The matrix,
\[\begin{pmatrix} 1 & -3 & 3 \\ 3 & -5 & 3 \\ 6 & -6 & 4 \end{pmatrix}\]The eigenvalues are \(\lambda= 4, -2, -2\) with corresponding bases for the eigenspaces,
\[\left\{ \begin{pmatrix}1/2 \\ 1/2 \\ 1\end{pmatrix} \right\}\]for \(\lambda = 4\) and,
\[\left\{ \begin{pmatrix}-1 \\ 0 \\ 1\end{pmatrix}, \begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix} \right\}\]for \(\lambda = -2\).
The matrix,
\[\begin{pmatrix} 9 & -8 & 6 & 3 \\ 0 & 9 & 0 & 1 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 7 \end{pmatrix}\]The eigenvalues are \(\lambda= 9, 9, 3, 7\) with corresponding bases for the eigenspaces,
\[\left\{ \begin{pmatrix}1 \\ 0 \\ 0 \\ 0\end{pmatrix} \right\}\]for \(\lambda = 9\),
\[\left\{ \begin{pmatrix}-1 \\ 0 \\ 1 \\ 0\end{pmatrix} \right\}\]for \(\lambda = 3\), and
\[\left\{ \begin{pmatrix}-7/2 \\ -1/2 \\ 0 \\ 1\end{pmatrix} \right\}\]for \(\lambda = 7\).
The second derivative \(\frac{d^2}{dx^2}\) is a linear operator on the space of functions. Let \(\omega > 0\). Show that \(\sin(\sqrt\omega x)\) and \(\cos(\sqrt\omega x)\) are eigenvectors of \(\frac{d^2}{dx^2}\), and find their eigenvalues.
To show that something is an eigenvector of a linear transformation, you just have to check:
\[\frac{d^2}{dx^2} \sin(\sqrt \omega x) = \frac{d}{dx}\sqrt \omega \cos(\sqrt \omega x) = -\omega \sin(\sqrt \omega x)\]So \(\sin(\sqrt \omega x)\) is an eigenvector with eigenvalue \(- \omega\), and
\[\frac{d^2}{dx^2}\cos(\sqrt \omega x) = \frac{d}{dx}{\left({-\sqrt \omega} \sin(\sqrt \omega x)\right)} = -\omega \cos(\sqrt \omega x)\]and \(\cos(\sqrt \omega x)\) is an eigenvector with eigenvalue \(- \omega\).
Find a matrix \(B\) that has eigenvalues 2, -1, 1, and for which,
\[\begin{pmatrix}1 \\ -1 \\ 0\end{pmatrix}, \begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}, \begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix},\]are their corresponding eigenvectors.
We know that \(B\) diagonalizes as,
\[\begin{pmatrix} 1 & 1 & 1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}^{-1}\]This is a fine answer. You can simplify it (if desired) to:
\[\begin{pmatrix} 0 & -2 & 1 \\ -3 & -1 & 3 \\ -2 & -2 & 3 \end{pmatrix}\]
If \(n\) is a positive integer, use diagonalizations to find \(A^n\). It’s fine (encouraged in fact!) to leave your answer factored:
\[\begin{pmatrix} 2 & -1 & 0 \\ -1 & 3 & -1 \\ 0 & -1 & 2 \end{pmatrix}\]We can find that \(A\) diagonalizes as:
\[A = \begin{pmatrix} 1 & -1 & 1 \\ 1 & 0 & -2 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix} \begin{pmatrix} 1 & -1 & 1 \\ 1 & 0 & -2 \\ 1 & 1 & 1 \end{pmatrix}^{-1}\]So \(A^n\) is,
\[A^n = \begin{pmatrix} 1 & -1 & 1 \\ 1 & 0 & -2 \\ 1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1^n & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 4^n \end{pmatrix} \begin{pmatrix} 1 & -1 & 1 \\ 1 & 0 & -2 \\ 1 & 1 & 1 \end{pmatrix}^{-1}\]
Let \(R\) be a transition matrix for a Markov chain,
\[R =\begin{pmatrix} \frac 12 & \frac 13 & 0 & \frac 15 \\ \frac 14 & \frac 13 & \frac 13 & \frac 25 \\ \frac 14 & \frac 16 & \frac 23 & \frac 15 \\ 0 & \frac 16 & 0 & \frac 15 \end{pmatrix}\]All entries of \(R^2\) are positive. Approximate,
\[R^{100,002,018}\begin{pmatrix}\frac 14 \\ \frac 14 \\ \frac 14 \\ \frac 14\end{pmatrix}.\]Since \(R^2\) has all positive entries, the approximation will just be the unique eigenvector of eigenvalue 1, scaled so that its entries sum to 1. So, we have to row reduce:
\[{\rm ref}(R - I) = {\rm ref}\begin{pmatrix} -1/2 & 1/3 & 0 & 1/5 \\ 1/4 & -2/3 & 1/3 & 2/5 \\ 1/4 & 1/6 & -1/3 & 1/5 \\ 0 & 1/6 & 0 & -4/5 \end{pmatrix}\]It’s easier if we clear denominators first by multiplying all entries by \(4(3)(5) = 60\) to get,
\[{\rm ref}\begin{pmatrix} -30 & 20 & 0 & 12 \\ 15 & -40 & 20 & 24 \\ 15 & 10 & -20 & 12 \\ 0 & 10 & 0 & -48 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & -18/5 \\ 0 & 1 & 0 & -24/5 \\ 0 & 0 & 1 & -57/10 \\ 0 & 0 & 0 & 0 \end{pmatrix}\]Whence we can see that an eigenvector for \(\lambda = 1\) is,
\[\begin{pmatrix}18/5 \\ 24/5 \\ 57/10 \\ 1\end{pmatrix}\]But this is too long, so we have to rescale (so that it sums to 1),
\[\frac{10}{151} \begin{pmatrix}18/5 \\ 24/5 \\ 57/10 \\ 1\end{pmatrix}.\]Then by our theorem,
\[R^{100,002,018}\begin{pmatrix}\frac 14 \\ \frac 14 \\ \frac 14 \\ \frac 14\end{pmatrix} \approx \frac{10}{151} \begin{pmatrix}18/5 \\ 24/5 \\ 57/10 \\ 1\end{pmatrix}.\]