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If I have an $m \times n$ matrix $A$ and an $n \times m$ matrix $B$ such that $AB=I_m$, how do I go about calculating the rank of $A$ and the rank of $B$? Any clues would be much appreciated!

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If $A$ is of order $m \times n$ and $B$ is of order $n \times m$, and $AB$ is nonsingular of order $m$, then both $A$ and $B$ are of full rank, i.e.,

$$\operatorname{rank} A = \operatorname{rank} B = m.$$

Furthermore, $m \le n$. If $m = n$, both $A$ and $B$ are nonsingular and, since $AB = I$, we see that $B = A^{-1}$.

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  • $\begingroup$ Thank you. Can you please explain as to how you got to this solution? $\endgroup$ – AYR Aug 10 '13 at 15:42
  • $\begingroup$ See here. $\endgroup$ – Vedran Šego Aug 10 '13 at 15:50
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A good way to think about it is to consider the subspace that results after $\mathbb{F}^m$ is transformed first by $B$, and then by $A$. Since $AB = I_m$, we have $AB(\mathbb{F}^m) = \mathbb{F}^m$. At no point in that process can $\mathbb{F}^m$ be multiplied by a matrix that has rank $< m$, because there is no way to get an $m$-dimensional subspace out of a lower dimensional subspace by multiplying by a matrix...once you're down to $m-1$ dimensions, you can't get back up.

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Let $S$, $T$ and $I$ be the linear transformation corresponding to matrices $A$, $B$ and $I_m$ respectively. Then, $S: \mathbb{R}^n \to \mathbb{R}^m$, $T: \mathbb{R}^m \to \mathbb{R}^n$ and so $I = S \circ T: \mathbb{R}^m \to \mathbb{R}^m$.

So we have $\mathbb{R}^m = I(\mathbb{R}^m) = (S \circ T)(\mathbb{R}^m) = S(T(\mathbb{R}^m)) = S(R(T))$ where $R(T) \subset \mathbb{R}^n$ is the range of $T$. So $S(R(T)) = \mathbb{R}^m$.But $R(T) \subset \mathbb{R}^n \implies S(R(T)) \subset S(\mathbb{R}^n) \subset \mathbb{R}^m$. Since $S(R(T)) = \mathbb{R}^m$, we must have $R(S) = S(\mathbb{R}^n) = \mathbb{R}^m$ which proves $\operatorname{rank}(A) = \operatorname{rank}(S) = m$.

As a side note, by rank-nullity theorem we must have $0 \leq \operatorname{nullity}(S) = n - \operatorname{rank}(S) = n - m \implies n \geq m$. When $n = m$, $S$ is of course invertible and $T = S^{-1}$ and also $\operatorname{rank}(B) = \operatorname{rank}(T)= \operatorname{rank}(S^{-1}) = m$.

Now in general, $x \in N(T) \implies x \in N(S \circ T) = N(I) = \{0\} \implies x = 0$, so $\operatorname{nullity}(T) = 0$ (where $N(T)$ is the null space of $T$). So by rank-nullity theorem, we have $\operatorname{rank}(T) = m$ and hence $\operatorname{rank}(B) = m$.

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