the trace of difference of two bounded projection operators is an integer

Question

I am solving this problem about trace class operators, but I got stuck:

Ler $P$ and $Q$ be bounded projections in a Hilbert space $\mathcal{H}$. Suppose $P-Q\in\mathcal{B}_1(\mathcal{H})$ (the set of trace class operators), then $tr(P-Q)\in\mathbb{Z}$.

Here are some thoughts that I have so far:

(1) Use these relations: Setting $A=P+Q$ and $B=P+Q-I$, I can get the relations $AB+BA=0$ and $A^2+B^2=I$.

(2) Show that if $\lambda_0\neq\{\pm{1},0\}$ is an eigenvalue of $P-Q$, then $AM(\lambda_0)=AM(-\lambda_0)$. $tr(A)=\sum_{i}\lambda_i$, where $\lambda_i$'s are the eigenvalues of $A$.

From here, I cannot proceed further. I am new to this part of functional analysis, so this question really challenges me. Can anyone connect my thoughts together and write a rather complete proof? Thank you!

Answer 1

Let $U = im P$, $V = im Q$. Come up with orthonormal bases for $A \cap B$ for $A = U$ or $U^\bot$, $B = V$ or $V^\bot$. The entire vector space is a direct sum of these four subspaces, so the union of these four orthonormal bases forms an orthonormal basis. It's easily seen that for each $v$ in this basis, the inner product of $v$ with $(P - Q)v$ will be an integer. Thus, so too will the trace be.

Answer 2

This should be a comment, but unfortunately I do not have enough reputation to comment so this'll have to do. The decomposition into intersections of the projections' kernels and ranges described in the previous answer unrfortunately does not exist in general. In fact, the intersections could all be empty. Take for example the orthogonal projections onto span((0,1)) and span((1,1)) in $\mathbb{R}^2$. The same can also happen in infinite dimensions.

Answer 3

Still not a proper answer, but a reference for one: The statement you are looking for is proved in "Why the circle is connected" by E. Effros as Lemma 4.1 (Link to the paper: https://link.springer.com/content/pdf/10.1007/BF03023772.pdf).