Find the eigenvalues and their corresponding eigenspaces for the matrix,
\[C = \begin{pmatrix} 2 & 0 & -2 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix}.\]\(\DeclareMathOperator{\span}{span}\) The characteristic polynomial of \(C\) is \((\lambda - 2)^2(\lambda - 3)\), so the eigenvalues are \(2, 2, 3\).
The eigenspace for eigenvalue 2 is two-dimensional and is spanned by the vectors found by solving the equation \((C - 2I) \vec v = \vec 0\); it is,
\[\span\left\{ \begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}, \begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix} \right\}\]The eigenspace for eigenvalue 3 is one-dimensional and is spanned by the solution to the equation \((C - 3I) \vec v = \vec 0\); it is,
\[\span\left\{ \begin{pmatrix} -2 \\ 2 \\ 1 \end{pmatrix} \right\}\]
Remember that the eigenvalues of the matrix,
\[A = \begin{pmatrix} 1&-3&3\\3&-5&3\\6&-6&4 \end{pmatrix}\]are \(\lambda = 4, -2, -2\), with eigenvector for \(\lambda = 4\),
\[\begin{pmatrix}1\\1\\2\end{pmatrix}\]and eigenvectors for \(\lambda = -2\),
\[\begin{pmatrix}-1\\0\\1\end{pmatrix} \quad\textrm{and}\quad \begin{pmatrix}1\\1\\0\end{pmatrix}.\]Find an invertible matrix \(P\) and a diagonal matrix \(D\) so that \(A = PDP^{-1}\).
Based on the known eigenvalues, one choice for \(D\) is,
\[D = \begin{pmatrix} 4&0&0 \\ 0&-2&0 \\ 0&0&-2 \end{pmatrix}.\]Based on this choice of \(D\), one valid choice for \(P\) is,
\[P = \begin{pmatrix} 1&-1&1 \\ 1&0&1 \\ 2&1&0 \end{pmatrix}.\]
Can the matrix
\[M = \begin{pmatrix}9 & 1 \\ -1 & 11\end{pmatrix}\]be diagonalized? Explain why or why not.
The characteristic polynomial of \(M\) is, \((\lambda - 9)(\lambda - 11) + 1 = \lambda^2 - 20\lambda + 100 = (\lambda - 10)^2\), so the eigenvalues are \(10, 10\). However, the only eigenvector of \(M\) is
\[\begin{pmatrix}1 \\ 1\end{pmatrix}.\]As there is only one eigenvector for the \(2\times 2\) matrix \(M\), it is not diagonalizable.
Can the matrix
\[N = \begin{pmatrix}2 & 3 \\ 1 & 2\end{pmatrix}\]be diagonalized? Explain why or why not.
The characteristic polynomial of \(N\) is, \(\lambda^2 - 4\lambda + 1\). This polynomial has discriminant “\(b^2 - 4ac\)” of \(12\). As the discriminant is greater than zero there are two unique real roots to the polynomial, and hence two different real eigenvalues for \(N\). As each eigenvalue has at least one eigenvector, \(N\) has at least two eigenvectors. So, as \(N\) is a \(2\times 2\) matrix, this means that yes, \(N\) is diagonalizable.
For \(n\) a positive integer and \(B\) the matrix,
\[B = \begin{pmatrix} 1 & 0 &-2 \\ 0 & -1 & 2 \\ -1 & 0 & 2 \end{pmatrix}\]find a formula for \(B^n\).
\(B\) diagonalizes as the matrix product,
\[B = PDP^{-1} = \begin{pmatrix} 0&2&-2 \\ 1&2&1 \\ 0&1&2 \end{pmatrix} \begin{pmatrix} -1&0&0 \\ 0&0&0 \\ 0&0&3 \end{pmatrix} \begin{pmatrix} 0&2&-2 \\ 1&2&1 \\ 0&1&2 \end{pmatrix}^{-1}.\]Since \(B^n = (PDP^{-1})^n = PD^n P^{-1}\), a formula for \(B^n\) is,
\[B = PDP^{-1} = \begin{pmatrix} 0&2&-2 \\ 1&2&1 \\ 0&1&2 \end{pmatrix} \begin{pmatrix} (-1)^n&0&0 \\ 0&0&0 \\ 0&0&3^n \end{pmatrix} \begin{pmatrix} 0&2&-2 \\ 1&2&1 \\ 0&1&2 \end{pmatrix}^{-1}.\]