# Linear operator T that commutes with every projection operator, infinite dimensional case

This is related to a question of Hoffman & Kunze, Linear Algebra (Section 6.7, #8, p. 219). The question in the text asks:

Let $T$ be a linear operator on $V$ which commutes with every projection operator on $V$. What can you say about $T$?

If $V$ is finite-dimensional, I think that $T$ is then a scalar multiple of the identity. (We know from an earlier exercise that if every subspace of a finite-dimensional vector space $V$ is invariant under $T$ then $T$ is a scalar multiple of the identity. We also know that $TE = ET \implies E(V)$ is invariant under $T$. Since we can construct a projection $E$ such that $E(V) = W$ for every subspace $W$ of $V$, we can then conclude that $T$ is a scalar multiple of the identity.)

I then have two questions: is the reasoning above for the finite-dimensional case accurate? How do we deal with the infinite-dimensional case? (It is possible the question implies that $V$ is finite-dimensional, but it does not state that, so I am uncertain.)

For background, I am self-studying linear algebra and have covered everything in Hoffman & Kunze up to this question.

Suppose that there exist $x \in V$ such that $Tx$ is not a multiple of $x$. Let $E_x$ be a projection operator whose kernel is precisely the span of $x$. Then $T E_x(x) = T(0) = 0$, but $E_x(Tx)$ is necessarily non-zero since $Tx$ is not in the span of $x$. So, $TE_x \neq E_x T$.
Thus, we conclude that for all $x$, $T x = \lambda_x x$ for some constant $\lambda_x$.
Now, suppose that there exist non-zero $x,y$ such that $\lambda_x \neq \lambda_y$. It follows that $T(x+y) = \lambda_xx + \lambda_yy$ is not a multiple of $x+y$, which is a contradiction of our premise.
Thus, there is some $\lambda$ so that for all $x \neq 0$, $Tx = \lambda x$.
Thus, $T$ is necessarily a multiple of the identity operator.