# Applications of the annihilator from linear algebra

I am currently assisting a course for future teachers at university level for a joint education in maths and physics in Germany and I have a question regarding the use and possible application of the following result.

We defined the annihilator for an element A of a K-algebra $$\mathcal A$$ as the set of polynomials over the field K, that return zero, i.e.

$$Ann(A):=\{f\in K[x]: \tilde f(A)=0\}$$

The tilda indicates the evaluation-homomorphism. We proved that any such f can be decomposed into a unique normed polynomial h and some rest, such that every $$f = g\cdot h$$. This can be used to show that a nilpotent matrix of order k, has $$M_A(x)=x^k$$ as minimal polynomial.

My problem: it is an interesting tool to find the minimal polynomial of say a real valued matrix. But I do not see the relevance for future teachers as this seems to be "just" a technical result without any applications that one could explain to someone without extensive mathematical education. I do know, that the subspace of linear functions with the same property is a subspace of the dual space.

So, are there any nice applications for this method in the realm of euclidean geometry? or physics? or maths?

If you feel this is not the right place to ask such an involved question, feel free to suggest an alternative. I have posted this very question to math.stackexchange, too.

• cross-posted: math.stackexchange.com/questions/4185503/…
– user507
Jun 29, 2021 at 0:42
• Does a normed polynomial mean one whose highest-order coefficient is 1? Does the dot mean multiplication (not composition)? I'm a physicist, and the only terms that "annihilator" evokes for me in English are creation and annihilation operators in quantum mechanics. I don't think they're related, unless there is some hidden similarity that isn't obvious to me.
– user507
Jun 29, 2021 at 0:52
• @BenCrowell, the answer to both questions is yes. Jun 30, 2021 at 13:20

The method of annihilators from constant coefficient ODEs is interesting. If $$L = P(D) \in \mathbb{R}[D]$$ where $$D = d/dt$$ and $$deg(P)=n$$ then $$L[y]=0$$ defines an $$n$$-th order homogeneous ODE. In this case, $$Ann(L)$$ is a finite dimensional subspace of the infinite dimensional space of smooth functions on $$\mathbb{R}$$. So, one interesting thing is just that annihilators give a formalism to characterize solution sets as they relate to the structure of an operator. This alone is not terribly interesting, the method of annihilators typically refers to the following mathematical slight of hand: given a nonhomogeneous $$n$$-th order ODE $$L[y] = g$$ we can apply the method of annihilators if there exists $$A \in \mathbb{R}[D]$$ with $$deg(A)=k$$ for which $$A[g]=0$$. Then, $$AL[y] = A[g] = 0$$. By the theory of ODEs, there exists a fundamental solution set for $$AL[y]=0$$ say $$y_1,y_2,\dots, y_n, y_{n+1}, \dots , y_{n+k}$$. Moreover, without loss of generality we may suppose $$y_1,y_2,\dots, y_n$$ is a fundamental solution set of $$L[y]=0$$ since solutions of $$L[y]=0$$ are also solutions of $$AL[y]=0$$. In fact, we can argue $$y=\displaystyle\sum_{j=1}^{n+k} c_jy_j$$ serves as the general solution of $$L[y]=g$$.

For example, $$y''-y = e^t$$ has $$L = D^2-1$$ and $$A=D-1$$ gives $$A[e^t]=0$$. Note, $$AL = (D-1)(D^2-1) = (D+1)(D-1)^2$$ thus $$AL[y]=0$$ has solution $$y = c_1e^{-t}+c_2e^t+c_3te^t$$. Observe $$L[y] = e^t$$ gives $$(D^2-1)(c_1e^{-t}+c_2e^t+c_3te^t) = e^t$$ hence $$c_3D^2[te^t] - c_3te^t = e^t$$ or $$c_3(te^t+2e^t) - c_3te^t = e^t$$ consequently, $$2c_3e^t = e^t$$ and we find $$c_3 = \frac{1}{2}$$ thus $$y = c_1e^{-t}+c_2e^t+ \frac{1}{2}te^t$$ is the general solution of $$y''-y=e^t$$

Notice the nonhomogeneous problem can be viewed as a finite dimensional linear algebra problem since $$ann(AL)$$ is an $$(n+k)$$-dimensional vector space by the existence and uniqueness theory for constant coefficient ODEs. I think this is fairly removed from direct matrix math. Of course, pick a basis, and we can be right back to the matrix.

Beyond this, the concept of annihilators play a large role in the study of differential forms and integral submanifolds. But, you might be looking for a very different sort of example, so I'll stop here.

• I actually first learned annihilators in this context in my introductory differential equations class, and only put it together when learning about annihilator subspaces in a self-study on linear algebra. Jun 28, 2021 at 23:43
• @user615 fair enough, although, notice that the choice of $A$ I made was minimal. You can take $A$ and multiply by whatever else and $AA'$ still serves as an annhilator of $g$. Moreover, $A$ is a polynomial in $D$ and $A[g]$ is akin to the evaluation homomorphism. In short, my "A" would be the analog of the "h" in the OP. But, I'm a bit sleepy, so correct me if I'm off base here. I was more responding to the request for examples of annihilators, that concept is more general than the narrow definition given by the OP. In fact, the context of the definition more or less indicates the applica... Jul 5, 2021 at 9:20
• continuing, the annihilator is all polynomials for which plugging in the given number returns zero. This allows us to characterize numbers in terms of their characteristic polynomial etc... that whole discussion is the real application of annihilator in this context in my estimation. For example, $i = \sqrt{-1}$ has $x^2+1$ or for $j$ such that $j\neq 1$ and $j^2=1$ we have $x^2-1$ which is reducible and hence adjoining $j$ does not give a field, indeed $\mathbb{R}[j]$ is an algebra with zero divisors which is isomorphic to $\mathbb{R} \times \mathbb{R}$. I shut up here. Jul 5, 2021 at 9:25

I'm not a theoretical math type, more of a technical guy. But I'm used to seeing the annihilator method and operator discussed in ODE courses. For example pages 193-195 of Spiegel, third edition Applied Differential Equations.

If there are other knuckle-draggers like I in your course (since the physics slant), perhaps they will find such a discussion (having to do with characteristic equation of familiar second order ODE with constant coefficients) easier to grok than the holomorphism and set notation above.