I would like to ask a question about the proof of the Thm in the title.
The proof is from "Linear Algebra Done Right" by Axler.
The statement is "Let $T \in L(V)$. Suppose $\lambda _1, …, \lambda _m$ are distinct eigenvalues of $T$ and $v_1, …, v_m$ are corresponding generalized eigenvectors. Then $v_1, …, v_m$ is linearly independent."
Axler starts proof like this.
Suppose $a_1, …, a_m$ are complex numbers s.t $$0= a_1v_1 + … + a_mv_m$$
We name this equation $\star $.
Let $k$ be the largest nonnegative integer s.t $(T-\lambda_1)^kv_1 \neq 0$. Let $$w=(T-\lambda_1)^kv_1$$
Thus, $$(T-\lambda_1I)w=(T-\lambda_1I)^{k+1}w=0$$
Hence, $Tw=\lambda _1w$. Thus $(T-\lambda I)w=(\lambda_1 – \lambda)^nw$ for every $\lambda \in \mathbb{F}$ and hence $$(T-\lambda I)^nw=(\lambda_1-\lambda)^nw$$ for every $\lambda \in \mathbb{F}$, where $n=\dim V$.
Apply the operator $$(T- \lambda_1I)^k(T-\lambda_2I)^n…(T-\lambda_mI)^n$$ to the equation $\star $. Then, we get
$$0=a_1(T- \lambda_1I)^k(T-\lambda_2I)^n…(T-\lambda_mI)^nv_1$$
And from this equation, he gets $a_1=0$.
I stopped copying his proof here because the last equation is the part I have a question about.
Intuitively, it makes sense to me that we only get $v_1$ part because $v_2 \in G(\lambda_2,T), …, v_m \in G(\lambda_m, T)$.
But, whenever I tried to decompose his works in the last equation, it is kind of confusing.
For example, if we are trying to compute $a_2(T- \lambda_1I)^k(T-\lambda_2I)^n…(T-\lambda_mI)^nv_2$, doesn't $v_2$ get mapped by $(T-\lambda_mI)^n$ first, and then the next linear map is applied, and so on.
So, how do we know for sure that we get $0$ when it comes to apply $(T-\lambda_2I)^n$ to the vector $u:=(T-\lambda_3I)^n…(T-\lambda_mI)^nv_2$?
Is this related to concept of "polynomials applied to operators"?
Best Answer
This is the consequence of the fact that for any two polynomials $p,q$ and a linear endomorphism $v$, the linear endomorphisms $p(v)$ and $q(v)$ commute.
The result is here applied to polynomials of the form $(x- \lambda)^k$.