Prove that the intersection of 2 generalised eigenspaces is the zero space

Question

I have searched for my above question and came across the "Trivial intersection of generalised eigenspaces" post on math stack exchange but I do not understand the proof using coprime polynomials. How do I proof such a statement (below) using just the definition of eigenvalues/generalised Eigenspaces?

I have seen/proven that if $\lambda \neq \mu $ . then the intersection between $ E_\lambda(T) \cap K_mu(T) = \{ \mathbf{0} \} $ (where $E_\lambda(T) $ is the Eigenspaces corresponding the eigenvalue $\lambda$. (not sure whether this information is required for the proof)

Let $ T: V \rightarrow V$ be a linear operator where $V$ is a finite dimensional vector space over $ \mathbb{C} $.

I want to prove that $$ \text{If } \lambda \neq \mu, \text{then } K_\mu(T) \ \cap \ K_\lambda(T) = \{\bf{0}\} $$ where $$ K_\lambda(T) = \{ \mathbf{v} \in V : (T-\lambda I_V)^m(\mathbf{v})=\mathbf{0}\} $$ Currently, the lecturer has only gone through the above definition of generalised Eigenspaces (he currently assumes that m need not be the same for different $\mathbf{v} \in K_\lambda(T)$, he has not gone through the prove that m can be chosen to satisfy all $\mathbf{v}$ in the generalised eigenspace yet)

Anyway,

I tried to prove the above statement by contradiction but I got stuck:

Let $ \lambda \neq \mu $ and assume $$ \exists_{non-zero \ vector \ \mathbf{v} \in V}\ \text{such that } v \in K_\mu(T) \cap K_\lambda(T) $$

Then $$ (T-\mu I_V)^m(\mathbf{v}) = \mathbf{0} = (T-\lambda I_V)^n(\mathbf{v}) $$ $$ (T-\mu I_V)^m(\mathbf{v}) = (T-\lambda I_V)^n(\mathbf{v})$$ $$ (T-\mu I_V)^m(\mathbf{v}) - (T-\lambda I_V)^n(\mathbf{v}) =\mathbf{0} $$

And I'm not sure how to proceed.

Thank you for your time!!

Answer 1

I think you indeed need to use Bezout's identity for the polynomials $f(x)=(x-\lambda)^n$ and $g(x)=(x-\mu)^m$ which are clearly coprime.

Bezout's identity says that there exist polynomials $p$ and $q$ such that $pf+qg=1$.

But then $p(T)(T-\lambda I)^n+q(T)(T-\mu I)^m=I$, so applying it to a hypothetical common generalized eigenvector $v$, we receive $$0=p(T)(T-\lambda I)^nv+q(T)(T-\mu I)^mv=Iv=v\,.$$

Answer 2

A simple minded approach would be to use Cayley Hamilton and Sylvester's Rank Inequality. The former tells you, for matrices in $\mathbb C^{n\times n}$, where $T$ has $m$ distinct eigenvalues
$\mathbf 0 =p\big(T\big) = \big(\lambda_1 I-T\big)^{k_1}\big(\lambda_2 I-T\big)^{k_2}...\big(\lambda_m I-T\big)^{k_m}$

we know for $Z:=\big(\lambda_j I-T\big)$ and any natural number $r$
$\dim \ker Z^{r} $
$=\text{geo multiplicity of eig 0 for }Z^r$
$\leq \text{alg multiplicity of eig 0 for }Z^r$
$=\text{alg multiplicity of eig 0 for }Z $
$= k_j$

now apply an equivalent form of Sylvester's Rank Inequality to get
$n$
$= k_1+k_2 + ....+k_m$
$\geq \dim\ker\Big(\big(\lambda_1 I-T\big)^{k_1}\Big) + \dim\ker\Big(\big(\lambda_2 I-T\big)^{k_2}\Big)+... +\dim\ker\Big(\big(\lambda_m I-T\big)^{k_m}\Big)$
$\geq \dim\ker\Big(\big(\lambda_1 I-T\big)^{k_1}\big(\lambda_2 I-T\big)^{k_2}...\big(\lambda_m I-T\big)^{k_m}\Big)$
$=\dim\ker\Big(\mathbf 0\Big)$
$=n$

This equality tells us
(i) $\dim\Big(\ker\big((\lambda_j I-T)^{k_j}\big)\Big) = k_j$ which is maximal
$\implies \dim\Big(\ker\big((\lambda_j I-T)^{k_j}\big)^2\Big) = \dim\Big(\ker\big((\lambda_j I-T)^{k_j}\big)^1\Big)$
$\implies \text{image}\Big(\big(\lambda_j I-T\big)^{k_j}\Big) \cap \ker\Big(\big(\lambda_j I-T\big)^{k_j}\Big) = \big\{\mathbf 0\big\}$
(ii) Checking the equality conditions of Sylvester's Rank Inequality tells us that for $j\neq i$
$\dim\Big(\ker\big((\lambda_j I-T)^{k_j}\big)\cap \text{image}\big((\lambda_i I-T)^{k_i}\big)\Big)=\dim\Big(\ker\big((\lambda_j I-T)^{k_j}\big)\Big)$
and by (i) this implies
$\dim\Big(\ker\big((\lambda_j I-T)\big)^{k_j}\cap \ker\big((\lambda_i I-T)^{k_i}\big)\Big) = 0$
as desired

note:
if for some reason we were concerned about other exponents, $r_j \neq k_j$ then
(a) if $k_j\lt r_j$, (i) tells us that
$ \ker\big((\lambda_j I-T)^{k_j}\big)=\ker\big((\lambda_j I-T)^{r_j}\big)$
(b) if $k_j\gt r_j$ we always have
$\ker\big((\lambda_j I-T)^{r_j}\big)\subseteq\ker\big((\lambda_j I-T)^{k_j}\big)$
so the former case is equivalent to setting $r_j := k_j$ and the latter case is implied by the $k_j$ case (i.e. the nesting of kernels gives an easy argument by contradiction for the latter case)