Question

Question: With A and D as in Example 2, find an invertible \({P_{\bf{2}}}\) unequal to the P in Example 2, such that \(A = {P_{\bf{2}}}DP_{\bf{2}}^{ - {\bf{1}}}\).

Short answer

The matrix \({P_2}\) is \(\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&2\end{array}} \right]\).

Step-by-step solution

Write the given information from example 2

The matrices from example 2 are,

\[A = \left[ {\begin{array}{*{20}{c}}7&2\\{ - 4}&1\end{array}} \right], D = \left[ {\begin{array}{*{20}{c}}5&0\\0&3\end{array}} \right]\]

Find the row reduced form of a matrix \(A - {\bf{3}}I\)

Find the matrix \(A - 3I\).

\[\begin{array}{c}A - 3I = \left[ {\begin{array}{*{20}{c}}7&2\\{ - 4}&1\end{array}} \right] - 3\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}7&2\\{ - 4}&1\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}3&0\\0&3\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}4&2\\{ - 4}&{ - 2}\end{array}} \right]\end{array}\]

The row-reduced form of the matrix is:

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{A - 3I}&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4&2&0\\{ - 4}&{ - 2}&0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}4&2&0\\0&0&0\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_2} \to {R_2} + {R_1}} \right)\\ = \left[ {\begin{array}{*{20}{c}}2&1&0\\0&0&0\end{array}} \right]\end{array}\]

The equations are:

\[2{x_1} + {x_2} = 0\]

\[{x_2} = - 2{x_1}\]

The general solution is:

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - t}\\{2t}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]t\end{array}\]

So, for the eigenvalues \(\lambda = 3\), the eigenvector is \(\left[ {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right]\).

Find the row reduced form of a matrix \(A - {\bf{5}}I\)

\[\begin{array}{c}A - 5I = \left[ {\begin{array}{*{20}{c}}7&2\\{ - 4}&1\end{array}} \right] - 5\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}7&2\\{ - 4}&1\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}5&0\\0&5\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}2&2\\{ - 4}&{ - 4}\end{array}} \right]\end{array}\]

The row reduced form of a matrix is:

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{A - 5I}&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}2&2&0\\{ - 4}&{ - 4}&0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}2&2&0\\0&0&0\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_2} \to {R_2} + 2{R_1}} \right)\\ = \left[ {\begin{array}{*{20}{c}}1&1&0\\0&0&0\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_1} \to \frac{{{R_1}}}{2}} \right)\end{array}\]

The equations are:

\[\begin{array}{c}{x_1} + {x_2} = 0\\{x_2} = - {x_1}\end{array}\]

The general solution is:

\[\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}t\\{ - t}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right]t\end{array}\]

So, for the eigenvalues \(\lambda = 5\), the eigenvector is \(\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right]\).

Find the matrix \({P_{\bf{2}}}\)

The matrix \({P_2}\) is formed using eigenvectors.

\[{P_2} = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&2\end{array}} \right]\]

The inverse of the matrix \({P_2}\) can be calculated as follows:

\[\begin{array}{c}P_2^{ - 1} = \frac{1}{{2\left( 1 \right) - \left( { - 1} \right)\left( { - 1} \right)}}\left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right]\\ = \frac{1}{{2 - 1}}\left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right]\end{array}\]

Verify matrix \({P_{\bf{2}}}\) using diagonalization theorem

Find the value of \({P_2}DP_2^{ - 1}\).

\[\begin{array}{c}{P_2}DP_2^{ - 1} = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}5&0\\0&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}2&1\\1&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&2\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{10}&5\\3&3\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}7&2\\{ - 4}&1\end{array}} \right]\\ = A\end{array}\]

So, the matrix \({P_2}\) is \(\left[ {\begin{array}{*{20}{c}}1&{ - 1}\\{ - 1}&2\end{array}} \right]\).