Let \(A = \begin{bmatrix} 5 & -2\\ 6 & -2 \end{bmatrix}\) and let \(x = \begin{bmatrix} 2\\ 3 \end{bmatrix}\). We will compute the following:
\(Ax\)
\(A^2x\)
\(A^{1000}x\)
The first two parts are straightforward to compute. However, computing \(A^{1000}x\) seems to involve a lot of work, especially if done by hand. Is there a better way than brute force?
Note that \(Ax = \begin{bmatrix} 5 & -2\\ 6 & -2 \end{bmatrix} \begin{bmatrix}2\\3\end{bmatrix} = \begin{bmatrix} 10 - 6\\12 -6\end{bmatrix} = \begin{bmatrix} 4\\ 6\end{bmatrix}. \)
But \(\begin{bmatrix} 4\\ 6\end{bmatrix} = 2\begin{bmatrix} 2\\ 3\end{bmatrix} = 2x\)! Hence, when we multiply \(x\) on the left by \(A\), it is the same as doubling \(x\). So \(A^2x = 4x=\begin{bmatrix} 8 \\12\end{bmatrix}\). And in general, \(A^kx = 2^kx\). So \(A^{1000}x = 2^{1000} \begin{bmatrix} 2 \\ 3 \end{bmatrix}.\)
However, if we had \(x = \begin{bmatrix} 1\\1\end{bmatrix}\), we wouldn't be able do something similar to the above since \(A\begin{bmatrix} 1\\1\end{bmatrix} =\begin{bmatrix} 3\\4\end{bmatrix}\) is not a scalar multiple of \(\begin{bmatrix} 1\\1\end{bmatrix}\).
Fortunately, there is another way. Let \(P = \begin{bmatrix} 2 & 1\\ 3 & 2 \end{bmatrix}\) and let \(D = \begin{bmatrix} 2 & 0\\ 0 & 1 \end{bmatrix}\). Then \(P^{-1} = \begin{bmatrix} 2 & -1\\ -3 & 2 \end{bmatrix}\). One can check that \(A = PDP^{-1}\).
Now, note that \[A^2 = PDP^{-1}PDP^{-1} = PDDP^{-1} = P D^2 P^{-1},\] and \[A^3 = AA^2 = PDP^{-1}PD^2P^{-1} = PDD^2P^{-1} = P D^3 P^{-1}.\] In general, we will have \(A^k = P D^k P^{-1}.\)
Since \(D\) is a diagonal matrix, \(D^k\) is obtained from \(D\) by raising each diagonal entry in \(D\) to the power \(k\). This makes computing \(A^k\) quite easy. In particular, \(A^{1000} = P \begin{bmatrix} 2^{1000} & 0 \\ 0 & 1 \end{bmatrix}P^{-1}\).
There remains one mystery. How does one find the matrices \(P\) and \(D\)? Answering that question will take us to the topic of eigenvalues and eigenvectors.
Let \(A\) be as above.
Compute \(A^{10000} \begin{bmatrix} 1 \\ 2\end{bmatrix}\).
Compute \(A^{10} \begin{bmatrix} 1 \\ 0\end{bmatrix}\).