Is Matrix $A^{-2}$ the Same as the Double Inverse, or A?

Question

I'm new to linear algebra and I couldn't find an answer to this question. If I have $A^{-2}$, is that the same as $A^{-1}×A^{-1}$ due to exponent laws, making it just the original matrix $A$? Any help would be greatly appreciated!

Answer 1

Yes, $A^{-2}=A^{-1}.A^{-1}$. However, in general $A^{-2}\ne A$. Perhaps that you have in mind the equality $\left(A^{-1}\right)^{-1}=A$.

Answer 2

$A^2 \cdot A^{-2}= I$ so it's the inverse of $A^2$, or which is the same, $(A^{-1})^2$, the square of the inverse of $A$. It's not $A$; $A$ can be written as $A^2 \cdot A^{-1}$ or $A^3 \cdot A^{-2}$, just as with ordinary numbers. Powers (also negative ones) behave like you're used to in any group. For a square matrix, if $A^{-2}$ is well-defined, so is $A^{-1}$ (which is the unique matrix such that $A \cdot A^{-1} = A^{-1} \cdot A = I$, $I$ being the identity for that dimension of matrix. From that latter fact we do have $(A^{-1})^{-1}=A$ which might be what you had in mind?

So mind you: $A^{-1} A^{-1} = A^{-2}$ is a valid exponential law, (comparable to $\frac12 \cdot \frac12 = \frac14$, which we can also write as $2^{-1} \cdot 2^{-1} = 2^{-2} = \frac{1}{2^2}$ etc.) (so adding the exponents) and so is $(A^{-1})^{-1}=A$ which corresponds to $\frac{1}{\frac{1}{2}} = 2$ or $(2^{-1})^{-1} = 2$. We multiply the two $-1$'s to $1$ in that case. Keep them well-distinguished. Multiplication of powers vs. powers of powers...