2

Using factorisation, $A=PJP^{-1}$ to compute $A^k$, where $k$ represents an arbitrary positive integer.

$$ \begin{bmatrix} \mathbf{0} & \mathbf{1} \\ \mathbf{-1} & \mathbf{2} \end{bmatrix} = \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{1} & \mathbf{1} \end{bmatrix} \begin{bmatrix} \mathbf{1} & \mathbf{1} \\ \mathbf{0} & \mathbf{1} \end{bmatrix} \begin{bmatrix} \mathbf{1} & \mathbf{0} \\ \mathbf{-1} & \mathbf{1} \end{bmatrix}. $$

Not sure how to finish this problem as the matrix $J$ is not diagonal, but $J^k$ can still be found somehow. How exactly is $J^k$ obtained in this case?

Rodrigo de Azevedo
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Mondli.K
  • 393

3 Answers3

2

You have basically put your matrix $A$ in Jordan normal form. This means that you can split your matrix $J$ as follows: $$J = D + N$$ where $D$ is diagonal and $N$ is nilpotent, i.e. there exists an a positive integer $n$ such that $N^n = 0$.

In your case $$J = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \underbrace{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}}_{=D} + \underbrace{ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}}_{=N} $$ and in your case you have $N^2 = 0$.

The basic idea is that when you expand your power $J^k = (D+N)^k$ using the binomial theorem, terms with $N^i$ with $i \geq n$ will disappear, since $N^n = 0$. In your case: $$(D+N)^k = {k \choose 0} D^k + {k \choose 1} D^{k-1}N + \dots$$ where the remaining terms will be zero, e.g. the next term will be $${k \choose 2}D^{k-2} \underbrace{N^2}_{=0} = 0.$$

More information can be found in this math.SE post.

Community
  • 1
Calle
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1

You can easily prove (for example with induction) that

$${\begin{pmatrix} 1 & 1 \\ 0 & 1 \\ \end{pmatrix}}^k=\begin{pmatrix} 1 & k \\ 0 & 1 \\ \end{pmatrix}$$

Now $A^k$ can be computed as follows:

$A^k=(S^{-1}BS)^{k}=\underbrace{S^{-1}BS\cdot S^{-1}BS\dots S^{-1}BS}_{k \text{ times }}=S^{-1}B^{k}S=S^{-1}\begin{pmatrix} 1 & k \\ 0 & 1 \\ \end{pmatrix}S$

Marm
  • 3,439
-1

Hint: It is related to the Fibonacci numbers.

Batman
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