Linear algebra, power of matrices

Question

$P^{-1}AP = \begin{pmatrix} -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} $ with $P= \begin{pmatrix} -1 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $ and $P^{-1}$ is the inverse of $P$

Find $A$, $A^{100}$?

I found $A$, $P^{-1}AP=B$ and multiplied by $P$ and $P^{-1}$. How can I find $A^{100}$? I have to use eigenvalues/vectors.

Answer 1

Note that $B=P^{-1}AP=(-I_3+N)\oplus2$. Therefore $B^{100}=P^{-1}A^{100}P=(-I_3+N)^{100}\oplus2^{100}$. Since $N^3=0$, by binomial theorem, $$(-I+N)^{100}=(-I)^{100}+\binom{100}1(-I)^{99}N+\binom{100}2(-I)^{98}N^2.$$

Answer 2

You've told us that you know $A$, so I'm assuming that this is the case.

Here's a general formula for finding $A^n$:

$A^n=\underbrace{(PB\overbrace{P^{-1})(P}^{I}B\overbrace{P^{-1})(P}^{I}B\overbrace{P^{-1})P}^{I}...\overbrace{P^{-1}(P}^{I}B\overbrace{P^{-1})(P}^{I}BP^{-1})}_{n \ \text{times}}$.

Simplifying this, we see that: $$\boxed{A^n=PB^nP^{-1}},$$ where $B$ is (diagonal) matrix of eigenvalues,and $P=[\vec r_1 \cdots\vec r_n]$ is the matrix of eigenvectors.

Subbing $n=100$ into this formula, we have: $$A^{100}=PB^{100}P^{-1}.$$

You should be able to take it from here.

Answer 3

So I kind of suspect that the point of this question is that $P$ and $P^{-1}$ diagonalise $B$. In the case of a diagonalised matrix,D, $D^{100}$ just means taking each element to the power of 100, since each multiplication just multiplies each diagonal element by itself. It is often the fastest way to take large powers of matrixes. First diagonalise, then note the trick by alex that $A^n = PB^nP^{-1}$ with B diagonal.

Answer 4

Because of the way that $P$ and $A$ are constructed, it is not hard to see that $PA=AP$. Consequently, $P^{-1}AP=P^{-1}PA=A$. So that makes life easier. You can then write $$ A = \left[\begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 2 \end{array}\right]+ \left[\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] = D+N, $$ where $D$ is the diagonal matrix and $N$ is nilpotent of order $3$ (meaning that $N^{3}=0$, but $N^{2} \ne 0$.) You have $DN=ND=-N$ and $$ N^{2} = \left[\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right],\;\;\; N^{3}=0. $$ This is how the Jordan canonical form works. $N$ works on the same part of the space corresponding to the columns with $-1$ entries. So $DN=ND=-N$. And $DN^{2}=N^{2}D=-N^{2}$. So, $$ (D+N)^{100} = D^{100}+(100)D^{99}N+(50)(99)D^{98}N^{2}=D^{100}-100N+50(99)N^{2}. $$ You can write down the matrix $A^{100}$ from this. Nothing further is needed.