Prove $(A^T)^{-1}$ = $(A^{-1})^T$ whenever $A$ is invertible.
Prove $(A^T)^{-1}$ = $(A^{-1})^T$ for any invertible matrix $A.$
I actually don't know where to start. I do not think I can just apply index laws.
Any help is cool! Thanks.
Try taking the transpose of the equation $$AA^{-1}=I.$$
first we define a "B" matrix like below :
$$B = (A^{-1})$$
so :
$$B^T=(A^{-1})^T *$$
and we can write :
$$AB = I$$
then
$$(AB)^T=I^T \textrm{ and }
B^T A^T =I$$
From this equation we can say that :
$$B^T=(A^T)^{-1}$$
and finally from * we can write :
$$B^T=(A^{-1})^T=(A^T)^{-1}$$
As we know $AA^{-1}=I$. Now taking transpose both sides, we get $$(AA^{-1})^T = I^T$$ which implies $$[ (A^{-1})^T ](A^T) = I$$ Now multiply both sides with $[(A^T)^{-1}]$ at right side, $$[(A^{-1})^T]{(A^T)(A^T)^{-1}} = (A^T)^{-1}$$ Here $(A^T)(A^T)^{-1}$ will form identity $I$, Since we know $AA^{-1} = I$, Therefore $$(A^{-1})^T = (A^T)^{-1}$$ Hence Proved!
Alternatively: $$A^{-1}=\frac{\text{adj} (A)}{|A|}$$ Transpose: $$\begin{align}(A^{-1})^T&=\left(\frac{\text{adj}(A)}{|A|}\right)^T=\\ &=\frac{(\text{adj} (A))^T}{|A|}=\\ &=\frac{\text{adj} (A^T)}{|A|}=\\ &=\frac{\text{adj} (A^T)}{|A^T|}=\\ &=(A^T)^{-1}.\end{align}$$ The following properties of adjoint and transpose were used: $$\begin{align}(\text{adj} (A))^T&=\text{adj} (A^T);\\ (cA)^T&=c(A)^T;\\ |A|&=|A^T|.\end{align}$$