If I am given a matrix $M=\begin{pmatrix} 13 & 1 \\ -16 & 5 \end{pmatrix}$
I need to find formulas for the entries of$ M^n$, where $n$ is a positive integer. I first calculated all the eigenvalues which are 9.So the eigenvector is: $$ \left[ \begin{matrix} -1/4\\ 1\\ \end{matrix} \right] $$.
I know that $M = PDP^{-1}$, so $ M^n = PD^nP^{-1}$.
So, $D=\begin{pmatrix} 9^n & 1 \\ 0 & 9^n \end{pmatrix}$
and $P=\begin{pmatrix} -1 & -1/4 \\ 4 & 0 \end{pmatrix}$
so is $P^{-1}$ $=\begin{pmatrix} 0 & 1/4 \\ -4 & -1 \end{pmatrix}$
so $M^n$ = \begin{pmatrix} -1 & -1/4 \\ 4 & 0 \end{pmatrix}\begin{pmatrix} 9^n & 1 \\ 0 & 9^n \end{pmatrix}\begin{pmatrix} 0 & 1/4 \\ -4 & -1 \end{pmatrix}
which gives \begin{pmatrix} 9^n+4 & 1 \\ -16 & 9^n-4\end{pmatrix}
But it is wrong. Does anyone know what I am doing wrong?
The characteristic equation is
$ | \lambda I - M | = (\lambda - 13)(\lambda - 5) + 16 = \lambda^2 - 18 \lambda + 81 =(\lambda - 9)^2$
It roots are $9$ with algebraic multiplicity of $2$.
Now we find the eigenvectors
Let $v = [v_1 , v_2]^T $ then we have to solve the system
$ (\lambda I - M) v = 0 $
which is
$ \begin{bmatrix} -4 && -1 \\ 16 && 4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$
The solutions is
$\begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = t \begin{bmatrix} -\frac{1}{4} \\ 1 \end{bmatrix} $
Since there is only one eigenvector, we have to generate a generalized eigenvector as follows
$ (M - \lambda I) u = v $
so that
$ \begin{bmatrix} 4 && 1 \\ -16 && -4 \end{bmatrix} \begin{bmatrix} u_1 \\u_2 \end{bmatrix} = \begin{bmatrix} -\frac{1}{4} \\ 1 \end{bmatrix} $
The system is consistent, and has the solution
$ \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \begin{bmatrix} \frac{1}{4}(-\frac{1}{4} - t) \\ t \end{bmatrix} $
Put $t = 1$ you get
$ u = \begin{bmatrix} -\dfrac{5}{16} \\ 1 \end{bmatrix} $
So now
$ M [ v, u] = [\lambda v , v + \lambda u ] = \begin{bmatrix} v, u \end{bmatrix} \begin{bmatrix} \lambda && 1 \\ 0 && \lambda \end{bmatrix} $
so that we can take
$P = [v, u] = \begin{bmatrix} -\frac{1}{4} && -\frac{5}{16} \\ 1 && 1 \end{bmatrix} $
then
$M = P J P^{-1}$ where
$J = \begin{bmatrix} \lambda && 1 \\ 0 && \lambda \end{bmatrix} = \begin{bmatrix} 9 && 1 \\ 0 && 9 \end{bmatrix} $
Note that $P$ is not unique, we could have chosen
$ v = \begin{bmatrix} -1 \\ 4 \end{bmatrix} $
and by setting $t = 0$ results in
$ u = \begin{bmatrix} -\frac{1}{4} \\ 0 \end{bmatrix} $
so that
$ P = \begin{bmatrix} -1 && -\frac{1}{4} \\ 4 && 0 \end{bmatrix} $
Now,
$M^n = P J^n P^{-1} $
You can show by induction, or otherwise, that
$J^n = \begin{bmatrix} 9^n && n (9)^{n-1} \\ 0 && 9^n \end{bmatrix}$
and from this you can compute $M^n$
$M^n = \begin{bmatrix} -\frac{1}{4} && -\frac{5}{16} \\ 1 && 1 \end{bmatrix} \begin{bmatrix} 9^n && n (9)^{n-1} \\ 0 && 9^n \end{bmatrix} \begin{bmatrix} 16 && 5 \\ -16 && -4 \end{bmatrix} $
Multiplying out the first two matrices
$M^n = \begin{bmatrix} - \dfrac{9^n}{4} && -\frac{9^{n-1}}{16} (4 n + 45 ) \\ 9^n && 9^{n-1} ( n + 9 ) \end{bmatrix} \begin{bmatrix} 16 && 5 \\ -16 && -4 \end{bmatrix} $
and this becomes
$M^n = \begin{bmatrix} 9^{n-1} ( 4n + 9 ) && 9^{n-1} n \\ -16 n (9)^{n-1} && 9^{n-1} (9 - 4n) \end{bmatrix} = 9^{n-1} \begin{bmatrix} 4 n + 9 && n \\ -16 n && 9 - 4n \end{bmatrix} $
I checked $M^2 $ by direct calculation and using the above expression, and they match.