# Confusion on commutativity of matrices/operators in a given proof: $(T-\mu I)(T-\lambda I)^{p-1}(x)$ = $(T-\lambda I)^{p-1}(T-\mu I)(x)$ = $0$

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:

1. by induction on $$p$$, this works
2. $$T$$ can be written as $$(T−λI)+λI$$ hence, again, they commute
3. $$T$$ commutes with both $$T$$ and $$λI$$, $$T$$ commutes with $$T−λId$$ and therefore it commutes with any power of $$T−λId$$.
4. 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 :)

• Because multiplication is associative, powers of $T$ commute with each other: $T^n T^m = T^{n + m} = T^m T^n$. In particular, this also works for $T^0 = I$. Put this together with addition and scalar multiplication to find that if $p$ and $q$ are any polynomials, then $p(T) q(T) = q(T) p(T)$ (which you can argue by expanding into powers of $T$ on either side). So any time you see expressions involving linear combinations of powers of $T$ and the identity, they will commute with other such expressions. This argument works for general linear operators just as well as it does for matrices. – Joppy Apr 1 at 3:14
• When we say that matrices don't commute, we mean that "$AB=BA$ for any two $n \times n$ matrices $A$ and $B$" is a false statement. I.e. there are two matrices such that $AB \neq BA$. This doesn't imply that $AB\neq BA$ for all such matrices (which would be wrong, see, e.g. the study of simultaneously diagonalizable matrices) – Brian Moehring Apr 1 at 3:20
• The more general intuition is if you have a collection of linear operators $A, B, C, \ldots$ which all commute pairwise, then you are free to rearrange any expression in these (or powers of these) however you like, treating it as a polynomial in variables $A, B, C, \ldots$. So for example $(AB + C - 2I)A^2 = A^2(BA + C - 2I)$ holds. – Joppy Apr 1 at 3:20