# Why are these functions commutatives?

1. Feb 16, 2013

### SqueeSpleen

I'm finishing studying for my Linear Algebra final, but there are two things I didn't understand (I'll make one thread per thing).
(It's a translation, the original one is in Spanish, second half of the page 174/184 book notation/acrobat reader notation if you're curious).
Let be V a K vector space with finite dimension. Let be f:V→V a linear map, with a $m_{f}$=P.Q and (P,Q)=1
Then...
(...)
$V = Nu(P(f)) \oplus Nu(Q(f))$
And one step of the proof (the only one I don't understand) say:
$x = (R(f) \circ P(f) )(x)+(S(f) \circ Q(f) )(x)$
(That's for the extended Euclid's algorithm for polynomials I think, this is not the problem).
The problem is just here:
"Now, having in account that:"
$Q(f) \circ R(f) = (Q.R)(f) = (R.Q)(f) = R(f) \circ Q(f)$

And I don't know why.
I understand that, if the composition is equal to the product, then the commutativity is obvious because the product of two polynomials is commutative. But why the composition is equal to the composition in this case? It must be related to extended Euclid's algorithm for polynomials but I think I never had it in a course =/
Sorry for my bad English, I learned most of it while gaming xD

Last edited: Feb 16, 2013
2. Feb 16, 2013

### micromass

Staff Emeritus
I have no idea what notations you are using, but I think it's clear to me what they mean.

For example, let $P(X)=X^3+X^2$ and $Q(X)=X^2+3X$.

Then $P(f)=f^3 + f^2$ and $Q(f)=f^3+3f$. And we define $f^2=f\circ f$ and $f^3=f\circ f\circ f$.

Then

$$(P\cdot Q)(X) = (X^3 + X^2)\cdot (X^2 + 3X) = X^5 + 4X^4 + 3X^3$$

So

$$(P\cdot Q)(f)=f^5 + 4f^4 + 3f^3$$

but

$$P(f)\circ Q(f) = (f^3 + f^2)\circ (f^2 + 3f) = f^3\circ f^2 + 3f^3\circ f + f^2\circ f^2 + 3f^2\circ f = f^5 + 4f^4 + 3f^3$$

Do you understand this?

3. Feb 16, 2013

### SqueeSpleen

I think it has to be composition because it's what follows:

$Q(f)((R(f) \circ P(f))(x)) = (Q(f) \circ R(f) \circ P(f))(x) = R(f)(Q(f) \circ P(f))(x) = R(f)(m_{f} (f)(x)) = R(f)(0) = 0;$
Then R(f).P(f) is in Q(f) kernel, beacuse Q(R(f).P(f))=0 (the idea is to prove that the sum of the kernels is equal to V).
The thing I don't understand if how Q(f) evaluated on R(f).P(f) is equal to R(f) evaluated on Q(f).P(f).
PD: Nu(Something) is the Kernel of something, I forget to translate it.
Probably I'm misunderstading something of the notation... again notation is my worst enemy.

Edit:
T_T Q(f) is a polynomial of a lineal transformation, so it can be represented as a matrix, then to evaluate it is to multiplicate it to the right for a vector... so yes they are all products, and then f(x)=f.x and the commutativity is obvious T_T I spend a couple of hours reading this, I don't know how I got SO stuck.