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We are given $A^p=A ...A$(p times)

And we are given matrix A:

$A=\begin{vmatrix}0.6&-0.4&0\\-0.4&0.6&0\\0&0&0.5\end{vmatrix}$

I need to compute $A^p$ as p approach Infinity.

By some calculations results that the values of matrix A get always close to $0$ as n grows to infinity, but i actually need a proof(something concrete).

In the web i found a person saying that:

The usual technique for doing this (in cases where the limit exists) is to diagonalise the matrix. If $A=P^{-1}DP$ , where $P$ is invertible and $D$ is the diagonal matrix whose entries are the eigenvalues of $A$, then $A^n=p^{-1}D^nP$. The matrix $D^n$ is easy to compute: it is diagonal, and its entries are those of $D$ raised to the power $n$. The limit matrix will only exist if all the eigenvalues are either $1$ or have absolute value less than $1$. In that case, $D^n$ will converge to a limit matrix $D_0$, and $A^n$ will converge to $P^{-1}D_0P$ .

Now i will compute the Eigenvalues of my matrix:

$(0.6-\lambda)^3-0.08+0.16\lambda=(\lambda^2-1.2λ+0.02)(0.5-\lambda)=(0.02λ-0.2)(5\lambda-1)(0.5-\lambda)$

Therefore:

$\lambda_1= 1.$

$\lambda_2 = 0.5 $

$\lambda_3 = 0.2 $

But from now on I'm stuck.

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    $\begingroup$ Please note that the matrix doesn't converge to $0$, as I just checked from matlab. $\endgroup$ – k99731 May 18 '15 at 10:44
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    $\begingroup$ If $1$ is indeed an eigenvalue, then the powers of the matrix cannot converge to $0$ -- suppose $v$ is the corresponding eigenvector, then $A^nv=v$ for all $n$, which cannot be the case if each of the entries in $A^n$ vanish towards $0$. On the other hand, your factorization of the characteristic polynomial doesn't like $1$ would be an eigenvalue. Why do you think it is? $\endgroup$ – Henning Makholm May 18 '15 at 10:44
  • $\begingroup$ Because i was making the "manual" computation of the matrix( i.e. $A^2,A^{10},A^{100}$ and the values inside the matrix were getting always smaller... but yes its my fault./ $\endgroup$ – BioShock May 18 '15 at 10:47
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    $\begingroup$ Actually, $1$ is an eigenvalue, with eigenvector $(1,-1,0)$ -- but your reasoning for finding the eigenvalues is strange. Neither of the three polynomials in your long formula seems to have $1$ as a root, so they can't be the characteristic polynomial. $\endgroup$ – Henning Makholm May 18 '15 at 10:51
  • $\begingroup$ @BioShock: Your computations must be in error: wolframalpha.com/input/… $\endgroup$ – Henning Makholm May 18 '15 at 10:53
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Everything is in the passage you quoted. THe matrix $A$ has three distinct real eigenvalues, hence it is diagonalisable: $A=PDP^{-1}$, where $D$ is a digonal matrix with eigenvalues of $A$ and $P$ is a matrix whose columns are eigenvectors of $A$.

In your case it easy to find these eigenvectors: $(1,-1,0)^T$ corresponds to $\lambda_1=1$, $(1,1,0)^T$ corresponds to $\lambda_2=0.2$ and $(0,0,1)^T$ corresponds to $\lambda_3=0.5$. In other words, $D=\begin{pmatrix}1&0&0\\ 0&0.2&0\\ 0&0&0.5\end{pmatrix}$ and $P=\begin{pmatrix}1&1&0\\ -1&1&0\\ 0&0&1\end{pmatrix}$.

Now we can say that $$A^N=PD^NP^{-1} = P\begin{pmatrix}1^N&0&0\\ 0&0.2^N&0\\ 0&0&0.5^N\end{pmatrix}P^{-1}\to P\begin{pmatrix}1&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}P^{-1}$$ $$=\frac 12\begin{pmatrix}1&1&0\\ -1&1&0\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\ 0&0&0\\ 0&0&0\end{pmatrix}\begin{pmatrix}1&-1&0\\ 1&1&0\\ 0&0&1\end{pmatrix}=\frac 12 \begin{pmatrix}1&-1&0\\ -1&1&0\\ 0&0&0\end{pmatrix}$$

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Doing the disorganization, you can find an invertible matrix $P$ s.t. $A=P^{-1} D P$, where $D=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0.5 & 0 \\ 0 & 0 & 0.2 \end{array} \right) $.

Then $A^p = P^{-1} D^p P = P^{-1} \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right) P$. Then you should be able to compute $A^{p}$.

By the way, if all absolute values of eigenvalues are strictly smaller than $1$, that is the spectral radius $< 1$, $A^p \rightarrow 0$. If I am not wrong, this is a iff. That is spectral radius $<1$ iff $A^p \rightarrow 0$

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