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Suppose that A is an n by n Matrix and there exist matrix B such that BA=Identity. Show that A is invertible! My attempt is to use linear mapping associated with a matrix L_A (x)=Ax and L_AB (x)=L_A L_B

Please help.

dyyyyssss
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3 Answers3

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$1=\det(AB)=\det(A)\det (B)\implies \det A\neq 0$

Asinomás
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Without invoking determinants, assume that there exists a square matrix $B \in M_n(\mathbb{F})$ such that $BA = I_n$. By considering the linear maps associated to $B,A$, we get $L_B \circ L_A = \operatorname{id}|_{\mathbb{F}^n}$. This implies that $L_A$ is one-to-one (for $L_A(\vec{x}) = \vec{0} \implies \vec{x} = L_B(L_A(\vec{x})) = L_B(\vec{0}) = \vec{0}$) and then, by the rank-nullity theorem, $L_A$ is also onto. .

dyyyyssss
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levap
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  • how do you know that L_B is the inverse of L_A? – dyyyyssss Oct 11 '16 at 03:52
  • @dyyyyssss: Prove the following: If $f \colon X \rightarrow Y$ and $g \colon Y \rightarrow X$ are functions, $f$ is one-to-one and onto and $g \circ f = \operatorname{id}|{X}$ then $f \circ g = \operatorname{id}|{Y}$. – levap Oct 11 '16 at 03:55
  • I think I got it! Thanks so much – dyyyyssss Oct 11 '16 at 03:57
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Let $BA=I$. By multiplying from the right by elementary matrices I can always column reduce $A$ to, say, $A'$ and get $BA'=P$. $P$, as a product of elementary matrices is invertible and hence has no zero column. This implies that $BA'$ has no zero column, which implies that $A'$ has no zero column, hence $A'=I$. This means that $P$ is the right inverse of $A$, and $AP=I$. Finally, $B=BI=BAP=IP=P$ and hence $A$ is invetible.

Artem
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  • Can I ask why the downvote? – Artem Oct 11 '16 at 14:06
  • yes: Your answer suggests that any time you multiply a matrix $A$ on the right by one or more elementary matrices, to get some other matrix $A'$, then $A'=I$. That is clearly not true so I felt no explanation was needed. The reasoning "$A'$ has no zero column and hence $A'=I$" does not make sense. Lots of non-identity matrices have no all-zero column. – Michael Oct 11 '16 at 18:38
  • I suppose your answer could be fixed if you assume you keep multiplying by elementary matrices to reduce $A$ to some special structure like "row echelon" (or perhaps "column echelon"?) form and you assert something about the structure of that form. You likely intended that, but it was not clear to me and the answer seemed to make strange claims. – Michael Oct 11 '16 at 18:44
  • @Michael Yes, of course, when I write to multiply by elementary matrices I mean to put in into "reduced column echelon form" It is an elementary exercise to prove that for a square matrix it is either an identity or has a column of zeros. – Artem Oct 11 '16 at 19:17