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.

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

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.
$\displaystyle (A^p)^{-1}$ is the multiplicative inverse of the matrix $\displaystyle (A^p)$. So we know that
$\displaystyle (A^p)^{-1} \cdot (A^p) = I$

Furthermore, this inverse is unique.

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

Therefore etc.

-Dan