Prove that if AB=I_m, then rk(A)=rk(B).

Question

If $A\in\mathfrak{M}_{m,n}(F)$ and $B\in\mathfrak{M}_{n,m}(F)$, prove that $AB=I_m$ implies $rk(A)=rk(B)$.

Here $A\in\mathfrak{M}_{m,n}(F)$ means $A$ is a ($m \times n$)-matrix with entries in $F$.

My approach:

I tried to consider $A$, $B$ as linear maps, $L_A:F^n\to F^m$, $L_B:F^m\to F^n$.

And we know $rk(A)=\dim{\mathrm{Im}~L_A}$, $rk(B)=\dim{\mathrm{Im}~L_B}$.

Then by the Dimension Theorem, it suffices to show that $n-\dim{\ker{L_A}}=m-\dim{\ker{L_B}}$.

However, I don't know how to prove this.

Answer 1

Consider the linear maps, $L_A:F^n\to F^m$ and $L_B:F^m\to F^n$. Note that $AB=I_m$ means $L_A \circ L_B : F^m \to F^m$ is an isomorphism. Therefore, $L_B$ must be an injection and $L_A$ must be a surjection. Combining all this information, we have

$rank (A) =m$ by the surjectivity of $L_A,$

and $m = rank(B)$ by the injectivity of $L_B.$

This proves the claim.

Answer 2

We have $L_B : \mathbb F^m \rightarrow \mathbb F^n$. No linear map can increase the dimension of a space, so $\text{rank}(B) \le m$. Furthermore, $m = \text{rank}(AB) \le \text{rank}(B) \le m$, hence $\text{rank}(B) = m$.

Conversely, $L_A : \mathbb F^n \rightarrow \mathbb F^m$, so $\text{rank}(A) \le m$. Also, $\text{rank}(A) \ge \text{rank}(A_{|\text{Im}(L_B)}) = \text{rank}(AB) = m$, so $\text{rank}(A) \ge m$ and thus $\text{rank}(B) = m = \text{rank}(A)$.