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 $\mathbb C^{n\times n}$ (or over any algebraically closed field), where $T$ has $m$ distinct eigenvalues
$\mathbf 0 =p\big(T\big) = \big(\lambda_1 I-T\big)^{k_1}\big(\lambda_2 I-T\big)^{k_2}...\big(\lambda_{m-1} I-T\big)^{k_{m-1}}\big(\lambda_m I-T\big)^{k_m}$

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

now apply equivalent forms of Sylvester's Rank Inequality twice to get
$n$
$= k_1+k_2 + ....+ k_{m-1}+k_m$
$\geq \sum_{j=1}^{m} \dim\ker\Big(\big(\lambda_j I-T\big)^{k_j}\Big)$
$=\Big(\sum_{j=1}^{m-1} \dim\ker\Big(\big(\lambda_j I-T\big)^{k_j}\Big)\Big)+ \dim \ker \Big(\big(\lambda_m I-T\big)^{k_m}\Big)$
$\geq \dim\ker\Big(\big(\lambda_1 I-T\big)^{k_1}\big(\lambda_2 I-T\big)^{k_2}\dots \big(\lambda_{m-1} I-T\big)^{k_{m-1}}\Big)+ \dim \ker \Big(\big(\lambda_m I-T\big)^{k_m}\Big)$
$\geq \dim\ker\Big(\big(\lambda_1 I-T\big)^{k_1}\big(\lambda_2 I-T\big)^{k_2}...\big(\lambda_{m-1} I-T\big)^{k_{m-1}}\big(\lambda_m I-T\big)^{k_m}\Big)$
$=\dim\ker\Big(\mathbf 0\Big)$
$=n$

The first inequality being met with equality tells us $\dim\ker\Big(\big(\lambda_m I-T\big)^{k_m}\Big)=k_m=\dim\ker\Big(\big(\lambda_m I-T\big)^{n}\Big)$

The final inequality is met with equality so Sylvester tells us
$\dim\ker\Big(\big(\lambda_1 I-T\big)^{k_1}\big(\lambda_2 I-T\big)^{k_2}\dots \big(\lambda_{m-1} I-T\big)^{k_{m-1}}\Big) $
$= \dim\left\{\ker\Big(\big(\lambda_1 I-T\big)^{k_1}\big(\lambda_2 I-T\big)^{k_2}\dots \big(\lambda_{m-1} I-T\big)^{k_{m-1}}\Big)\bigcap \text{image }\big(\lambda_m I-T\big)^{k_m}\Big)\right\}$
$\implies \dim\left\{\ker\Big(\big(\lambda_1 I-T\big)^{k_1}\big(\lambda_2 I-T\big)^{k_2}\dots \big(\lambda_{m-1} I-T\big)^{k_{m-1}}\Big)\bigcap \ker \big(\lambda_m I-T\big)^{k_m}\Big)\right\} = 0$
since $V = \ker \big(T-\lambda_m\big)^n\oplus \text{image } \big(T-\lambda_m\big)^n$

Now suppose $v \in W_m$ and $v \in W_1 + W_2 + \dots + W_{m-1}$
$\implies v \in \ker \Big(\big(\lambda_m I-T\big)^{k_m}\Big)$ and $v \in \ker \Big(\big(\lambda_1 I-T\big)^{k_1}\big(\lambda_2 I-T\big)^{k_2}\dots \big(\lambda_{m-1} I-T\big)^{k_{m-1}}\Big)$
$\implies v = \mathbf 0 $
$\implies V = W_1 \oplus W_2 \oplus \dots \oplus W_{m-1} \oplus W_m$
since the labeling of eigenvalues is arbitrary.
(That is: for $w_j \in W_j$, if $\sum_j w_j = \mathbf 0$ with some $w_i\neq 0$, set $v:= w_i$ and then re-label $\lambda_i$ and $\lambda_m$, to get a contradiction.)

This is a stronger result than what the OP asked for. I.e. we can restrict to the vector (sub)space $V':= W_i \oplus W_j = W_i + W_j$ for $i\neq j$ to recover $\dim V' = \dim W_i + \dim W_j = \dim W_i + \dim W_j + \dim\big(W_i \cap W_j\big)\implies\dim\big(W_i \cap W_j\big)=0$

Answer 3

For the case when $\mu$ and $\lambda$ are eigenvalues of $T$, there is a nice one-line proof (if you know that generalised eigenvectors corresponding to distinct eigenvalues are linearly independent):

Suppose for a contradiction that $0\neq v\in K_\mu(T) \cap K_\lambda(T)$ with $\lambda \neq \mu$. Then $v$ is a generalised eigenvector corresponding to $\mu$ as well as a generalised eigenvector corresponding to $\lambda$, so $v$ and $v$ are linearly independent - contradiction. So from $\{0\} \subset K_\mu(T) \cap K_\lambda(T)$ we conclude that $K_\mu(T) \cap K_\lambda(T) = \{0\}$. $\square$

(The general case was very nicely proved in earlier answers :)