Proving two linear operators are equivalent iff they have the same rank

Question

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.

Answer 1

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.

Answer 2

The linear transformations $A$ and $B$ are equivalent if there exist invertible transformations $P$ and $Q$ so that (Halmos page 87, Exercise 6, (b)): $$ A = P^{-1}BQ. $$

If this is the case, then the rank of $A$ and $B$ are the same (Halmos, $\S50$, THM3, (10)): $$ \rho(A) = \rho(P^{-1}BQ) = \rho(P^{-1}B)=\rho(B). $$

To prove the converse, we do the following. Suppose $k=\rho(A)=\rho(B)$ and that we are working in a vector space $V$ with dimension $n$. Let $\{x_{k+1},\dots,x_n\}$ be a basis of the null space of $A$. Add vectors $\{x_1,\dots,x_k\}$ to get a basis of $V$. Do the same thing with $B$ to get a basis $\{y_1,\dots,y_k,y_{k+1},\dots,y_n\}$ of V, in which $\{y_{k+1},\dots,y_n\}$ is a basis of the null space of $B$.

Define the linear map $Q$ by: $$ Qx_i = y_i, $$ $i=1,\dots,n$. It is clearly linear. It is also easy to show that it is invertible (show that the null space of $Q$ is the zero subspace of $V$).

Now we will define $P$. Set: $$ z_i = Ax_i, $$ and $$ w_i = By_i, $$ for $I=i,\dots,k$. These sets span the ranges of $A$ and $B$, respectively. We stop at $k$ because for $i=k+1,\dots,n$ the vectors $x_i$ and $y_i$ are mapped to zero. We add to each set $n-k$ linear independent vectors to make two new bases of $V$, say $\{z_1,\dots,z_k,z_{k+1},\dots,z_n\}$ and $w_1,\dots,w_{k},w_{k+1},\dots,w_n$, respectively. Make the linear, invertible map: $$ w_i = Pz_i. $$

We are done. Observe that: $$ PAx_i = Pz_i = w_i = By_i = BQx_i. $$