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Does a matrix have only one inverse matrix (like the inverse of an element in a field)? If so, does this mean that

$A,B \text{ have the same inverse matrix} \iff A=B$?

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5 Answers 5

up vote 17 down vote accepted

More generally, in any situation where the associative law holds, if some $x$ has both a left-inverse $l$ and a right inverse $r$, then $l=r$. The reason is that $l=l(xr)=(lx)r=r$. In particular, if $x$ has a $2$-sided inverse, then that's unique. On the other hand, it is entirely possible for some $x$ to have many different left-inverses if it has no right-inverse. It is also possible for some $x$ to have many right-inverses if it has no left-inverse. Both of these possibilities actually happen in the case of non-square matrices.

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(+1) It can also happen if instead of matrices we consider linear transformations $V\to V$. If $V$ is infinite-dimensional, then there are transformations with many left-inverses and no right-inverse, or vice versa. –  Henning Makholm Jul 27 at 15:26

If $A$, $B$ are square matrices with same inverse $C$, then $AC=CA=I$ and $BC=CB=I$. Therefore, $$ A =AI= A(CB)= (AC)B = IB = B. $$ The odd thing about matrices is this: If $A$, $B$ are $n\times n$ matrices over a field, then $AB=I$ iff $BA=I$. This is a direct consequence of the fact that the $N\times N$ matrices form a finite-dimensional linear space.

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Note that $GL(n, \mathbb{F})$, the set of invertible $n\times n$ matrices over the field $\mathbb{F}$, is a group. In any group, inverses are unique, so if $a^{-1} = b^{-1}$, by taking inverses it follows that $a = b$. In particular, this applies to the group $GL(n, \mathbb{F})$.

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What GL stands for? –  gbox Jul 27 at 14:58
    
@gbox the set of all the invertible matrices –  Ale Jul 27 at 15:04
    
General linear. The group $GL(n, \mathbb{F})$ is often called the general linear group. –  Michael Albanese Jul 27 at 15:07
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But how do we know that $GL(n,\mathbb F)$ is a group in the first place? That's because we already know matrix inverses behave like they ought to. –  Henning Makholm Jul 27 at 15:11
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@HenningMakholm: In verifying that $GL(n, \mathbb{F})$ is a group, we do not need to prove that if $A^{-1} = B^{-1}$ then $A = B$. My point was that this implication is true in any group, in particular $GL(n, \mathbb{F})$. –  Michael Albanese Jul 27 at 15:29

Yes, it is unique. To show this, assume a matrix $A$ has two inverses $B$ and $C$, so that $AB=I$ and $AC=I$. Therefore $AB=AC \implies BAB=BAC \implies B=C$. So the inverse is indeed unique. For the second question, note that $(A^{-1})^{-1}=A$ so that if $A$ and $B$ both have inverse $A^{-1}$, then $A^{-1}$ has a unique inverse as well. Since $A$ and $B$ are both inverses, therefore $A=B$.

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Your calculation depends on $B$ being a two-sided inverse (or at least on $A$ having some left inverse), which seems to deserve a proof of its own. (E.g., here) –  Henning Makholm Jul 27 at 15:08
    
(Ignore the link in the above comment. It assumes that every right inverse has a right inverse itself). –  Henning Makholm Jul 27 at 15:17

Specific counterexample to the non-square case: Let $A=\left(\begin{smallmatrix}1&0&0\\0&1&0\end{smallmatrix}\right)$.

Then there is a matrix $B$ such that $AB=I$, namely $B=\left(\begin{smallmatrix}1&0\\0&1\\x&y\end{smallmatrix}\right)$. Note that $x,y$ can each be anything, so this right inverse is not unique. Also note that $I$ is $2\times 2$.

On the other hand, there is no matrix $C$ such that $CA=I$, where now $I$ would have to be $3\times 3$. The reason is that the last column of $A$ is all zeroes, so the last column of $CA$ would be all zeroes as well. Thus $A$ has no left inverse at all.

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