Evaluate $A^n$ with $A$ is a matrix that: $ A=\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix} $

Question

Evaluate $A^n$ with $A$ is a matrix that: $ A=\begin{bmatrix} 3 & 5\\ 1 & -1 \end{bmatrix} $

This is kinda tough question for me because normally I can use mathematic induction to find the recursive form of the answer, but to evaluate this with induction theorem is not so effective, and I can not find the answer, so can anyone give me a suggestion or a clear answer for this. I'm grateful for this.

Answer 1

The eigenvalues of $A$ are $4$ and $-2$. An eigenvector of $A$ with eigenvalue $4$ is $(5,-1)$ and an eigenvector of $A$ with eigenvalue $-2$ is $(1,1)$. So, if you take $$ B=\begin{bmatrix} 5&1\\ -1&1 \end{bmatrix} $$ then $$ B^{-1}AB=\begin{bmatrix} 4&0\\ 0&-2 \end{bmatrix}. $$ So, for each $n\in\mathbb{N}$ \begin{align} A^n&=B.\begin{bmatrix} 4&0\\ 0&-2 \end{bmatrix}^n.B^{-1}\\ &=\begin{bmatrix} 5&1\\ -1&1 \end{bmatrix}.\begin{bmatrix} 4^n&0\\ 0&(-2)^n \end{bmatrix}.\frac16\begin{bmatrix} 1&1\\ -1&5 \end{bmatrix}\\ &=\frac{1}{6}\begin{bmatrix} 2^{n} \left((-1)^n+5\times 2^n\right) & 5\times 2^{n} \left((-1)^{n+1}+2^n\right) \\ 2^{n} \left((-1)^{n+1}+2^n\right) & 2^{n} \left(5 (-1)^n+2^n\right) \end{bmatrix}. \end{align}

Answer 2

1. Find matrix eigenvalues $λ_1, λ_2$ by solving the equation $\det\left(A-\lambda I\right) = 0 \Leftrightarrow \begin{vmatrix}3-λ & 5 \\ 1 & -1-λ\end{vmatrix} = 0$. Given $λ_1$ and $λ_2$, make a matrix $\mathbf{\Lambda} = \begin{pmatrix}λ_1 & 0 \\ 0 & λ_2\end{pmatrix}$.

2. For each eigenvalue $\lambda_1$ and $λ_2$, find a corresponding eigenvector $\mathbf{v}^{(1)}$ and $\mathbf{v}^{(2)}$: $A\mathbf{v}^{(i)} = λ_i\mathbf{v}^{(i)}$, $i = 1,2$. Then, make a matrix $\mathbf{V} = \begin{pmatrix}\mathbf{v}^{(1)} & \mathbf{v}^{(2)}\end{pmatrix} = \begin{pmatrix}\mathbf{v}^{(1)}_{1} & \mathbf{v}^{(2)}_1 \\ \mathbf{v}^{(1)}_2 & \mathbf{v}^{(2)}_2\end{pmatrix}$.

3. Use the formula $A = \mathbf{V}\mathbf{Λ}\mathbf{V}^{-1} \Rightarrow A^n = \mathbf{V}\mathbf{Λ}^n\mathbf{V}^{-1}$, where

$$ \mathbf{Λ}^n = \begin{pmatrix} λ_1^n & 0 \\ 0 & λ_2^n \end{pmatrix}\text{ and }\mathbf{V}^{-1}\text{ is an inverse matrix of matrix}\mathbf{V}. $$

Answer 3

I think the best way to go here is to diagonalize $A$! Notice that $A$ is full rank, and can be written as $A = SDS^{-1}$ where $D$ is diagonal and has as its diagonal elements the eigenvalues of $A$ and $S$ is the matrix whose columns correspond to the eigenvectors of $A$. That way, $A^{n} = (SDS^{-1})^{n} = SD^{n}S^{-1}$. But $D^{n}$ is especially easy to evaluate...

Answer 4

If you are allowed to use a computer to diagonalize the matrix, then it's easy to find a closed form answer.

As $A = T\Lambda T^{-1}$ with eigenvectors $T = \begin{bmatrix}5&-1\\1&1\end{bmatrix}$ and eigenvalues $\Lambda = \begin{bmatrix}4&0\\0&-2\end{bmatrix}$, you can calculate the $n$th power as \begin{equation} A^n = T\Lambda^n T^{-1} = \frac{1}{6}\begin{bmatrix} 5&-1\\1&1\end{bmatrix}\begin{bmatrix}4^n&0\\0&(-2)^n\end{bmatrix}\begin{bmatrix}1&1\\-1&5\end{bmatrix} \end{equation}

Answer 5

Here is the general method to find the power of a diagonalizable matrix.

Let $A$ be a matrix, the problem is to find $A^n$, for some $n\in \mathbb{N}$.

If $A$ is diagonalizable, then exist in particula a diagonal matrix $D$ and an invertible matrix $S$ such that: $$D= S^{-1}AS$$ Now: $$D^n=(S^{-1}AS)^n=\overbrace{(S^{-1}AS)(S^{-1}AS)\dots(S^{-1}AS)}^{n\,\,\, \text{times}}=S^{-1}A^nS $$ FInally you have: $$A^n=SD^nS^{-1} $$ Note that since $D$ is a diagonal matrix, to find $D^n$ you just have to raise to the $n$ power the elements of the diagonal (this can be easily proven by induction).