A problem with the geometric series and matrices?

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

$\displaystyle\sum_{k=0}^\infty A^k=(I-A)^{-1}$ 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
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
I think deletion is appropriate. – dfeuer Dec 12 '13 at 23:34
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

First... It's nonstandard notation to divide by a matrix. You should write it as $(A-I)^{-1}(A^{n+1}-I)$ or $(A^{n+1}-I)(A-I)^{-1}$; I believe both are valid, it just depends on how you multiply $A$ to the series when you are getting to the simplified form.

Sometimes you will just not be able to solve systems of equations. It can be frustrating, but it actually has an interesting geometric interpretation. Linear algebra is the study of transformations on vectors. Matrix multiplication is simply rotating and scaling a vector, and matrix inversion is a way to reverse that scaling and rotation. Let's consider a simple example, the system of equations

$Mx=b$.

If the matrix $M$ has a nonzero determinant, that means it is invertible. What it really means is that for every single $b$ there is exactly one unique $x$ that will get you there (that will satisfy the equation), which means if you know $M$ and $b$ you can find the original $x$ every time.

If $M$ has a determinant of $0$ it means for a given $b$ there might be more than one $x$ that will get you there. That means if you try to take a direct inverse, it will (and should) fail because even if you know $M$ and $b$ there is no way to know exactly what $x$ is.

Sometimes it might suffice to find a solution for $x$, but you have to remember that it won't be unique. There are a few ways to do this: the Pseudoinverse is one of them, and another involves using optimization methods. If you are interested in learining more about optimization I recommend starting with Boyd's Convex Optimization (the book and lectures are available for free online).

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I need more details. Sorry im no expert. – mick Apr 10 '15 at 15:26