403/503 – Eigenvectors for operators on real odd dimensional spaces

February 28, 2010

The goal of this note is to give a proof of the following result:

Theorem 1 Let {V} be an odd dimensional vector space over {{\mathbb R},} and let {T:V\rightarrow V} be linear. Then {T} 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 {{\rm dim}(V)=3,} that we discussed in lecture. It was found by David Milovich in Facebook.

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403/503 – The fundamental theorem of algebra via linear algebra

February 26, 2010

The argument we gave in class for the existence of eigenvectors for operators on finite dimensional complex vector spaces (and for the existence of invariant planes for operators on finite dimensional real vector spaces) uses the fundamental theorem of algebra. One can actually prove the existence of eigenvectors without appealing to this result, although the argument is more complicated.

As a corollary, one obtains a linear algebra proof of the fundamental theorem of algebra, which seems like a nice outcome.

The details can be found in a nice paper by Harm Derksen, currently available through his website or in JSTOR (American Mathematical Monthly, Vol. 110 (7) (2003), 620-623). A variation of the proof (perhaps more accessible) is in this paper by Keith Conrad, currently available through his website. 

There is a slight disadvantage to both papers (which is perhaps the reason why I am not presenting their result in class) if we want to follow the approach of the textbook, and avoid introducing determinants at this stage. The problem is Corollary 4 in Conrad’s paper or Lemma 4 in Derksen’s, that operators on odd dimensional real vector spaces admit eigenvectors. Their proofs use determinants. The proof we gave (or are in the midst of giving) in lecture avoids determinants, but of course uses the fundamental theorem (so we can find an invariant plane and then argue by induction). 

Can you find a way of obtaining this result without appealing to either determinants or the fundamental theorem, so we have a proof of the existence of eigenvectors compatible with the philosophy of the textbook and entirely self-contained?

(Note that an odd degree polynomial with real coefficients has a real root, and this can be proved very easily. From this, the argument for operators on {\mathbb R}^3 does not require the fundamental theorem, and we can extend this to operators on {\mathbb R}^5, 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 {\mathbb R}^7?)


187 – Quiz 4

February 26, 2010

Here is quiz 4. 

Problem 1 asks to determine (with brief justifications) the truth value of the following statements about integers:

  1. \forall x\,\forall y\,(x>y).
  2. \exists x\,\forall y\,(x>y).
  3. \forall x\,\exists y\,(x>y).
  4. \exists x\,\exists y\,(x>y).

1. is False. To show this we provide a counterexample: Specific integers x,y such that x\not>y. For example, 1\not>23.

2. is False. To show this we need to exhibit for each integer x an integer y such that x\not> y. For example, x\not>x+1. Note that, although y is a fixed integer once we know x, we are not giving a fixed value of y that serves as a simultaneous counterexample for all values of x.

3. is True. To show this we exhibit for each integer x a specific integer y such that x>y. For example: x>x-1. Note that, although y is a fixed integer once we know x, we are not giving a fixed value of y that works simultaneously for all x.

4. is True. To show this, we exhibit specific values of x,y such that x>y. For example: 1777>-52451256.

Problem 2 asks to show by contradiction that no integer can be both odd and even. Here is the proof: Suppose otherwise, i.e., there is an integer, let’s call it x, such that x is both odd and even. This means that there are integers y,z such that x=2y+1 (since x is odd) and x=2z (since x is even). 

Then we have that 2y+1=2z, or 1=2(z-y). 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 x ought to be false. This means that no integer can be both odd and even, which is what we wanted to show.

Note that we have not shown that every integer is either odd or even. We will use mathematical induction to do this.

Problem 3 asks for symbolic formulas stating Goldbach’s conjecture and the twin primes conjecture (both are famous open problems in number theory).

Goldbach’s conjecture asserts that every even integer larger than 2 is sum of two primes:

\forall x\,([{\tt Even}(x)\land(x>2)]\Rightarrow\exists p\,\exists q\,[x=p+q\,\land {\tt Prime}(p)\land{\tt Prime}(q)]).

Here, {\tt Even}(x) is the formula asserting that x is even, namely, \exists y\in{\mathbb Z}\,(x=2y), and {\tt Prime}(n) is the formula (given in the quiz) asserting that n 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 x.

The twin primes conjecture asserts that there are infinitely many primes p such that p+2 is also prime. 

The difficulty here is in saying “there are infinitely many,” since the quantifier \exists only allows us to mention one integer at a time, and writing something of infinite length such as \exists x_1\,\exists x_2\,\exists x_3,\dots is not allowed.

We follow the suggestion given in the quiz, and represent “there are infinitely many n with [some property]” by saying “for all m there is a larger n with [some property].”

\forall m\in{\mathbb Z}\exists n\,(n>m\land {\tt Prime}(n)\land {\tt Prime}(n+2)).


403/503 – Homework 3

February 26, 2010

This homework is due Friday, March 5.

Solve at least 10 of the following problems from the textbook: Chapter 4: 2, 4. Chapter 5: 1, 2, 6, 7, 8, 14, 15, 20, 21, 22, 23, 24.


187 – Midterm 1

February 18, 2010

Here is the midterm.

Solutions follow.

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403/503 – Homework 2

February 15, 2010

This homework is due Wednesday, February 24.

Solve the following problems from Chapter 3 of the textbook: 2, 4, 5, 7, 9, 10, 26. 

In addition, solve the following problem:

Let T:V\to W be a linear map between finite dimensional vector spaces over {\mathbb F}={\mathbb R}. For S\subseteq W, we denote by T^{-1}[S] the preimage of S under T. This is the set of all vectors v\in V such that Tv\in S. 

Recall that a subset C of a vector space U over {\mathbb R} is convex iff whenever u_1,u_2\in C and \alpha\in[0,1], then also \alpha u_1+(1-\alpha)u_2\in C. Show that T^{-1}[S] is convex iff S\cap{\rm ran}(T) is convex. Is it necessary that T is linear for this to hold, or does it suffice that T is continuous?

Assume that S is a subspace of W, and show that T^{-1}[S] is a subspace of V.


187 – Quiz 3

February 15, 2010

Here is quiz 3. 

Problem 1 requests a counterexample to the statement: An integer x is positive if and only if x+1 is positive.

To find a counterexample to a statement of the form “an integer x has property A if and only if it has property B, we need an integer x such that one of the properties is true of x while the other is false. Note that if x is positive, i.e., x>0, then x+1>1>0, i.e., x+1 is also positive. This means that the only way the statement is going to fail is if we have that x+1 is positive yet x is not. 

We can now easily see that x=0 is a counterexample. (And, in fact, it is the only counterexample.)

Problem 2 asks to show that C\subseteq D, where C=\{x\in{\mathbb Z}\colon x|12\} and D=\{x\in{\mathbb Z}\colon x|60\}. 

By definition, C\subseteq D iff any element of C is also an element of D. Let us then consider an element x of C. By definition of C, this means that x is an integer, and x|12. We need to show that x\in D, i.e., that x\in{\mathbb Z} and that x|60. The first requirement is automatic, so we only need to verify that x|60. 

To see this, we use that x|12. By definition, this means that there is an integer, let us call it a, such that xa=12. To show that x|60, we need (again, by definition) that there is an integer b such that xb=60. Now, since xa=12, then 5(xa)=5\times 12=60, or x(5a)=60. We immediately see that we can take b=5a. This is an integer, and xb=60, as needed.

Problem 3 asks for the cardinality of the set 2^{2^{\{1,3,7,11\}}}. 

Recall that for any set A, then |2^A|=2^{|A|}, as shown in lecture. In particular, if A=\{1,3,7,11\}, then |2^A|=2^4=16. Now let A'=2^{\{1,3,7,11\}}=2^A. Then |2^{A'}|=2^{|A'|}=2^{16}=65536.