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  • Let $A$ be an $n\times n$ matrix. The $n\times n$ matrix $B(=A^{-1})$ is an inverse for $A$ if $AB=BA=I$.
  • Let $A$ be a $k\times n$ matrix.The $n\times k$ matrix $B$ is a right inverse for $A$ if $AB=I$.The $n\times k$ matrix $C$ is a right inverse for $A$ if $CA=I$.
  • Here $I$ is identity matrix of appropriate order.


Theorem 2: For any square matrix $A$, the following statements are equivalent:

  1. $A$ has inverse.

  2. $A$ has a right inverse.

  3. $A$ has a left inverse.

I know that

Theorem 1: If $A$ has a right inverse $B$ and a left inverse $C$, then $A$ is invertible and $B=C=A^{-1}$.

Proof: $C=CI=C(AB)=(CA)B=IB=B.$

But I can't prove Theorem 2 .


marked as duplicate by user1551, user61527, Martin Argerami, Cameron Buie, Hanul Jeon Nov 9 '13 at 6:03

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  • 1
    $\begingroup$ When you write that you can't prove Theorem 1, do you mean that you can't prove Theorem 2? If so, can you edit your question accordingly, please? $\endgroup$ – Gerry Myerson Nov 9 '13 at 4:05
  • $\begingroup$ Oh very sorry . $\endgroup$ – Silent Nov 9 '13 at 4:16

To see that Theorem 2 holds, note that $(1) \Rightarrow (2), \, (3)$ since, by the given definition of the inverse of $A$, $B$ is such if $BA = AB = I$, showing $B$ is both a left and right inverse of $A$. To see that $(2) \Rightarrow (1)$, note that, since $A$ is $n \times n$, so is $B$ such that $AB = I$; then $\det(B)$ is well-defined, as is $\det(A)$, and we have $\det(A) \det(B) = \det (AB) = I$, whence $\det(A), \, \det (B) \ne 0$. Thus $A^{-1}$ and $B^{-1}$ exist, and by $AB = I$ we have $B = IB = (A^{-1}A)B = A^{-1}(AB) = A^{-1}I = A^{-1}$. Thus $A$ has a two-sided inverse which is in fact $B$. The demonstration that $(3) \Rightarrow (1)$ is similar.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

  • $\begingroup$ Thanks a lot. Will you just explain these two things: Cheerio and Fiat Lux ? :) $\endgroup$ – Silent Nov 10 '13 at 6:32
  • 1
    $\begingroup$ @Sush: But of course! Word play. Cheerio is a British variant of the salutation "Cheers" which many folks use in MSE answers and comments, etc. Sort of means, "signing off with best regards", as in the phrase, "Cheerio, Old Chap!" Fiat Lux is latin for the Biblical phrase from Genesis, "Let there be light." The motto of Harvard, incidentally. Since I feel like enlightenment, of a sort, is what we do around here, I adopted it for my personal signature, as it were. By the way, thanks for the kind words of gratitude, and of course, for the "acceptance". Dare I say it? Fiat Lux!!! $\endgroup$ – Robert Lewis Nov 10 '13 at 6:43
  • $\begingroup$ I live in a non-English nation and thus got new knowledge(or music of words?).Thanks again. $\endgroup$ – Silent Nov 10 '13 at 9:02
  • $\begingroup$ To Sush: again, my pleasure, sir! $\endgroup$ – Robert Lewis Nov 10 '13 at 9:08

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