An endomorphism maps a vector space onto itself.
\(end (V)=\hom (V, V)\)
An endomorphism maps a vector space onto itself.
\(end (V)=\hom (V, V)\)
Need to show that endomorphism is a vector space
Essentially
\(v\in V\)
\(f\in F\)
\(av = f\)
\(bv = g\)
\((a\oplus b)v=f+g\)
\((a\oplus b)v=av + bv\)
so there is some operation we can do on two members of endo
linear in addition. That is, if we have two dual "things", we can define the addition of functions as the operation which results int he outputs being added.
what about linear in scalar? same approach.
Well we define
\(c\odot a)=cav\)
There is a unique endomorphism which results in two other endomorphisms being added together. define this as addition
\(\dim (end(V))=(\dim V )^2\)
A projection is a linear map which if applied again returns the original result.
A projection can drop a dimension for example.
The kernel of a linear operator is the set of vectors such that:
\(Mv=0\)
The kernel is also called the nullspace.
This can be shown as \(\ker (M)\)
The image of a linear operator is the set of vectors \(w\) such that:
\(Mv=w\).
This can be shown as \(\Im (M)\)
We also know that:
\(span (M)=\ker (M)+\Im (M)\)
We previously discsussed morphisms on vector spaces. We can write these as matrices.
Matrices represents transformations of vector spaces
We can represent vectors as row or column matrices.
\(v=\begin{bmatrix}a_{1} & a_{2}&...&a_{n}\end{bmatrix}\)
\(v=\begin{bmatrix}a_{1}\\a_{2}\\...\\a_{m}\end{bmatrix}\)
For any two bases, there is a unique linear mapping from of the element vectors to the other.
The general linear group, \(GL(n, F)\), contains all \(n\odot n\) invertible matrices \(M\) over field \(F\).
The binary operation is multiplication.
We can view each member of the group \(g\) as a homomorphim on \(s\).
Where \(s\) is a vector space \(V\), the representation on each group member is an invertible square matrix.
If the set we use is the vector space \(V\), then we can represent each group element with a square matrix acting on \(V\).
Faithful means \(a\ne b\) holds for repesentation too.
Representation theory. groups defined by \(ab=c\). if we can match each eleemnt to amatrix where this holds we have represented the matrix.
Finite groups can all be represented with square matrices.