# Universality and Characteristic Polynomials

Today I’m going to describe a simple, yet powerful, trick to deduce specific problems from a “universal” case. The construction is common in algebra; given an object ${O}$ and a property ${P}$ that we would like to deduce about this object, we construct a certain “universal” object ${U}$ such that

1. ${U}$ has property ${P}$;
2. there exists a surjective homomorphism ${\varphi : U \rightarrow O}$ that “preserves” ${P}$.

In order to justify this abstraction, it should be easier to show that ${U}$ satisfies ${P}$ than it is to show that ${O}$ satisfies ${P}$. However, ${U}$ should not be a “trivial” object that vacuously satisfies ${P}$. Instead, ${U}$ should have several “nice” properties that ${O}$ does not have, but it should not have properties that you feel are irrelevant to the current problem. Terry tao discusses a similar trick in analysis (trick 10) in his “problem solving strategies,” post.

One of my favorite examples of this strategy is the following standard qual problem:

Problem 1 Let ${R}$ be a commutative ring with unity. Let ${M}$ and ${N}$ be ${n\times n}$ matrices with entries in ${R}$. Then ${MN}$ and ${NM}$ have the same characteristic polynomial.

The first thing we notice is that if ${M}$ or ${N}$ is invertible, the problem is easy. Indeed, if ${M}$ is invertible, then ${MN = MNMM^{-1}}$ so $\displaystyle \begin{array}{rcl} \det(I X- MN) &=& \det(IX-MNMM^{-1}) \\ &=& \det(MM^{-1}X - M\left(NM\right)M^{-1}) \\ &=& \det(M)\det(IX - NM)\det(M)^{-1} \\ &=& \det(IX - NM). \end{array}$

This observation shows that a possible candidate for a universal space ${U}$, should contain a lot of invertible matrices. This may not be the case for ${R}$ because it may have many zero divisors and few units. Even if ${M}$ has a non-zero determinant, it might not be invertible. Another possible reduction could involve working over a field. Indeed, in this case, a proof involving row and column operations exists, but it is tedious.

A clever way of proceeding, is to consider the following ring: $\displaystyle \begin{array}{rcl} U &:=& \mathbf{Z}[X_{ij}, Y_{kl} : i,j, k, l \in \{1, \cdots, n\}], \end{array}$

where ${X_{ij}}$ and ${Y_{lk}}$ are indeterminates. Define the matrices ${M' = (X_{ij})_{i,j \in \{1, \cdots, n\}}}$ and ${N' = (Y_{kl})_{k, l \in \{1, \cdots, n\}}}$. We consider the the quotient field $\displaystyle \begin{array}{rcl} K = \text{quot}(U) = \mathbf{Q}(X_{ij}, Y_{kl} : i,j, k, l \in \{1, \cdots, n\}) \end{array}$

Observe that over ${K}$, ${\det(M')}$ and ${\det(N')}$ are non-zero. Hence, ${M'}$ is invertible. Our remark before shows that the characteristic polynomial of ${M'N'}$ is the characteristic polynomial of ${N'M'}$. Moreover, the characteristic polynomial of ${M'N'}$ has integer coefficients by the definition of the determinant. Thus, the remark is almost trivially true over ${U}$. Further, we can define a map $\displaystyle \begin{array}{rcl} \varphi : U &\rightarrow& R \\ X_{ij} &\mapsto& a_{ij} \\ Y_{kl} &\mapsto& b_{kl} \\ 1 &\mapsto& 1, \end{array}$

where ${M = (a_{ij})_{i, j \in \{1, \cdots, n\}}}$ and ${N = (b_{kl})_{k, l \in \{1, \cdots, n\}}}$. Because the coefficients of ${P(X)}$, the characteristic polynomial of ${M'N' }$, has coefficients that are polynomials in the entries of ${M'N'}$, with a bit of thought, we see that the characteristic polynomial of ${MN}$ is equal to ${P^\varphi}$. The same statement holds for ${N'M'}$ and ${NM}$. Thus, ${MN}$ and ${NM}$ have the same characteristic polynomial.

Because ${\mathbf{Z}}$ is the initial object in the category of commutative rings with unity, we can always define such a map. If ${R}$ does not contain an identity element, is it possible to apply this proof to that case as well? The map ${\varphi}$ is not so well defined in this case. However, this technique exposes this problem for what it really is, (symbol manipulation), so it appears that the result should continue to hold.