From Halmos's Finite-Dimensional Vector Spaces, #7 section 51, rank and nullity. Prove that two linear operators are equivalent iff they have the same rank.
Typically when Halmos phrases exercises like that it means that they are provable, but I don't see how it's true.
What if two operators on $\mathcal{V}$ both have disjoint ranges of dimension $\rho$? They have the same rank but they are certainly not equivalent because they map any $v \in \mathcal{V}$ to different subspaces. Also, what if the two operators map to the same ranges but one is a scalar multiple of the other ($A = \alpha A$)? They aren't equivalent but they have the same rank.
Could someone explain what Halmos might mean or how this might be true?
Edit: There's no definition of equivalence in the past sections or in the exercises. The closest thing is "similarity", where $B$ and $C$ are similar if $C = A^{-1}BA$ for some linear transformation A.
Best Answer
For any linear transformation $T:X \to Y$, select a basis as follows:
Let $x_{r+1},\dots,x_{n}$ be a basis for the null-space of $T$. Extend to a basis $x_1,\dots,x_n$ of $X$.
Let $y_1,\dots,y_r$ be such that $y_i = T(x_i)$. Extend to a basis $y_1,\dots,y_m$ of $Y$.
With respect to this basis, we write the linear transformation as $$ \pmatrix{ 1\\ &1\\ &&\ddots\\ &&&1\\ &&&&0\\ &&&&&\ddots\\ &&&&&& 0 } $$ where the number of $1$s is equal to $r$, the rank of the transformation.