# Use the factorization A=PDP^{-1} to compute A^{k}.

Polynomial factorization

Use the factorization $$A=PDP^{-1}$$ to compute $$A^{k}$$. where k represents an arbitrary integer.
$$\begin{bmatrix}a & 11(b-a) \\0 & b \end{bmatrix}=\begin{bmatrix}1 & 11 \\0 & 1 \end{bmatrix}\begin{bmatrix}a & 0 \\0 & b \end{bmatrix}\begin{bmatrix}1 & -11 \\0 & 1 \end{bmatrix}$$
$$A^{k}=$$?

2021-05-28
Step 1
Given
$$\begin{bmatrix}a & 11(b-a) \\0 & b \end{bmatrix}=\begin{bmatrix}1 & 11 \\0 & 1 \end{bmatrix}\begin{bmatrix}a & 0 \\0 & b \end{bmatrix}\begin{bmatrix}1 & -11 \\0 & 1 \end{bmatrix}$$
To find
$$A^{k}=$$?
Step 2
We know that,
$$A^{k}=PD^{k}P^{-1}=\begin{bmatrix}1 & 11 \\0 & 1 \end{bmatrix}\begin{bmatrix}a^{k} & 0 \\0 & b^{k} \end{bmatrix}\begin{bmatrix}1 & -11 \\0 & 1 \end{bmatrix}$$
$$=\begin{bmatrix}a^{k} & 11b^{k} \\0 & b^{k} \end{bmatrix}\begin{bmatrix}1 & -11 \\0 & 1 \end{bmatrix}$$
$$=\begin{bmatrix}a^{k} & -11a^{k}+11b^{k} \\0 & b^{k} \end{bmatrix}$$
$$=\begin{bmatrix}a^{k} & 11b^{k}-11a^{k} \\0 & b^{k} \end{bmatrix}$$
Hence,
$$A^{k}=\begin{bmatrix}a^{k} & 11(b^{k}-a^{k}) \\0 & b^{k} \end{bmatrix}$$