# Quaternions as eigenvalues of rank 3 tensors

Let us consider a matrix $$M^{(a)}$$ of size $$N \times N$$, having $$N$$ eigenvalues $$\lambda_i \in \mathbb{C}$$.

Considering a rank-3 tensor, we can informally think of it as a sequence of $$N$$ matrices $$M^{(a)}$$ (where $$1\leq a\leq N$$) of size $$N\times N$$. Then, if we fix one of the components we recover a matrix $$M^{(a)}$$ with complex eigenvalues.

Has there been any literature on extending eigenvalues for rank-3 tensors where the eigenvalues are represented as quaternions? Where now the eigenvalues $$\lambda_i=a+b\mathrm{i}+c\mathrm{j}+d\mathrm{k}$$.

I know there exist different generalizations on eigenvalues of tensors but I am particularly keen on knowing if there exists one linked with quaternions.

Edit: I am aware the notation is not very rigorous, I still have trouble formulating my problem in a rigorous way: we can represent the components of a rank-3 tensor $$\Lambda$$ by $$\Lambda_{ijk}$$ where $$1\leq i,j,k \leq N$$. Then if we fix one of the components we technically recover the components of a matrix: for example for $$k=1$$ we obtain a matrix $$\Lambda_{ij1}$$ for $$1\leq i,j \leq N$$ and we can attribute an eigenvalue to it. Depending which index we fix, we will get a sequence of matrices. There are three combination possible so I was thinking of something in those lines:

fixing the first index will give us complex eigenvalues with imaginary part $$\mathbf{i}$$, fixing the second index will give us a matrix with eigenvalues whose imaginary part are $$\mathbf{j}$$ and fixing the third index will yield eigenvalues with imaginary part $$\mathbf{k}$$. For example fixing the third index at $$k=4$$ will yield a quaternionic eigenvalue with imaginary part $$\mathbf{k}$$: \begin{align*} \sum_j \Lambda_{ij4}v_j&=\lambda_i v_i\\ &=(a+\mathbf{k} b)v_i \quad \text{where } a,b\in\mathbb{R} \end{align*} Where $$\vec{v}$$ would be an eigenvector of the resulting matrix.

Therefore could we define $$N$$ general eigenvalues $$\lambda_i=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$$ for the tensor $$\lambda$$ without having to consider fixing indices?

It seems an intuitive generalization of complex eigenvalues for rank-3 tensors and this would enable us to keep definitions and hence nice properties of the resolvent or distribution of eigenvalues.

If there are obvious inconsistencies in this generalization please let me know and I will close this topic.

• The order seems a bit strange: you say first that you consider a matrix $M^{(a)}$, and then later define $M^{(a)}$ as one of a sequence of matrices. Probably the first paragraph does not belong? It really serves only to bind the names of the eigenvalues $\lambda_i$, but, since the only later occurrence of those eigenvalues has a different meaning, it's probably not necessary to do so. (Also, when you are "extending eigenvalues", are the matrices now quaternionic, or do you somehow want to use the rank-3 structure to give complex matrices quaternionic eigenvalues?) – LSpice Jan 22 at 14:47
• Could you formulate it more clearly? I don't quite see how does it typecheck. If you would have two complex numbers in place of one, you could indeed try to invoke quaternions to encode pairs of complex numbers. But in your situations you have an $N$-tuple of complex numbers in place of one, and then I don't see how quaternions might help. If I am not mistaken, you rather would need some $N$-dimensional algebra for that. – მამუკა ჯიბლაძე Jan 23 at 10:12

Because quaternions do not commute, there are two types of eigenvalues of an $$n\times n$$ quaternion matrix $$A$$: left eigenvalues solve $$Av=\lambda v$$ for some nonzero quaternion vector $$v$$, while right eigenvalues solve $$Av=v\lambda$$. Left quaternion eigenvalues are more closely analogous to complex eigenvalues, in particular, the existence of a left eigenvalue $$\lambda$$ implies that the matrix $$\lambda I-A$$ is not invertible.
A theorem by Wood [1] says that there exists at least one left eigenvalue. For complex eigenvalues $$c$$ the determinant of $$cI-A$$ provides an algebraic way to find the eigenvalues, but no such general algebraic approach exists in the quaternion case: For $$2\times 2$$ matrices, see [2], for larger $$n$$ see [3].