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Prove if $A$, $B$, and $C$ are square matrices and $ABC = I$, then $B$ is invertible and $B^{-1}= CA$.

I know that this proof can be done by taking the determinant of $ABC=I$ and showing that $A$, $B$, and $C$ are invertible and then finding the inverse of $B$. However, in this chapter of the book, we have not yet learned determinants so I would like to solve the problem without determinants. My proof method involves using a contradiction and is as follows:

Assume ${C^{-1}}$ does not exist, then $\exists$ $x$ $\neq$ $0$ such that $Cx = 0$.
$ABCx =Ix$

$AB0 = x$

$0=x$, which is a contradiction since we know that $x$ $\neq$ $0$, and therefore ${C^{-1}}$ exists.

$AB$${C^{-1}}$ $=I$${C^{-1}}$

$AB=$${C^{-1}}$

WLOG, B is invertible

$CAB =C$${C^{-1}}$

$CAB = I$

$CAB$${B^{-1}}$ $=I$${B^{-1}}$

$CA=$${B^{-1}}$

My question is if it is correct to assume ${C^{-1}}$ does not exist since the proof does not mention anything about ${C^{-1}}$ existing or not.

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  • $\begingroup$ possible duplicate? If ABC non-singular prove that A, B and C non-singular too. $\endgroup$ – AccidentalFourierTransform Mar 25 '18 at 14:08
  • $\begingroup$ What do you know about inverses, at that point in the book? For instance, what definition and/or characterizations of inverses and invertible matrices do you have available? $\endgroup$ – Federico Poloni Mar 25 '18 at 20:05
  • $\begingroup$ @FedericoPoloni I know An n × n matrix A is invertible when there exists an n × n matrix B such that AB = BA = I and if A is an invertible matrix, then the system of linear equations Ax = b has a unique solution x = A^(-1)b. I used the second fact in my proof, where I made vector b a zero vector. So at this point in the book, I know nothing of elementary matrices or determinants. $\endgroup$ – Jay Mar 25 '18 at 21:10
  • $\begingroup$ @Jay Do you know that checking only one of those two equations ($AB=I$, for instance) is sufficient for invertibility, or do you have to check both conditions? Most of the answers build on this fact. $\endgroup$ – Federico Poloni Mar 26 '18 at 6:35
  • $\begingroup$ @FedericoPoloni Yes, there is a proof in the textbook that shows this theorem $\endgroup$ – Jay Mar 26 '18 at 23:51
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It can be shown, via elementary methods, that if $M$ and $N$ are square matrices such that $MN = I$, then $NM= I$.

Thus, if $ABC = A(BC) = I$, then $(BC)A = B(CA) = I$, which shows that $B$ is invertible and $B^{-1}=CA$.

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A square matrix $A$ is invertible if and only if there is another matrix $A^{-1}$ such that $A^{-1}A=I$. In the express$ABC=I$, chose $X=AB$ and we have $XC=I$. Thus $C^{-1}=X$. Similarly show $A$ is invertible. Now, $$ABC=I$$ $$AB=C^{-1}$$ $$CAB=I$$ $$CA=B^{-1}$$

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  • $\begingroup$ The first statement is only true for square matrices, since $A^{-1}A=I$ implies $AA^{-1}=I$. However, for matrices in general, it should be "A matrix $A$ is invertible iff. there is another matrix $A^{-1}$ which satisfies $A^{-1}A=AA^{-1}=I$. $\endgroup$ – Erlend Graff Mar 25 '18 at 12:53
  • $\begingroup$ @ErlendGraff noted and edited! $\endgroup$ – Prathyush Poduval Mar 25 '18 at 14:15
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A square matrix is invertible if and only if its rank is $n$.

Also, we know that $\operatorname{rank}(AB) \le \min(\operatorname{rank}(A),\operatorname{rank}(B) )$

In this question

$$ABC=I$$

Hence $\operatorname{rank}(ABC)=n$

$$n \le \min(\operatorname{rank}(A),\operatorname{rank}(B), \operatorname{rank}(C) )$$

Hence $\operatorname{rank}(A)=\operatorname{rank}(B) =\operatorname{rank}(C)=n$ and they are all invertible.

Hence $B=A^{-1}C^{-1}$ and $B^{-1}=(A^{-1}C^{-1})^{-1}=CA$

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    $\begingroup$ +1 for no-nonsense proof that they must all be invertible. (Another way to phrase this: if one of the matrices were noninvertible, then there would have to be some subspace which is at some point in the composition chain mapped to zero, but then $ABC$ couldn't be the identity.) — However the question did also ask about $B^{-1} = CA$, which can't be shown that easily. $\endgroup$ – leftaroundabout Mar 25 '18 at 15:51
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Since, $ABC = I$, then $A$ is invertible (see below). Same with $C$ since $(AB)C=I$. The inverse of $C$ is $AB$. So, $B = A^{-1}C^{-1}$ and $BCA= I$. So, the inverse of $B$ is $CA$.

Proof-sketch of AB=I $\implies$ A and B are invertible and they are inverses of each other

Consider the linear transformations induced by matrices $A$ and $B$ on $\mathbb R^n$. For, convenience I will represent the corresponding linear transformations by $A$ and $B$ as well. Since $AB=I$, it implies the composition of linear transformations $A$ and $B$ is the identity linear transformation. So, $B$ is an injective linear transformation and $A$ is a surjective linear transformation. Since, we are in a finite-dimensional vector space this implies both $A$ and $B$ are invertible linear transformation (by rank-nullity theorem). So, $A$ and $B$ are invertible as matrices. Also not that since $AB=I$, we get $BABA=BA$. Now by multiplying inverse of $A$ and $B$ from right on both sides, we get $BA=I$ as well. Now, from the definition inverse of matrices, we get $A$ and $B$ are inverses of each other.

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  • $\begingroup$ The first statement is not just the definition of invertible matrices. It requires a non-trivial proof. $\endgroup$ – Adayah Mar 25 '18 at 11:57
  • $\begingroup$ I edited my answer. $\endgroup$ – grontim Mar 25 '18 at 14:17

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