1. ## Proof

Hello,
For a nonsingular matrix A and a nonnegative integer p, show what (A^p)^-1 = (A^-1)^P
How to proof it.

2. Originally Posted by patrick
Hello,
For a nonsingular matrix A and a nonnegative integer p, show what (A^p)^-1 = (A^-1)^P
How to proof it.
You should also say A is a square.

$(A^p)^{-1} = (AA...A)^{-1}$
In general (even for a group) for matrices (invertible):
$(AB)^{-1} = B^{-1}A^{-1}$.
Use this rule repeatedly here to get,
$A^{-1}A^{-1}...A^{-1} = \left(A^{-1}\right)^p$.

3. Originally Posted by patrick
Hello,
For a nonsingular matrix A and a nonnegative integer p, show what (A^p)^-1 = (A^-1)^P
How to proof it.
$(A^p)^{-1}$ is the multiplicative inverse of the matrix $(A^p)$. So we know that
$(A^p)^{-1} \cdot (A^p) = I$

Furthermore, this inverse is unique.

So take $(A^{-1})^p$ and multiply it by $(A^p)$. If the result is I, then the proposition is true.
$(A^{-1})^p \cdot (A^p) = (A^{-1} \cdot A)^p = (I)^p = I$

Therefore etc.

-Dan