Question

(M)Use a matrix program to diagonalize

\(A = \left( {\begin{aligned}{*{20}{c}}{ - 3}&{ - 2}&0\\{14}&7&{ - 1}\\{ - 6}&{ - 3}&1\end{aligned}} \right)\)

If possible. Use the eigenvalue command to create the diagonal matrix \(D\). If the program has a command that produces eigenvectors, use it to create an invertible matrix \(P\). Then compute \(AP - PD\) and \(PD{P^{{\bf{ - 1}}}}\). Discuss your results.

Short answer

\(A\)is not diagonalizable.Use command eig in MATLAB to get invertible matrix \(P\) and diagonal matrix \(D\) if possible.

Step-by-step solution

Determine whether \(P\) is invertible or not

First, we need to define a matrix\(A\)in MATLAB and use command eig to get an invertible matrix\(P\)and a diagonal matrix\(D\).

\(\left( {\begin{aligned}{*{20}{c}}P&D\end{aligned}} \right) = {\rm{eig}}\left( A \right)\)

Therefore,

\(P = \left( {\begin{aligned}{*{20}{c}}{0.3244}&{ - 0.3244}&{0.3333}\\{ - 0.8111}&{0.8111}&{ - 0.6667}\\{0.4867}&{ - 0.4867}&{0.6667}\end{aligned}} \right)\)

And

\(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&2&0\\0&0&1\end{aligned}} \right)\)

As the matrix \(A\)has two eigenvalues\(2\)and\(1\).

Since the first two columns of the matrix are linearly dependent that means the eigenvalue\(2\)is dimensional that shows\(P\)is not invertible.

Therefore\(A\)is not diagonalizable.

Compute \(AP - PD\) and \(PD{P^{{\bf{ - 1}}}}\)

Since the matrix\(P\)is not invertible therefore\(PD{P^{ - 1}}\)cannot be computed.

However, \(AP - PD\) can be computed.

\(AP\)

\(PD\)

\(AP - PD\)

As \(AP - PD\) is equal to zero matrix.

Since \(P\) consists of eigenvectors of \(A\) therefore columns of \(AP - PD\) are eigenvectors of \(A\) which are multiplied by its eigenvalues.

Also, \(D\) consists of eigenvalues of \(A\) and columns of \(PD\) are eigenvectors of \(A\) multiplied by its eigenvalues.

Thus, use command eig in MATLAB to get invertible matrix \(P\) and diagonal matrix \(D\) if possible.