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I'm trying to learn how to show a series of matrices is convergent and find the limit. However my numerical analysis books fail to mention this and I cannot find any relevant material online! Anyway say I had

$\sum_{k=0}^ \infty A^K$, where A = $\begin{bmatrix}\frac{1}{2} & -2 & -1\\0 & \frac{1}{3} & 0\\ 0 & 0 & -\frac{1}{2}\end{bmatrix}$

How would i show that the series of matrrices converges and compute its limit?

Would I use some kind of induction? Some help would be amazing, many thanks!

user67411
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3 Answers3

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Hints:

  1. How would you first write the matrix power $A^K$?

  2. Once you have an expression for $A^K$, you can take the sum to each $a_{i,j}$ in your 3x3. Of course four of the terms are immediately known! Also, two of terms in the diagonal are the same, so you would be finding the sum of four expressions that are sums over $k$.

Update

So, here our sum is:

$$\displaystyle \sum_{k=0}^\infty A^k = \sum_{k=0}^\infty \begin{bmatrix}-\frac{1}{2} & -2 & -1\\0 & \frac{1}{3} & 0\\ 0 & 0 & -\frac{1}{2}\end{bmatrix}^k = \sum_{k=0}^\infty \begin{bmatrix}(-1/2)^k & \left(\frac{4}{5}\right) 3^{1-k} ((-3/2)^k-1) & (-1)^k 2^{1-k} k \\ 0 & 3^{-k} & 0 \\ 0 & 0 & (-1/2)^k)\end{bmatrix}$$

Now, how do you evaluate each of these sums and do they each converge?

Update 2

You should get:

$$\displaystyle \sum_{k=0}^\infty A^k = \begin{bmatrix}\frac{2}{3} & -2 & -\frac{4}{9}\\0 & \frac{3}{2} & 0\\0 & 0 & \frac{2}{3}\end{bmatrix}$$

I would also recommend taking this answer and seeing how the other answers apply.

Amzoti
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Choose any algebra norm $||.||$ and if $||A||<1$ then the geometric series $$\sum_{k=0}^\infty ||A||^k$$ is convergent so the series $\displaystyle\sum_{k=0}^\infty A^k$ is normal convergent and then it's convergent.

One possible method to calulate the limit: let $\pi_A$ the minimal polynomial of $A$ and by the Euclidean division we have $$x^k=Q(x)\pi_A(x)+R_k(x)$$ with $\deg(R_k)<\deg(\pi_A)$ and we have $$\sum_{k=0}^\infty A^k=\sum_{k=0}^\infty R_k(A)$$ and the last sum is more simple to calculate.

Added In our case, the matrix $A$ is triangular and since $(A+\frac{1}{2}I_3)(A-\frac{1}{3}I_3)\neq0$ so we can see easly that the minimal polynomial is $$\pi_A(x)=\frac{1}{12}(3x-1)(2x+1)^2$$ hence in the Euclidean division the remainder polynomial is $$R_k(x)=a_k x^2+b_k x+c_k$$ and we have $$\begin{align}\\ \left(\frac{1}{3}\right)^k&=a_k/9+b_k/3+c_k\\ \left(\frac{1}{2}\right)^k&=a_k/4-b_k/2+c_k\\ k\left(\frac{-1}{2}\right)^{k-1}&=-a_k+b_k\\ \end{align}$$ so we solve the above system for $a_k,b_k$ and $c_k$ and finally we find $$\sum_{k=0}^\infty A^k=\left(\sum_{k=0}^\infty a_k\right)A^2+\left(\sum_{k=0}^\infty b_k\right)A+\left(\sum_{k=0}^\infty c_k\right)I_3$$

  • but i litterally have no idea about convergent matrices, i have done convergence in analysis with normal series and sequences and can prove a series or sequence is convergent using an epsilon proof, but i really dont get this at all, but only as i have no reference for it! could you please suggest a page that may have the desired information? – user67411 Apr 18 '13 at 00:19
  • You can see http://en.wikipedia.org/wiki/Series_(mathematics) and particulary the item series in banach spaces. –  Apr 18 '13 at 00:41
  • im using this http://en.wikipedia.org/wiki/Convergent_matrix as a guide but i am getting a different value for the limit of each series in the matrice! should i not be doing it using this? – user67411 Apr 18 '13 at 01:07
  • @user67411 I added the method of calculus of the sum and I think you can complete all necessary calculus. –  Apr 18 '13 at 01:45
  • okay i think i get it now! the limits of all of the k series is 0 thus convergent and do i need to find the eigen values also? thank you! – user67411 Apr 18 '13 at 02:00
  • I think you've said minimal polynomial when you meant chracteristic polynomial. Otherwise, you need a little extra argument to deduce it from simply looking at the diagonal. Note: I know that minimal=characteristic polynomial in this special case. – Julien Apr 18 '13 at 02:00
  • @julien I know that the use of characteristic polynomial gives the same result but in general the use of minimal polynomial is better thanks to its degree which's generally less than $chi_A$, and obviously I didn't deduce it from simply looking at the diagonal but also $(A+\frac{1}{2}I)(A-\frac{1}{3}I)\neq0$ –  Apr 18 '13 at 02:13
  • I know that you know and that you did not deduce it like that, of course. If you really meant minimal, it would probably be good to add that $(A+1/2)(A-1/3)\neq 0$ in your argument for the sake of the OP. – Julien Apr 18 '13 at 03:04
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A special solution to this problem:

You may write $A = D+N$, where $D$ is diagonal and $N$ is nilpotent in an obvious way. Then it is easy to see (either by a direct calculation or by the celebrated Cayley-Hamilton theorem) that $N^3 = O$ and thus we can simply $A^k$ as

$$ A^k = D^k + kD^{k-1}N + \frac{k(k-1)}{2}D^{k-2}N^2. $$

Now both the convergence and the summation value can be easily analyzed from the knowledge on power series in $\Bbb{R}$.

Sangchul Lee
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  • $D$ and $N$ do not commute, if by $N$ you meant what I think you meant. – Julien Apr 18 '13 at 01:54
  • @julien, you're right. I was careless. Maybe we can use Cayley-Hamilton theorem to establish a 3rd order linear homogeneous recurrence relation of $(A^k)$ to obtain a formula for $A^k$, though the solution now becomes much complex. – Sangchul Lee Apr 18 '13 at 04:24
  • That's what Sami Ben Romdhane did. It is a pretty good way, indeed. – Julien Apr 18 '13 at 05:02