The proof is given in a linear algebra textbook for the proposition:
Let T be a linear operator on a vector space $V$, and let $λ$ be an eigenvalue of $T$. Then for any scalar $μ \ne λ$, the restriction of $T − μI$ to $K_{λ}$ is one-to-one.
And the proof is provided:
Let $x ∈ K_{λ}$ and $(T − μI)(x) = 0$. By way of contradiction, suppose that $x \ne 0$. Let $p$ be the smallest integer for which $(T−λI) p (x) = 0$, and let $y = (T − λI)^{p−1} (x)$.
Then $(T − λI)(y)$ = $(T − λI)^{p}(x) = 0$, and hence $y ∈ E_{λ}$ . Furthermore,
$(T-\mu I)(T-\lambda I)^{p-1}(x) = (T-\lambda I)^{p-1}(T-\mu I)(x)$ = $0$
To be fair I've read a few other duplicate questions on the site and people provide very reasonable and quite obvious explanations that:
- by induction on $p$, this works
- $T$ can be written as $(T−λI)+λI$ hence, again, they commute
- $T$ commutes with both $T$ and $λI$, $T$ commutes with $T−λId$ and therefore it commutes with any power of $T−λId$.
- just binomial expansion of $LHS$ and $RHS$ would make this apparent
And I am convinced at this point, but decided to still ask because I've never seen so many similar questions on this formulation( on this book's Stephen H. Friedberg).
I think most of my confusion came from the fact that when I see $ (T - \lambda I)(x)$ I implicitly think of it as a matrix and we were often cautioned in the previous courses that matrices do NOT commute.
What is the most correct way to get out of this bind: should I tell myself that both $T$ and $I$ really do not have to be matrices in this formulation and are just arbitrary operators? Or is it less incorrect to say that matrices don't commute in general, but some particular -- do.
I realize this might be pretty solipsistic for such a simple question, but would still appreciate if somebody with the right intuitions weighed in. Non-esoteric (counter)examples are especially appreciated :)