Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I calculate this limit:

$\displaystyle\lim_{n\to\infty}\begin{bmatrix}0.9 & 0.2\\0.1 & 0.8\end{bmatrix}^n$

What is the tool that i need to aply? eigenvalues and eigenvectors? diagonalization? canonical form?

(This came in a contest and was the only problem i cannont have an idea for solve it).

share|improve this question
After diagnolizing the matrix it should be fairly easy to see how to proceed - Raising the matrix to the $n^{th}$ power then becomes very easy and then you can proceed by analyzing the sequences formed within the diagonal matrix. –  DanZimm Jun 22 '14 at 21:25
That matrix is diagonalizable (with distinct (real) eigenvalues). However, one eigenvalue is exactly $1.0$. So this is very "sensitive" to rounding errors. So don't calculate the matrix powers with binaray floating point computer arithmetic. –  Jeppe Stig Nielsen Jun 22 '14 at 21:43
Note that if you have any square matrix $A$ than existence of the limit $L$ determines that $A\cdot L=L$, so if $\det(A−E)\neq 0$ then $L=0$, you can also consider the Cayley–Hamilton theorem.Here matrix $A$ has only $2$ dimensions so, it's possible using a vertex form of the characteristic polynomial and deducing suitable form for any $2\times2$ matrix with remembering that $A^n$ can be expressed as $\sum_{i=0}^n b_i(A−x_0 E)^i$. –  Darius Jun 22 '14 at 23:11

4 Answers 4

Because that is a stochastic matrix you just need to fine the fixed probability vector or steady state vector.

$$(t1,t2)\cdot\left( \begin{array}{cc} 0.9 & 0.1 \\ 0.2 & 0.8 \\ \end{array} \right)=(t1,t2)$$

Notice I have transposed A for convenience and this yields the simultaneous set of equations

$$.9 \cdot t1+.2 \cdot t2=\text{t1}$$

$$.1 \cdot t1+.8 \cdot t2=\text{t2}$$

$$t1 + t2 = 1$$

this is easily solved to get



Because this is a regular Markov chain all the rows of $A^{\infty} $ are equal to $(t1,t2)$, so

$$A^{\infty} = \left( \begin{array}{cc} \frac{2}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{1}{3} \\ \end{array} \right)$$

Transpose A back:

$$A^{\infty} = \left( \begin{array}{cc} \frac{2}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{1}{3} \\ \end{array} \right)$$

share|improve this answer

Diagonalization is precisely the tool you need.

If you can write $A = PDP^{-1}$, where $D$ is a diagonal matrix, then $A^n = PD^nP^{-1}$, where $D^n$ is also diagonal, and its entries are just the $n$-th power of the entries in $D$.

Then, $\displaystyle\lim_{n\to\infty}A^n = \lim_{n\to\infty}PD^nP^{-1}$ is easy to compute.

share|improve this answer
is the limit so $\begin{bmatrix}2/3 & 2/3\\1/3 & 1/3\end{bmatrix}$ ? –  Julio Jun 22 '14 at 21:29
The limit should not have $n$ in it. –  JimmyK4542 Jun 22 '14 at 21:30
sorry, a finger error –  Julio Jun 22 '14 at 21:31
Yes, that is what Wolfram Alpha gets: wolframalpha.com/input/… –  JimmyK4542 Jun 22 '14 at 21:33

Note that your matrix is not an arbitrary matrix --- it is a column stochastic matrix and thus a Markov transition matrix. Hence, from the Perron-Frobenius theorem you will know that each column of the limit matrix will be the normalized eigenvector of your matrix corresponding to the eigenvalue $1$, and as you can check

$$\begin{bmatrix}0.9 & 0.2\\0.1 & 0.8\end{bmatrix}\begin{bmatrix}2/3\\1/3\end{bmatrix}=\begin{bmatrix}2/3\\1/3\end{bmatrix},$$

so that, indeed, $\lim_{n\rightarrow\infty}\begin{bmatrix}0.9 & 0.2\\0.1 & 0.8\end{bmatrix}^n=\begin{bmatrix}2/3 & 2/3\\1/3 & 1/3\end{bmatrix}$.

This methodology applies generally: If you have a column stochastic matrix $B$ with strictly positive entries, its (power) limit exists and each column of $\lim_n B^n$ is given by the normalized eigenvector of $B$ corresponding to the eigenvalue $1$.

share|improve this answer

$$A=\begin{bmatrix}\frac{9}{10} & \frac{2}{10}\\\frac{1}{10} & \frac{8}{10}\end{bmatrix}$$

$$A=PDP^{-1}$$ where $$P=\begin{bmatrix}2 & -1\\1 & 1\end{bmatrix}$$

$$D=\begin{bmatrix}1 & 0\\0 & 7/10\end{bmatrix}$$

So the limit is given by:

$$\lim_{n\to\infty} A^n=\lim_{n\to\infty} PD^nP^{-1}=PD_{\infty}P^{-1}......(1)$$

where $$D_{\infty}=\begin{bmatrix}1 & 0\\0 & \lim_{n\to\infty}(7/10)^n\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}......(2)$$

Substitution of $D_{\infty}$ into (1) leads to:

$$\lim_{n\to\infty} A^n=PD_{\infty}P^{-1}=\begin{bmatrix}\frac23 & \frac23\\\frac13 & \frac13\end{bmatrix}$$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.