# Minimal polynomial of operator with all eigenvalues are 1

I'm trying to understand proof of this lemma.

I have next questions:

1)If $$\tau$$ acts as the identity on the subspace $$\mathbb{R}\alpha$$ and on the $$E/\mathbb{R}\alpha$$ doesn't it mean that $$\tau$$ acts as the identity on the $$E$$? And by the way, why it acts as the identity on quotient space?

2)If all eigenvalues are 1, why minimal polynomial divides $$(T-1)^l$$. Isn't is possible that minimal poynomial looks like $$(T-1)^kp$$(T) where $$k?

• How is it meant that $\sigma$ 'leaves $\Phi$ invariant'? I would suppose it means $\sigma(\Phi)=\Phi$ (as is written for $\tau$ in the proof). But then, how could an $\alpha\in\Phi$ be sent to its negative? It seems that we have to assume that $-\alpha\in\Phi$ as well, is it correct? – Berci Oct 11 '18 at 10:45
• Yeah, this is about root systems. That is with every $alpha$ it contains $-\alpha$ – Kirill Losev Oct 11 '18 at 11:11
• Which book is this? This is a truly lousy proof. – Marc van Leeuwen Oct 17 '18 at 11:50
• @Marc van Leeuwen Introduction to Lie Algebras and Representation Theory Author: HUMPHREYS – Kirill Losev Oct 17 '18 at 11:58

1. Consider the linear map $$\varphi:(x,y)^T\mapsto (x+y,y)^T$$ on $$E=\Bbb R^2$$, which has matrix $$\pmatrix{1&1\\0&1}$$, and $$\alpha=(1,0)^T$$.
Then $$\varphi$$ acts as the identity on $$\Bbb R\alpha$$, and also $$\varphi(x,y)^T=(x+y,y)^T=(x,y)^T+(y,0)^T\ \in (x,y)^T+\Bbb R\alpha$$, so it acts as the identity on $$E/\Bbb R\alpha$$.
Now in the exercise, both $$\sigma$$ and $$\sigma_\alpha$$ fix $$E/\Bbb R\alpha$$ pointwise: by condition, $$\sigma$$ fixes a hyperplane $$P$$ pointwise, which must be complementary to $$\Bbb R\alpha$$ (since $$\alpha$$ is not fixed), i.e. $$E=P\oplus\Bbb R\alpha$$, thus $$E/\Bbb R\alpha$$ is represented by $$P\$$ (we have a natural isomorphism $$P\cong E/\Bbb R\alpha$$).
The same goes with $$\sigma_\alpha$$ and its pointwise fixed hyperplane $$P_\alpha$$ (which is also known to be orthogonal to $$\Bbb R\alpha$$).

2. The roots of the minimal polynomial are exactly the eigenvalues. Since $$0$$ is not an eigenvalue, the minimal polynomial can't have $$T$$ as a factor.
Saying in other way: note that the characteristic polynomial of $$\tau$$ is exactly $$(T-1)^l$$, if $$\tau$$ has only $$1$$ as eigenvalue, and the minimal polynomial divides it.

• Couldn't characteristic polynomial be $(T-1)^{l-2}(T^2+1)$ for example. In other words, why there is no complex roots? – Kirill Losev Oct 11 '18 at 14:36
• No. Being identity on $E/\Bbb R\alpha$ induces $l-1$ elements $1$ in the diagonal.. Note that, by conditions on $\sigma$, it must be a (not necessarily orthogonal) reflection through $P$, in the direction of $\alpha$, i.e. $\sigma(p+c\alpha)=p-c\alpha$ for $p\in P$. Similarly, $\sigma_\alpha$ with $P_\alpha\ \perp\alpha$. You can calculate their composition by hand.. – Berci Oct 11 '18 at 22:13
• The matrix will be like this $\pmatrix{A&0\\0&1}$? The last basis vector is $\alpha$ and A is unipotent. Right? – Kirill Losev Oct 12 '18 at 18:16

I don't understand why there are all those hypotheses, which end up not defining what $$\sigma_\alpha$$ and $$P_\alpha$$ are (except that the former is some reflection), so that it will be hard to justify the conclusion. OK, I get that this is from a book that probably defined $$\sigma_\alpha$$ and $$P_\alpha$$ earlier, but still no need for all those hypotheses. If a linear operator $$\sigma$$ on an $$\Bbb R$$-vector space$$~V$$ leaves a hyperplane $$P$$ point-wise fixed and sends some nonzero vector $$\alpha$$ to $$-\alpha$$, then obviously $$\alpha\notin P$$, so $$V=P\oplus\langle\alpha\rangle$$, and $$\sigma$$ has $$P$$ as eigenspace for eigenvalue $$1$$ and $$\langle\alpha\rangle$$ as eigenspace for $$-1$$, which completely describes$$~\sigma$$. Therefore if $$\sigma_\alpha$$ has the same properties (which I presume was established before) then $$\sigma=\sigma_\alpha$$. This is just the general fact that a linear operator that stabilises each subspace in a direct sum decomposition of the whole space is determined by its restrictions to each of those subspaces (here those restrictions are scalar multiplications by $$1$$ respectively by $$-1$$).