( {{\rm{0}}{\rm{.7 1}}{\rm{.1 2 1}}{\rm{.7; - 2 - 4 - 8}}{\rm{.6 - 7}}{\rm{.4; 0 - 0}}{\rm{.5 - 1 - 1; 1 2}}{\rm{.8 6 5}}{\rm{.3}}} \right)\)

Get the expression of command for getting the Eigenvalue(s) of A 

We use the following command to get the eigenvalue of the matrix \(A\).

\({\rm{ev}} = {\mathop{\rm eig}\nolimits} (A) = (.2 + .5i,.2 - .5i,.3 + 1i,.3 - 1i)\)

Operate on the matrix to get the eigenvalue. 

For \(\lambda = .2 - .5i\), we find the value of the operand in this case to find the eigenvalues.

\({\mathop{\rm nulbasis}\nolimits} ({\bf{A}} - {\mathop{\rm ev}\nolimits} ({\bf{2}})*{\mathop{\rm eye}\nolimits} ({\bf{4}})) = \left( {\begin{aligned}{}{.5 - .5i}\\{ - 2}\\0\\1\end{aligned}} \right)\)

\({{\bf{v}}_1} = \left( {\begin{aligned}{}{.5 - .5i}\\{ - 2}\\0\\1\end{aligned}} \right)\)

For \(\lambda = .3 - .1i\), we find the value of the operand in this case to find the eigenvalues.

\({\mathop{\rm nulbasis}\nolimits} ({\bf{A}} - {\mathop{\rm ev}\nolimits} ({\bf{2}})*{\mathop{\rm eye}\nolimits} (4)) = \left( {\begin{aligned}{}{ - .5}\\{ - .5i}\\{ - .75 - .25i}\\1\end{aligned}} \right)\)

\({{\bf{v}}_2} = \left( {\begin{aligned}{}{ - .5}\\{ - .5i}\\{ - .75 - .25i}\\1\end{aligned}} \right)\)

Now, by theorem 9:

\(P = \left( {\begin{aligned}{}{{\mathop{\rm Re}\nolimits} {v_1}}&{{\mathop{\rm Im}\nolimits} {v_1}}&{{\mathop{\rm Re}\nolimits} {v_2}}&{{\mathop{\rm Im}\nolimits} {v_2}}\end{aligned}} \right) = \left( {\begin{aligned}{}{.5}&{}&{ - .5}&{}&{ - .5}&{}&0\\{ - 2}&{}&0&{}&0&{}&{.5}\\0&{}&0&{}&{ - .75}&{}&{ - .25}\\1&{}&0&{}&1&{}&0\end{aligned}} \right)\)and \(C = \left( {\begin{aligned}{}{.2}&{}&{ - .5}&{}&0&{}&0\\{.5}&{}&{.2}&{}&0&{}&0\\0&{}&0&{}&{.3}&{}&{.1}\\0&{}&0&{}&{.1}&{}&{.3}\end{aligned}} \right)\).

Hence, \(P = \left( {\begin{aligned}{}{.5}&{}&{ - .5}&{}&{ - .5}&{}&0\\{ - 2}&{}&0&{}&0&{}&{.5}\\0&{}&0&{}&{ - .75}&{}&{ - .25}\\1&{}&0&{}&1&{}&0\end{aligned}} \right),C = \left( {\begin{aligned}{}{.2}&{}&{ - .5}&{}&0&{}&0\\{.5}&{}&{.2}&{}&0&{}&0\\0&{}&0&{}&{.3}&{}&{.1}\\0&{}&0&{}&{.1}&{}&{.3}\end{aligned}} \right)\).