Let $A\in\mathbb{C}^{m\times n}$, then a generalized inverse matrix $B$ of $A$ satisfies the following $$ABA = A \ \text{and} \ BAB = B.$$

I am to show that $B$ is unique if $A$ is square and invertible. I also want to know if my approach is correct.

From the first criterion, we get $B=A^{-1}$. My first question is then, does this imply $A=B^{-1}$?

Second, if there exists a $C\neq B$ such that $$ACA = A \ \text{and} \ CAC = C,$$

then clearly $C=A^{-1}$, and since $A^{-1}$ is unique, then $C=B$, and so we have a contradiction.

Is this a correct way to prove the generalized inverse $B$ of $A$ is unique? And can I deduce that $A=B^{-1}$ from the fact that $B = A^{-1}$? I also feel that I am not really using the second criterion ($BAB=B$) and I feel that I should.


  • $\begingroup$ If $\mathbf{A}$ is square and invertible, then $\mathbf{B}$ should be square, given $\mathbf{A}\mathbf{B}\mathbf{A}=\mathbf{A}$. Initially you said that $\mathbf{A}$ is $m\times n$ with complex entries. Regarding your question: "My first question is then, does this imply A=B−1? ", yes. But you are imposing, extra assumptions. $\endgroup$ – PabloG. Oct 2 '18 at 4:24
  • $\begingroup$ Yes, I gave the definition but then specified below that in this problem, $A$ is square and invertible $\endgroup$ – sadlyfe Oct 2 '18 at 7:24

Suppose that $B$ and $C$ are generalised inverses of $A$. In particular, we have $ABA=ACA$. Therefore, $$B=(A^{-1}A)B(AA^{-1})=A^{-1}(ABA)A^{-1}=A^{-1}(ACA)A^{-1}=(A^{-1}A)C(AA^{-1})=C.$$ Note that the equations $BAB=B$ and $CAC=C$ are not needed to derive the result.


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