# Stable linear transformations under composition

1. Jul 15, 2013

### burritoloco

Hi,

Let $f$ be a linear transformation over some finite field, and denote $f^{n} := f \circ f \circ \cdots \circ f$, $n$ times. What do we know about the linear maps $f$ such that there exist an integer $n$ for which $f^{N} = f^n$ for all $N \geq n$? Also, how about linear maps $g$ satisfying $g = g \circ f^i$ for any $i\geq 0$? Something tells me that I've seen this before in my undergrad years but my memory is very vague on this. Thanks!

2. Jul 15, 2013

### burritoloco

Remark: In the 2nd question, f doesn't have to satisfy the property in the 1st question.

3. Jul 15, 2013

### lurflurf

That does not have to do with finite fields particularly. Both cases are related. The transformation decomposes the vector space.
Suppose if g=gf
then we can write x=y+z where
gfy=0
gfz=gz
in other words f can change x to another vector with the same value under g
f thus defines a partition or equivalence class

4. Jul 15, 2013

### burritoloco

Thanks lurflurf. I'm not sure I understand why it defines a partition. But I was particularly interested in knowing whether there is a general formula for the linear maps $g$ satisfying $g = g \circ f^i$, in terms of $f$. For instance, the fact that we have a finite field guarantees that we can find two powers of $f$ that are the same, say$f^a = f^b$ with $a < b$. Then if we let $g = \sum_{j=a}^{b-1} f^j$, it satisfies our property. I am wondering if there exists a more general expression for $g$. Cheers.

Last edited: Jul 16, 2013
5. Jul 16, 2013

### burritoloco

If not a general formula, it would be nice to have as many examples as possible, like the above. OK, I understand what you mean now (after a good night sleep! hehe). Thanks. But as you can see from the previous post I'm more interested in, given f, what are the g's satisfying this? It might be of use to know that the set of powers of f on a finite field forms a finite semigroup and hence has at least one idempotent.

Last edited: Jul 16, 2013
6. Jul 16, 2013

### Office_Shredder

Staff Emeritus
If g = gf then g(I-f) = 0, which means that the image of I-f must lie in the kernel of g. Similarly the image of I-f2, I-f3 etc lie in the kernel of g. So let K be the sum of the images of all I-fk for all k (of which you only need to check finitely many), then g can be any linear transformation such that K lies in the kernel of g.

7. Jul 16, 2013

### burritoloco

Thank you. Now, suppose that we restrict $g$ so that its kernel is the image of $id - f$. One can show that this $g$ also satisfies $g = g \circ f^i$ for each $i \geq 0$. Any ideas whether it would be possible to obtain an expression for $g$ in terms of $f$ or so?

Last edited: Jul 16, 2013
8. Jul 16, 2013

### burritoloco

For instance, the example that I gave for $g$ has the image of $id - f$ contained in its kernel, but I'm not sure that we can show the converse is true in all cases of $f$.