- #1
VinnyCee
- 489
- 0
Given that [itex]A^2[/itex] is invertible, does it neccesarily mean that
[tex]\left(A^2\right)^{-1}\,=\,\left(A^{-1}\right)^2[/tex]?
I know that this is true, but I have no idea on where to even start a proof of this!
Maybe:
[tex]\left(A^2\right)^{-1}\,=\,\left(A\,A\right)^{-1}[/tex]
But how would I operate on infinite matrices (i.e. - [itex]a_{i\,j}[/itex])?
[tex]\left(A^2\right)^{-1}\,=\,\left(A^{-1}\right)^2[/tex]?
I know that this is true, but I have no idea on where to even start a proof of this!
Maybe:
[tex]\left(A^2\right)^{-1}\,=\,\left(A\,A\right)^{-1}[/tex]
But how would I operate on infinite matrices (i.e. - [itex]a_{i\,j}[/itex])?
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