What causes a trace of a product of two projection operators to not be a whole number

Question

Generally it should mean that their product is not a projector itself(correct me if I'm wrong). But is the trace in this case indicative of something else? What does this mean intuitively? Also, if the trace lies between two numbers, is the dimension of the intersection the lower one and why?

Answer 1

The trace of the product can be nicely interpreted in terms of the Frobenius norm (AKA Hilbert-Schmidt) distance between them. For an operator $A$, we have $$ \|A\|_F^2 = \operatorname{tr}(A^*A) = \operatorname{tr}(AA^*), $$ where $A^*$ denotes the adjoint of $A$ (the transpose for real matrices, Hermitian transpose for complex). Let $P_1,P_2$ be orthogonal projection operators with ranges $U_1$ and $U_2$ respectively. Note that $P_j = P_j^2 = P_j^*$ for $j = 1,2$. We compute $$ \|P_1 - P_2\|_F^2 = \operatorname{tr}[(P_1 - P_2)(P_1 - P_2)^*]\\ = \operatorname{tr}[P_1^2 - P_1P_2 - P_2 P_1 + P_2^2]\\ = \operatorname{tr}(P_1^2) + \operatorname{tr}(P_2^2) - \operatorname{tr}(P_1P_2) -\operatorname{tr}(P_2P_1)\\ = \operatorname{tr}(P_1) + \operatorname{tr}(P_2) - 2\operatorname{tr}(P_1P_2) \\ = \dim(U_1) + \dim(U_2) - 2 \operatorname{tr}(P_1P_2). $$ Note also that $\operatorname{tr}(P_1P_2)$ will always be non-negative (generally true for the trace of a product of positive semidefinite operators). The larger $\operatorname{tr}(P_1P_2)$ is, the closer $P_1$ and $P_2$ are, which equivalently means that $U_1$ and $U_2$ are closer to each other. The smallest possible trace occurs if and only if $U_1$ and $U_2$ are mutually orthogonal subspaces, and the largest possible trace occurs if and only if one subspace contains the other. Note that $$ \operatorname{tr}(P_1P_2) \leq \sigma_\max(P_1)\operatorname{tr}(P_2) = \operatorname{tr}(P_2), $$ so that $\operatorname{tr}(P_1P_2) \leq \min\{\operatorname{tr}(P_1), \operatorname{tr}(P_2)\}$.

Note that in general, $P_1P_2$ will be an orthogonal projector if and only if $P_1P_2 = P_2P_1$, which occurs if and only if $U_1$ can be decomposed as $U_1 = W_1 + W_2$ with $W_1 \subset U_2$ and $W_2 \subset U_2^\perp$. In this case, we can guarantee that the trace of a product is an integer, but this is not the only case where the trace of the product is an integer. As a counterexample, consider the projections in $\Bbb C^{4 \times 4}$ given by $$ P_1 = \pmatrix{I_2 & 0\\0&0}, \quad P_2 = \frac 12 \pmatrix{ I_2 & I_2\\I_2 & I_2}, $$ where $I_2$ denotes the size-2 identity matrix. Here, we have $\operatorname{tr}(P_1P_2) = 1$, but $U_1 \cap U_2 = \{0\}$. Also, $\dim(U_1) = \dim(U_2) = 2$.

If the trace lies between two numbers, is the dimension of the intersection the lower one and why?

The answer to this is no. In fact, we can have any value for the trace $0 < \text{tr}(P_1P_2) < \dim(U_1)$ while having $U_1 \cap U_2 = \{0\}$. Generalizing the above, for any real $\theta$ with $0 < \theta < \pi/2$ and integer $k$, consider $$ P_1 = \pmatrix{I_k & 0\\0 & 0}, \quad P_2 = \pmatrix{\cos^2 \theta I_k & \cos \theta \sin \theta I_k\\ \cos \theta \sin \theta I_k & \sin^2 \theta I_k}. $$ We have $\dim(U_1) = \dim(U_2) = k$, $U_1 \cap U_2 = \{0\}$, and $\operatorname{tr}(P_1P_2) = k\cos^2\theta$.

For small values of $\theta$, $\operatorname{tr}(P_1P_2)$ is close to its theoretical maximum of $k$ despite the spaces having an empty intersection because these subspaces are "close" in the sense that the canonical angles between them are small.