# What is the intuition behind “states with support on orthogonal subspaces”?

I'm sure I don't fully understand support, but I am having trouble seeing how it connects to things like density operators. I have an idea that it means, according to Wikipedia:

"In mathematics, the support of a real-valued function f is the subset of the domain containing those elements which are not mapped to zero."

First of all, what would support be used for in quantum theory? What is the meaning of the support of a density operator or quantum state? Perhaps someone could explain what "states with support on orthogonal subspaces means" and what implications it has.

Consider a vector space $$V$$ with an inner product and a linear operator $$A:V\rightarrow V$$.
The definition of support that you have posted from Wikipedia can be a bit confusing. It says $$\text{supp}(A) = \{u\in V|Au\neq 0\}$$. This is the complement of the kernel where the kernel of $$A$$ is $$\text{ker}(A) = \{v\in V| Av = 0\}$$. However, this definition of the support leaves you with a set that is not a vector space (for example, the zero vector is not in $$\text{supp}(A)$$).
The definition of the support of $$A$$ which is used in quantum information is the orthogonal complement of the kernel i.e. $$\text{supp}(A) = \{u\in V| \langle u, v\rangle = 0, v\in \text{ker}(A)\}$$. See for example page 14 of this textbook where it is introduced. Some useful properties are
1. For self-adjoint $$A$$, the image of $$A$$ is the same as the support of $$A$$. To see this, pick any $$u\in \ker(A)$$ and $$v\in V$$. Then you have $$0 = \langle Au, v \rangle = \langle u, Av \rangle \implies Av \in \text{supp}(A)$$
If $$\rho$$ and $$\sigma$$ have support on orthogonal subspaces, then it means that the $$\text{supp}(\sigma)\subseteq \ker{(\rho)}$$ and $$\text{supp}(\rho)\subseteq \ker{(\sigma)}$$. You can then construct a projective measurement $$\{P_{\text{supp}(\rho)}, I - P_{\text{supp}(\rho)}\}$$ or $$\{P_{\text{supp}(\sigma)}, I - P_{\text{supp}(\sigma)}\}$$ and this is guaranteed to distinguish perfectly between $$\rho$$ and $$\sigma$$.