Find a formula in terms of k for the entries of $A^k$, where A is the diagonalizable matrix This is my 2x2 matrix (sorry for formatting):
[$-5$ $8$]
[$-4$ $7$]
I've tried this question a million times and I always get the answer the wrong. To start off, I found my eigenvalues and eigenvectors which are: eigenvalue1: -1 eigenvalue2: 3, eigenvector1: (1,1), (2,1).
Can someone please show me a step by step without ommitting any calculations? I have the answer, but I do not know how to arrive at it. Thanks a lot.
We write the Jordan Normal Form (that is, write it as a diagonal matrix) as:
$$A = \begin{bmatrix}-5 & 8\\-4 & 7\end{bmatrix} = PJP^{-1} = \begin{bmatrix}2 & 1\\1 &1\end{bmatrix}\begin{bmatrix}-1& 0\\0 & 3\end{bmatrix}\begin{bmatrix}1 & -1\\-1& 2\end{bmatrix}$$
Now
$$A^k = PJ^kP^{-1}= \begin{bmatrix}2 & 1\\1 &1\end{bmatrix}\begin{bmatrix}(-1)^k& 0\\0 & (3)^k\end{bmatrix}\begin{bmatrix}1 & -1\\-1& 2\end{bmatrix}$$
You can multiply this out as:
$$A^k = \begin{bmatrix} 2 (-1)^k-3^k & -2 (-1)^k+2\ 3^k \\ (-1)^k-3^k & -(-1)^k+2\ 3^k \\ \end{bmatrix} $$