# What is the simplified form of the generalized eigen space when the characteristic polynomial does not split in the given field.

Let $$V$$ be a finite dimensional vector space over a field $$\mathbb{F}.$$ Let $$T$$ be a inear operator on $$V$$ and $$\lambda \in \mathbb{F}$$ be an eigenvalue of $$T$$ of algebraic multiplicity $$m.$$ Now suppose that the characteristic polynomial does not split in $$F.$$ Then is it true that the generalized eigen space corresponding to the eigenvalue $$\lambda$$ is given by $$S=\{x \in V : (T-\lambda I)^mx=0\},$$ where the Generalized eigen space of $$\lambda$$ $$G(\lambda,T)$$ is defined as $$\{x \in V : (T-\lambda I)^ix=0\;\text{for}\;\text{some}\;i \in \mathbb{N}\cup \{0\}\}.$$

I have seen the proof when the Characteristic polynomial splits in the field $$\mathbb{F}.$$ In our case to start with we should assume that $$m < \text{dim}(V)$$ so that $$c_{T}(X)=(X-\lambda)^mP(X).$$ Clearly for all $$v \in V$$ $$P(T)(v) \in G(\lambda,T).$$ And also $$S \subseteq G(\lambda,T).$$ How can I prove the converse ? Any help will be appreciated. Many Thanks.

• What is the proof when the polynomial splits? Maybe only some minor changes are needed to make it work in general. – Francesco Milizia Dec 21 '18 at 8:46
• If the characteristic polynomial does not split, there is no such $\lambda\in\mathbb F$ is there? Instead $\lambda$ is an element of the splitting field. – Klaas van Aarsen Dec 22 '18 at 12:26
• What if the ch polynomial is $(X-1)^2(X^2+1) \in \mathbb{R}[X],$ where $\lambda =1$ and $\mathbb{F}=\mathbb{R}.$ – user371231 Dec 22 '18 at 12:51
• By split of a polynomial I mean split completely into linear factors. – user371231 Dec 22 '18 at 12:52

When the characteristic polynomial does not split into linear factors I would approach this as follows.

A way to prove the existence of most of the canonical forms is the observation that having a linear transformation $$T:V\to V$$ allows us to turn $$V$$ into an $$\Bbb{F}[x]$$-module by letting $$x$$ act as $$T$$. More precisely, if $$v\in V$$ and $$p(x)=\sum_{i=0}^na_ix^i\in\Bbb{F}[x]$$ are arbitrary, we define $$p(x)\cdot v=\sum_{i=0}^na_i T^i(v).$$ Anyway, let's see where this approach takes us.

As $$V$$ was assumed to be finite dimensional, it becomes a finitely generated $$\Bbb{F}[x]$$-module. A polynomial ring over a field is a PID (a Euclidean domain even), so the structure of f.g. modules over a PID allows us to write $$V$$ as a direct sum oof cyclic modules $$V=\bigoplus_{j=1}^rV_j,$$ where each summand $$V_j\simeq \Bbb{F}[x]/\langle p_j(x)\rangle$$ for some non-constant polynomial $$p_j(x)\in\Bbb{F}[x]$$.

If $$p_j(x)=\prod_\ell q_\ell(x)^{a_\ell}$$ is a factorization of $$p_j(x)$$ into a product of powers of irreducible polynomials $$q_\ell(x)$$, then the polynomial variant of the Chinese Remainder Theorem allows us to further split the submodules $$\Bbb{F}[x]/\langle p_j(x)\rangle=\bigoplus_\ell\Bbb{F}[x]/\langle q_\ell(x)^{a_\ell}\rangle.$$ Putting this all together we see that $$V$$ can be written as a direct sum of modules of the form $$\Bbb{F}[x]/\langle r(x)^a\rangle$$ for some irreducible polynomial $$r(x)\in\Bbb{F}[x]$$ and positive integer $$a$$. Such a decomposition (into submodules) is automatically also a decomposition into $$T$$-stable subspaces.

Let's extract one more thing from the theory of modules over a PID. To each irreducible polynomial $$r(x)$$ we can define the $$r$$-power torsion submodule $$V_r:=\{v\in V\mid r(x)^k\cdot v=0\ \text{for some integer k>0}\}.$$ In my opinion the subspaces $$V_r$$ are the natural analogue of the generalized eigenspaces. After all, should $$\Bbb{F}$$ be algebraically closed, we would necessarily have $$r(x)=x-\lambda$$ for some $$\lambda\in\Bbb{F}$$, and the above definition of $$V_r$$ becomes the definition of a generalized eigenspace.

The submodule $$V_r$$ is the direct sum of all those earlier summands $$\Bbb{F}[x]/\langle q_\ell(x)^{a_\ell}\rangle$$ where the irreducible polynomial $$q_\ell(x)=r(x)$$.

My main point was that when the characteristic polynomial doesn't split into linear factors, it feels natural to lump together conjugate eigenvalues, i.e. the zeros of an irreducible factor of the characteristic polynomial. The above subspaces $$V_r$$ become, upon the appropriate extension of scalars, the sum of generalized eigenspaces belonging to all those conjugate eigenvalues.

It is easy to build an analogue of a Jordan block in this setting. A submodule of the form $$\Bbb{F}[x]/\langle r(x)^a\rangle$$ then corresponds with an $$ab\times ab$$ matrix, $$b=\deg r(x)$$. Along its diagonal we have $$b\times b$$-blocks representing $$r(x)$$. We can arrange those diagonal blocks to become, say, companion matrices of $$r(x)$$. Then, on top of the main diagonal we have blocks of the form $$I_b$$, analogous to those extra $$1$$ on top of the diagonal of a Jordan block in the case $$b=1$$.

• In 2ndlast paragraph do you wanted to say that it is enough to consider that the ch polynomial splits completely in the base field ? In my understanding If $K$ be the splitting field of the ch polynomial then one may consider the vector space $K \otimes_{\mathbb{F}} V$ over K (By extension of scalars). I know that the induced linear operator on $K \otimes_{\mathbb{F}} V$ also has the same ch polynomial. Then the generalized eigen space corresponding to the eigenvalue $\lambda$ has same dimension in both the cases. If this is it please confirm me so that I may try to prove this. Thank you. – user371231 Dec 23 '18 at 9:52
• Yes. That's what happens. Any $\Bbb{F}$-automorphism of $K$ (acting on that tensor product on the left factor) leaves the spaces $K\otimes V_r$ invariant, and permutes the generalized eigenspaces belonging to conjugate eigenvalues. – Jyrki Lahtonen Dec 23 '18 at 10:16