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Let $n$ be a positive integer. Let $A$ be a square matrix. Let $I$ be the identity matrix with the same size as $A$.

I want to simplify $f_n(A) = I + A + A^2 + A^3 + A^4 + \cdots + A^n$

Now I know that for a complex number $z$ we have $1 + z + z^2 + z^3 + z^4 + \cdots + z^n= \dfrac{z^{n+1}-1}{z-1}$ when $z$ is not equal to $1$ ... and even if $z=1$ then by computing the limit we get the right answer : $n+1$. This is well known as the geometric series and leads to the so-called q-analogue ideas.

But for matrices $A$ this does not work if the determinant of $A-I$ is $0$ , because then $\dfrac{A^{n+1} - I}{A - I}$ is not well defined.

This frustrates me a lot. Now I have heard about pseudoinverses but I do not know much about them and I even wonder if they can be of any help here.

I tried some things from calculus and continued fractions too, but nothing worked for me.

I also considered this problem for infinite (square) matrices but I assume that is analogue to this problem and so is its solution.

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The fact that a matrix doesn't have an inverse does not rule out the possibility of dividing by it; it just means only some matrices can be divided by it. Whether it will work out right is another question. –  dfeuer Dec 12 '13 at 23:17
    
If $A$ is diagonalizable, it will be easy to compute $f_n(A)$. Otherwise, you can still Jordanize it, but it will be more tedious. –  1015 Dec 12 '13 at 23:18
    
@julien Intresting. But that will fail for infinite matrices most of the time not ? –  mick Dec 12 '13 at 23:19
    
Infinite matrices? Then you want to be more specific. Are you interested in, say, $B(H)$, bounded linear operators on Hilbert space (these are well-behaved infinite matrices). Then you want your operator $A$ to be normal to do that nicely with the help of the spectral theorem. –  1015 Dec 12 '13 at 23:21
    
@julien I was thinking about carleman matrices. –  mick Dec 12 '13 at 23:26

1 Answer 1

$\displaystyle\sum_{k=0}^\infty A^k=(I-A)^{-1}$ provided each eigenvalue $\lambda_i$ of $A$ satisfies $|\lambda_i|<1$.

The argument for this is analogous to the scalar case: If the eigenvalue condition above holds, then $I-A$ is invertible and $$ S_n:=\sum_{k=0}^{n-1} A^k=(I-A)^{-1}(I-A^n). $$ Thus, $$\sum_{k=0}^\infty A^k=\lim_{n\to\infty}S_n=(I-A)^{-1}, $$ since the eigenvalue condition guarantees $\lim_{n\to\infty} A^n=0$.

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Isn't the OP specifically asking about the case where $I-A$ has no inverse? –  dfeuer Dec 12 '13 at 23:24
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Notice Im intrested in finite sums , not the infinite case. Thanks anyway , intresting stuff. –  mick Dec 12 '13 at 23:28
    
Ok, perhaps I misunderstood where the question lay in the OP: I focused on the "simplify" part and when he wrote a quotient of matrices I thought (s)he was trying to make sense of that expression. Apologies. If you all think my reply muddies the water unnecessarily, I am happy to delete. –  JohnD Dec 12 '13 at 23:30
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I think deletion is appropriate. –  dfeuer Dec 12 '13 at 23:34
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I think it should not be deleted. If the eigenvalues are not less than 1 (in abs terms) there is no point of talking about geometric series (infinite many powers) in the first place. It is a needed prereq to find a sum of infinite powers of a matrix –  imranfat Dec 13 '13 at 1:54

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