Let V, W be vector spaces over a field k and M = {A ∈ Hom(V, W) rank A < ∞} the the set of all linear maps V → W of finite rank. Recall that the rank of a linear map is the dimension of its image, i.e rank A = dim A(V ) for A ∈ Hom(V, W).
Prove that the function d: M × M → R + 0 given by d(A, B) = rank(A − B) is a distance function on M.
Is d induced from any distance on Z, i.e. is there a map f : M → Z such that d(A, B) = rank(A − B) = |f(A) − f(B)| ?
My attempt at a solution so far is...
1) d(A,B)=Rank(A-B)=dim(A-B)=dim(A)-dim(B)
let d(A,B)=0
=>Rank(A-B)=dim(A-B)=dim(A)-dim(B)=0
=>dim(A)=dim(B) but as every vector space has a unique basis
=> A=B if d(A,B)=0
2) [this part I'm not really sure of it makes sense intuitively to me that a negative distance is just a positive distance in the other direction but i don't think its formal enough]
d(A,B)=Rank(A-B)=dim(A)-dim(B)=-1(dim(B)-dim(A))=(-1)(d(B,A))=-d(B,A)
=> d(A,B)=d(B,A)
3) d(A,C)=dim(A)-dim(C) d(A,B)+d(B,C)=dim(A)-dim(B)+dim(B)-dim(C)=dim(A)-dim(C) =>d(A,C)=dim(A,B)+dim(B,C) and so triangle inequality holds.
for the metric induced from any distance on Z , i think that because the rank(A-B) is an element of the natural numbers then there would be an isometry , as there is a bijective function f:N-->Z such as
f(n) = \begin{cases} \frac{n}{2} & n\text{ is even} \\ -\frac{n + 1}{2} & \text{else} \end{cases}
You are right: this is wrong, because the equality $\operatorname{dim}(A-B)= \operatorname{dim}A-\operatorname{dim} B$ is wrong. In particular, if $\operatorname{dim} B> \operatorname{dim} A$, it yields $\operatorname{dim} (A-B)<0$, which is impossible. On the other hand, for a function $d(A, B) =\operatorname{rank}(A − B)$ all axioms of a metric are obvious, but the triangle inequality.