Diagonalization method for k'th power of a matrix

Question

I am having problems with a question.

The question is: Using the diagonalisation method, find the formula for the k-th power of the matrix

$$A=\begin{bmatrix}-8 & -6 \\ 15 & 11 \end{bmatrix}^k$$

check your formula for k = 1 and k = -1

Any help would be appreciated thanks.

Answer 1

Hint

It seems that the problem is to apply the diagonalization method to the specific matrix. If so verify that: $$A=MDM^{-1}=\begin{bmatrix}-8 & -6 \\ 15 & 11 \end{bmatrix}=\begin{bmatrix}-2 & -3 \\ 3 & 5 \end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 2 \end{bmatrix}\begin{bmatrix}-5 & -3 \\ 3 & 2 \end{bmatrix}$$

so you can write: $$ A^k=\begin{bmatrix}-2 & -3 \\ 3 & 5 \end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 2^k \end{bmatrix}\begin{bmatrix}-5 & -3 \\ 3 & 2 \end{bmatrix}= $$ $$ =\begin{bmatrix}10 & 6 \\ -15 & -9 \end{bmatrix}+2^k\begin{bmatrix}-9 & -6 \\ 15 & 10 \end{bmatrix} $$

Now show that for $k=1$ this gives $A$ and for $k=-1$ gives the inverse of $A$.

Answer 2

Hint: if $M=B^{-1} J B$, then $$ M^2 = M\cdot M = B^{-1} J B B^{-1} J B = B^{-1} J J B = B^{-1} J^2 B $$ and in general $M^k = B^{-1} J^k B $ for any $k\in\mathbb{Z}$.

Answer 3

$$A = P D P^{-1}$$ The diagonal matrix $D$ is composed of the eigenvalues: $$D = \begin{pmatrix} 2 & 0\\0&1 \end{pmatrix}$$ The eigenvectors corresponding to the eigenvalues in $D$ compose the columns of $P$: $$P = \begin{pmatrix} -\frac{3}{5} & -\frac{2}{3}\\1&1 \end{pmatrix}$$ And $$P^{-1} = \begin{pmatrix} 15 & 10\\-15&-9 \end{pmatrix}$$ Therefore $$A^k = P D^k P^{-1}= \begin{pmatrix} -\frac{3}{5} & -\frac{2}{3}\\1&1 \end{pmatrix} \begin{pmatrix} 2^k & 0\\0&1 \end{pmatrix} \begin{pmatrix} 15 & 10\\-15&-9 \end{pmatrix} $$ For $k = 1$ : $A^1 = A$.

For $k = -1 :$ $$A^{-1} = \begin{pmatrix} -\frac{3}{5} & -\frac{2}{3}\\1&1 \end{pmatrix} \begin{pmatrix} \frac{1}{2} & 0\\0&1 \end{pmatrix} \begin{pmatrix} 15 & 10\\-15&-9 \end{pmatrix} = \begin{pmatrix} \frac{11}{2}& 3\\-\frac{15}{2}&-4 \end{pmatrix}. $$