2
$\begingroup$

If $A$ is a square matrix such that it is not singular, then $(A^{-1})^{-1} = A$ How can I prove this property? I would appreciate it if somebody can help me.

$\endgroup$
  • 2
    $\begingroup$ Just use the definition of an inverse. In any group one has $(a^{-1})^{-1}=a$ for each element. $\endgroup$ – Janko Bracic Feb 23 '15 at 15:15
  • 6
    $\begingroup$ The statement that $B$ is the inverse of $A$ is that $AB=I=BA$. It is literally identical to the statement that $A$ is the inverse of $B$; there is strictly nothing to prove. (Well, one could invoke the symmetry of the equality relation, if forced to provide some argument.) $\endgroup$ – Marc van Leeuwen Feb 23 '15 at 17:05
  • 1
    $\begingroup$ That's one definition of inverse. For self-enlightenment and fun, though, relax the inverse definition to say that B is only a right-inverse of A (that is, $AB = I$) and then prove $BA = I$. For example, prove that if $Z$ has a right inverse then $ZZ = Z \implies Z = I$ (and carefully observe that if $Z$ doesn't have an inverse this doesn't hold, since $0 0 = 0$). Then prove that $BA$ is non-singular. $\endgroup$ – Steve Jessop Feb 23 '15 at 18:04
  • $\begingroup$ Janko, you actually do not need any uniqueness to prove this. In any context where some $A$ has a left-, right-, or two-sided-inverse $B$, you can always conclude that $B$ has $A$ as a right-, left-, or two-sided-inverse (respectively) from the definition. No uniqueness is required. $\endgroup$ – Kellen Myers Mar 2 '15 at 7:14
14
$\begingroup$

Of course we know that for invertible $A$, we have that there exists an invertible $A^{-1}$ such that $AA^{-1} = A^{-1}A = I.$ We also use the fact that $(AB)^{-1} = B^{-1}A^{-1}$.

$$ (AA^{-1})^{-1} = I^{-1} = I$$

$$(A^{-1})^{-1}A^{-1}= I$$

$$[(A^{-1})^{-1}A^{-1}]A = IA$$

$$(A^{-1})^{-1}(A^{-1}A) = A$$

$$(A^{-1})^{-1} = A$$

$\endgroup$
  • $\begingroup$ Great proof, so simple! $\endgroup$ – egarro Feb 23 '15 at 15:39
  • 8
    $\begingroup$ @egarro: rather funny, this is the most complicated proof among all answers and it is the only one to require the property about the inverse of a product! $\endgroup$ – Yves Daoust Feb 23 '15 at 21:05
  • $\begingroup$ Usually in textbooks that I've read, they use Uniqueness Property of Inverse of a Matrix: A*A^(-1) = I & (A^(-1))^(-1)*A^(-1) = I, so A = (A^(-1))^(-1). $\endgroup$ – Bhaskar Feb 24 '15 at 4:53
18
$\begingroup$

By the definition of the inverse $$AA^{-1}=A^{-1}A=I.$$

This reads both ways: $A^{-1}$ is the inverse of $A$, or $A$ is the inverse of $A^{-1}$.

$\endgroup$
8
$\begingroup$

By definition, $C$ is the inverse of the matrix $B=A^{-1}$ if and only if $BC = CB = I$.

Therefore, you can prove your property by showing that a product of a certain pair of matrices is equal to $I$.

$\endgroup$
  • $\begingroup$ I already know that, but i don't know how begin the proof using that. $\endgroup$ – egarro Feb 23 '15 at 15:16
  • 1
    $\begingroup$ @egarro I extended my answer. $\endgroup$ – 5xum Feb 23 '15 at 15:16
  • $\begingroup$ Can you help me showing the complete proof? $\endgroup$ – egarro Feb 23 '15 at 15:21
  • 1
    $\begingroup$ @egarro Yes, I can help you. I did. Yes, I can show you a complete proof. But I will not. I gave you a hint. I advise you to use it. You need to prove that a certain product of matrices is equal to $I$. Which two matrices should you multiply? $\endgroup$ – 5xum Feb 23 '15 at 15:27
  • 2
    $\begingroup$ @egarro There you go, it's that simple. Yes, it may be deceptively simple, but it is a pretty basic property of all groups. Sometimes, things in mathematics are really that simple. $\endgroup$ – 5xum Feb 23 '15 at 15:46
5
$\begingroup$

If $A$ is an invertible matrix, then a matrix $B$ is its inverse iff $AB=I=BA$. Since $A^{-1}A=I=AA^{-1}$, the inverse of $A^{-1}$ is $A$.

$\endgroup$
4
$\begingroup$

Suppose that there were two different inverse matrices: $B$ and $(A^{-1})^{-1}$.

Then we have $A^{-1}B = I = A^{-1}(A^{-1})^{-1}$. If we multiply by $B$ then we have:

$$BA^{-1}B = BI = BA^{-1}(A^{-1})^{-1}$$

However, $BA^{-1} = I$, so we have $B=B=(A^{-1})^{-1}$. Thus the inverse is unique.

Finally we need to find some inverse of $A^{-1}$. We know that $A A^{-1} = A^{-1} A = I$ by definition. Thus by the uniqueness above $A=(A^{-1})^{-1}$.

$\endgroup$
3
$\begingroup$

There are really three possible issues here, so I'm going to try to deal with the question comprehensively. First, since most others are assuming this, I will start with the definition of an inverse matrix. Given a matrix $X$ ($n\times n$), a matrix $Y$ ($n\times n$) is an inverse for $X$ if and only if: $$XY=YX=I_n$$

This definition says "an inverse" and not "the inverse." That is an important distinction. We could prove one or more of the following statements:

1. The matrix $A$ is an inverse of the matrix $A^{-1}$.

This is proved directly from the definition. Assuming only that some matrix $A^{-1}$ is an inverse of $A$, we have by definition ($A$ plays the role of $X$, $A^{-1}$ plays the role of $Y$): $$AA^{-1}=A^{-1}A=I$$ and by the symmetric property of equality, we may write: $$A^{-1}A=AA^{-1}=I$$ which is the definition of $A$ being an inverse of $A^{-1}$ (where the roles of $A$ and $A^{-1}$ are now reversed, so that $A$ is in the place of $Y$ and $A^{-1}$ in that of $X$).

That is the proof. You seem to be concerned that you don't know how to begin, proving this from the definition of inverse, but that's literally all there is to it. You're just parsing the definition, applying a very simple property of equality, and then parsing the definition again to draw a slightly different conclusion.

It is foolhardy to assume that a more complicated proof is better here, especially one that relies on lemmas and properties that are themselves not axioms or definitions. That's just good practice, but in fact, it might be worse than that.

Although it is not necessarily the case here, depending on how you proved the statement $(AB)^{-1}=B^{-1}A^{-1}$, you could find yourself proving these lemmas in a circular fashion. For example, the proof from Wolfram MathWorld of $(AB)^{-1}=B^{-1}A^{-1}$ seems to rely on the fact that $A$ is an inverse for $A^{-1}$ when it left-multiplies both sides by $A$ in order to knock out an $A^{-1}$. (It is also assuming inverses are unique, which is not necessary, see part 2 below.)

Of course, it is trivial to note that this rule "works both ways" so it's just as easy to "cancel out" an $A$ by right- or left-multiplying by $A^{-1}$ as it is to do the opposite, but that's the whole point of what you're trying to prove here in the first place. Longer proofs requiring more "tools" are usually not better when there is a simple alternative.

2. For any two inverses $B$ and $C$ of a matrix $A$, $B=C$.

Because you are using the notation $A^{-1}$, I am under the impression you already know this, since that notation seems to presume the uniqueness of an inverse.

This is proved fairly easily. From the definition, we have: $$AB=BA=I$$ $$AC=CA=I$$ From this, we use the transitive property of equality to obtain something like: $$AC=AB$$ and right-multiplying both sides by $A$ and applying the associative law gives $C=B$.

Thus we may now refer to the inverse of $A$, which means that we can upgrade statement 1 to say "The matrix $A$ is the inverse of the matrix $A^{-1}$."

3. The right- and left-inverses of a matrix $A$ are unique and equal.

I won't prove this, since it's very clear you don't mention left- and right-inverses, but repeating part 2 for each side proves each is unique, and a bit more work proves they are in fact equal.

$\endgroup$
0
$\begingroup$

$XA^{-1}=A^{-1}X=I$ is known to have the solution $X=A$, hence $(A^{-1})^{-1}=A$.

$\endgroup$
0
$\begingroup$

$$(A^{-1})^{-1} A^{-1}=I=AA^{-1}$$ The second factors are the same in the RHS and the LHS, and they are invertible. Hence, multiplying both sides on the right by $(A^{-1})^{-1}$ yields: $$(A^{-1})^{-1}=A$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.