The goal of this note is to give a proof of the following result:
Theorem 1 Let
be an odd dimensional vector space over
and let
be linear. Then
admits an eigenvector.
The proof that follows is in the spirit of Axler’s textbook, so it avoids the use of determinants. However, I feel it is easier than the argument in the book, and it has the additional advantage of not depending on the fundamental theorem of algebra. In fact, the motivation for finding this argument was to avoid the use of the fundamental theorem.
The proof we present of Theorem 1 can be seen as an elaboration of the argument in the case when 
Posted by andrescaicedo 
does not require the fundamental theorem, and we can extend this to operators on 
, again avoiding the theorem, because we have explicit formulas that allow us to factor a quartic into the product of two quadratics. Can we find an argument for operators on 
?)
such that 
For example, 
an integer 
such that 
For example, 
Note that, although 
we are not giving a fixed value of 
For example: 
Note that, although 
such that 
(since 
(since 
or 
But this is impossible, since 1 is not divisible by 2. We have reached a contradiction, and therefore our assumption that there is such an integer ![\forall x\,([{\tt Even}(x)\land(x>2)]\Rightarrow\exists p\,\exists q\,[x=p+q\,\land](https://s0.wp.com/latex.php?latex=%5Cforall+x%5C%2C%28%5B%7B%5Ctt+Even%7D%28x%29%5Cland%28x%3E2%29%5D%5CRightarrow%5Cexists+p%5C%2C%5Cexists+q%5C%2C%5Bx%3Dp%2Bq%5C%2C%5Cland&bg=ffffff&fg=333333&s=0&c=20201002)
![{\tt Prime}(p)\land{\tt Prime}(q)]).](https://s0.wp.com/latex.php?latex=%7B%5Ctt+Prime%7D%28p%29%5Cland%7B%5Ctt+Prime%7D%28q%29%5D%29.&bg=ffffff&fg=333333&s=0&c=20201002)
is the formula asserting that 
and 
is the formula (given in the quiz) asserting that 
is prime. Note we had to add existential quantifiers in order to be able to refer to the two prime numbers that add up to 
such that 
is also prime. 
only allows us to mention one integer at a time, and writing something of infinite length such as 
is not allowed.
there is a larger 
be a linear map between finite dimensional vector spaces over 
For 
we denote by ![T^{-1}[S]](https://s0.wp.com/latex.php?latex=T%5E%7B-1%7D%5BS%5D&bg=ffffff&fg=333333&s=0&c=20201002)
the preimage of 
under 
This is the set of all vectors 
such that 
of a vector space 
over 
is convex iff whenever 
and ![\alpha\in[0,1],](https://s0.wp.com/latex.php?latex=%5Calpha%5Cin%5B0%2C1%5D%2C&bg=ffffff&fg=333333&s=0&c=20201002)
then also 
Show that 
is convex. Is it necessary that 
is linear for this to hold, or does it suffice that 
and show that 
is positive.
if and only if it has property 
we need an integer 
then 
i.e., 
is a counterexample. (And, in fact, it is the only counterexample.)
where 
and 
iff any element of 
Let us then consider an element 
By definition of 
this means that 
We need to show that 
i.e., that 
and that 
The first requirement is automatic, so we only need to verify that 
such that 
To show that 
we need (again, by definition) that there is an integer 
such that 
Now, since 
then 
or 
We immediately see that we can take 
This is an integer, and 
as needed.
then 
as shown in lecture. In particular, if 
, then 
Now let 
Then 
and 
then 
Note that 
so we can take 
and we see that 
where 
Then 
By definition, this means that 
are integers and we have that 
and 
then also 
By definition, this means that there are integers 
and 
Then 
But 
Let 
Then 
is an integer, and 
By definition, this means that 
is composite.
is a positive integer. However, 
is not composite according to the definition given in the book: 
and 
then 
Now, if 
and 
so 
Also, since 
then 
and therefore 
We have shown that 
and that 
By definition, this means that 
of subsets of 
is called an independent family iff for any pairwise distinct 
in 
is infinite.
of infinite sequences of zeros and ones.)
recall that its characteristic function is the map 
such that 
if 
and 
if 
We can think of each 
as a vector in the space 
Suppose now that 
is a linearly independent set. (Thus any basis for 
over 
and show that any basis must have the same size as the reals.
satisfying the axioms listed in pages 2, 3 of the textbook as properties of 
see also 
and Nim-multiplication 
on 
as follows:
is the result of adding without carrying the binary expansions of 
and 
. For example, 
is computed by applying the following rules:
we have 
and 
for 
is a field. 
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