# Thread: Inverse of a Squared Matrix

1. ## Inverse of a Squared Matrix

This was a question on one of my prelims for linear algebra that I'm reviewing for the final:

If A has an inverse, then so does A squared and the inverse of A squared = A inverse, squared:

(A^(2))-1 = (A^(-1))2

I started to prove that A squared has an inverse but didn't make much headway. What should I have done?

2. Use the definition of inverse.

The inverse of A^2, call this B, has the property B * A^2 = I
Now let B = (A^-1)^2

3. $\displaystyle (A*A)^{-1}=A^{-1}*A^{-1}=(A^{-1})^2$

4. reposted in proper formal below, sorry.

5. Originally Posted by dwsmith
$\displaystyle (A*A)^{-1}=A^{-1}*A=I$
$\displaystyle (A*A)^{-1}= A^{-1}* A^{-1}$?

6. Ah, yes. Sorry.

$\displaystyle (A*A)^{-1}=A^{-1}*A^{-1}=(A^{-1})^2$

7. I corrected the original post too.

8. $\displaystyle (A*A)^{-1}=A^{-1}*A^{-1}=(A^{-1})^2$ ....

$\displaystyle (A^{2})^{-1}= (A*A)^{-1}= A^{-1}*A^{-1}=(A^{-1})^2$

Is that all?? Still confused about the first part, a proof that if A is invertible, so is A squared. Also, not seeing how the Identity works here..

9. Our goal is to show that both sides are equal. Once we have shown that we are done.

10. Always, if you already have a guess for the inverse, just multiply the inverse guess by the original matrix, if the result is the identity matrix, then you have shown conclusively it is the inverse. A more low level proof would be something like this:

We know that A^{-1} exists.
This means that A^{-1} * A^{-1} = (A^{-1})^2 exists.

Now we have
(A^2) * (A^{-1})^2 = A * A * A^{-1} * A^{-1} = A * I * A^{-1} = A * A^{-1} = I

Thus, (A^{-1})^2 satisfies the definition of the inverse of A^2 or (A^2)^{-1}
so (A^2)^{-1} = (A^{-1})^2

(Every thing posted before is correct, but I just thought it also good to provide a proof using basic definitions)