General Information

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Project Description

Let H be the set of quaternions. The quaternions can be thought of as a four-dimensional vector space over the real numbers with the basis {1,i,j,k} where
i² = j² = k² = -1
ij = -ji = l
jk = -kj = i
ki = -ik = j
The quaternions form a division ring. That is, the quaternions satisfy all of the axioms for a field except that multiplication is not commutative.

The Fundamental Theorem of Algebra states that an nth-degree polynomial has exactly n solutions (counting multiplicities) over the complex numbers. This is not true over the quaternions. A quadratic equation over the quaternions has 1, 2, or infinitely many solutions. We can also consider polynomial equations over matrices. Let M₂(C) be the set of 2 by 2 matrices over the complex numbers.The number of solutions of a quadratic over M₂(C) is either 0, 1, 2, 3, 4, 5, 6, or infinitely many. Let M₂(H) be the set of 2 by 2 matrices over the quaternions. It is not true that the number of solutions a quadratic has over M₂(C) and over M₂(H) is the same. For example, x²-diag{-1,-4}=0 has no solutions over M₂(C) but has infinitely many solutions over M₂(H). My goal this summer is to investigate the number of solutions of quadratic equations over M₂(H).

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Weekly Log

Week 1-2:

The past two weeks, I read over notes that Professor Wilson gave me on the number of solutions of quadratic equations over M₂(C) , and a paper by Niven "Equations in Quaternions" that proves a quadratic equation over the quaternions has 1, 2 or infinitely many solutions. I also prepared and gave a ten-minute presentation on my project.

I looked at quadratic equations over the quaternions with real coefficients. I found that such an equation, if it has a real solution, has no non-real quaternion solutions, and if it does not have a real solution, it has infinitely many non-real quaternion solutions. I also tried to work out how many solutions X²-diag{1,4}=0 has over M₂(H), but I was unable to get an answer.

Week 3:

I met with my mentor, Professor Wilson, this week, and we were able to find a proof that X²-diag{1,4}=0 has exactly 4 solutions. Unfortunately, the proof is very specific to this example, but it does show that quadratices over M₂(H) can have a finite number of solutions.

Professor Wilson and I discussed a method to find diagonalizable solutions to quadratics. Any quaternion can be represented as a 2 by 2 matrix over the complexes. Therefore, if we have a 2 by 2 matrix over the quaternions, we can look at this as a 4 by 4 matrix over the complexes. From Professor Wilson's notes, we know how to find the diagonalizable solutions to quadratics over matrices over the complexes. Therefore, we can take a quadratic over M₂(H), look at it as a quadratic over M₄(C) and find our diagonalizable solutions. Some of these solutions will not have the right form to be a solution for the equation over the quaternions, but any solution to the quaternions will correspond to a solution over the complexes.

Suppose our quadratic over M₂(H) is X²+AX+B=0, where A and B are 2 by 2 matrices with corresponding coefficients. Let A' and B' be 4 by 4 matrices over the complexes such that every 2 by 2 block corresponds to the quaternion entry in A or B. Then any solutions to X²+AX+B=0 can be represented as solutions to X²+A'X+B'=0. From Professor Wilson's notes, we know that this equation has seventy or less solutions if det[t²+tA+B] has no double roots. Thus, the number of solutions to X²+AX+B=0 is finite if det[t²+tA+B] has no double roots. We tried a few examples to see how this works out. Our examples mainly involved quaternion coefficients that were real, and we found that we also got repeated roots. I plan to look into mathematica and try more random examples. It would be nice to find conditions on A and B such that det[t²+tA+B]=0.

Week 4:

Given an equation X²+AX+B=0, we can easily find the corresponding matrix A' and B', and therefore, the det[t²+tA+B]. However, once we have this determinant, it is difficult to figure out if it has repeated roots or not. Professor Wilson explained the concept of a resultant to me, and how to calculate it for two arbitrary matrices. A polynomial has a repeated root if and only if it has a root in common with it's derivative. Therefore, taking the resultant of our determinant and it's derivative will tell us if it has a repeated root. It would be nice to use this method to find conditions on A and B such that the corresponding determinant does not have repeated roots, and hence, the number of solutions to the quadratic is finite. The only way I could think to approach this is to symbolically represent A and B, and therefore, A' and B' and find the corresponding determinant and resultant. In order to do this, I spent a lot of time this week learning mathematica since I have not used it before, and it seems better suited to symbolic computation than mathematica. I used it to calculate det[t²+tA+B]. Unfortunately, the results were not encouraging. The determinant was about half a page long, and therefore, too long to analyze. Mathematica ran out of memory trying to compute the resultant.

Since this approach didn't work, Professor Wilson and I decided to focus on number of solutions when det[t²+tA+B] has no roots. I used mathematica to create several random quadratics over M₂(H), convert it to an equation over M₄(C), and use the method described in Professor Wilson's notes to find the diagonalizable solutions. Because I focused on the case where the determinant has no repeated roots, all of the solutions to X²+A'X+B'=0 are diagonalizable, so this method found all of the solutions. I found that in the examples I tried, none of the solutions are in a form corresponding to a 2 by 2 matrix over the quaternions, and thus, X²+AX+B=0 has no solutions. However, I was not entirely careful with the program I wrote for mathematica, so this result is likely (and hopefully) to be incorrect.

Professor Wilson and I also discussed whether there was a way to relate the eigenvectors for a 2 by 2 matrix over the quaternions and the corresponding 4 by 4 matrix over the complexes. We were not sucessful in finding a relation and plan to continue trying to find one next week.

Week 5:

I re-checked my work and found that I had made a mistake in the mathematica program. After correcting it, I found that the random cases had 6 quaternion solutions, each with four eigenvalues that are two pairs of complex conjugate pairs. We were able to prove that the roots of det[t²+At+B]=0 always occur in complex conjugate pairs, and that quaternions solutions have four eigenvalues that consist of two sets of conjugate pairs.

Week 6:

We found a necessary and sufficient condition for a 4 by 4 complex matrix to correspond to a 2 by 2 quaternion matrix. X corresponds in this way iff Conjugate[X] = diag{1,-1,1,-1}*X*diag{1,-1,1,-1}. Additionally, we proved that real roots of the roots of det[t²+At+B]=0 always have mulitplicity greater than 1, and that a 4 by 4 matrix representing a quaternion solution must have eigenvalues that consist of two pairs of complex conjugates, even if the eigenvalues are real. In short, we have basically finished that case where det[t²+At+B]=0 has distinct roots. We began focusing on the case where this equation does not have distinct roots by working out examples that we have done over complex matrix. We found one example in which there is four solutions, the case where X²=diag{1,4}, which we had previously worked out by other methods.

Week 7:

We continued with the case where the roots of det[t²+At+B]=0 are not distinct. To do this, we began by look at equations over H, and treating them as equations over M₂(C). This case seems to be like the case we worked out except for one large problem, which is that there has not been any comprehensive solution to how to find all the solutions to a quadratic over M₂(C).

I prepared and gave my final presentation.

Week 8:

I met with my mentor to review and fix up all of the proofs we've done over the summer. I wrote my final report and met with my mentor to go over it and edit it.

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Presentations

• First Presentation

• Second Presentation

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Additional Information

• My Mentors
  • Dr. Robert Wilson