Question

Question: A factorization \(A = PD{P^{ - {\bf{1}}}}\) is not unique. Demonstrate this for the matrix A in Example 2. With \({D_{\bf{1}}} = \left[ {\begin{array}{*{20}{c}}{\bf{3}}&{\bf{0}}\\{\bf{0}}&{\bf{5}}\end{array}} \right]\), use the information in Example 2 to find the matrix \({P_{\bf{1}}}\) such that \(A = {P_{\bf{1}}}{D_{\bf{1}}}P_{\bf{1}}^{ - {\bf{1}}}\).

Short answer

The matrix \({P_1}\) is \(\left[ {\begin{array}{*{20}{c}}1&1\\{ - 2}&{ - 1}\end{array}} \right]\).

Step-by-step solution

Write the information given

The matrices from example 2 are,

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

Find the matrix \({P_{\bf{1}}}\)

For eigenvalue 5, the eigenvector is \(\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\end{array}} \right]\) and for eigenvalue 3, the eigenvector is \(\left[ {\begin{array}{*{20}{c}}1\\{ - 2}\end{array}} \right]\).

The matrix \({P_1}\) is such that \(A = {P_1}{D_1}P_1^{ - 1}\), so the matrix \({D_1}\) is:

\[{D_1} = \left[ {\begin{array}{*{20}{c}}3&0\\0&5\end{array}} \right]\]

Interchange the columns of P to find the matrix \({P_1}\).

\[{P_1} = \left[ {\begin{array}{*{20}{c}}1&1\\{ - 2}&{ - 1}\end{array}} \right]\]

Find the product \({P_{\bf{1}}}{D_{\bf{1}}}P_{\bf{1}}^{ - {\bf{1}}}\)

The product \({P_1}{D_1}P_1^{ - 1}\) can be calculated as follows:

\[\begin{array}{c}{P_1}{D_1}P_1^{ - 1} = \left[ {\begin{array}{*{20}{c}}1&1\\{ - 2}&{ - 1}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}3&0\\0&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 1}\\2&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}7&2\\{ - 4}&1\end{array}} \right]\\ = A\end{array}\]

So, the matrix \({P_1}\) is \(\left[ {\begin{array}{*{20}{c}}1&1\\{ - 2}&{ - 1}\end{array}} \right]\).