0
$\begingroup$

Can $(A B A^T)^{-1}$ be simplified to $A^{-T} B^{-1} A^{-1}$? If yes, why?

Edit: Would $A^T(A B A^T)^{−1}AB^{−1}$ always give me $I$?

Assuming $B$ is invertible and $A$ and $B$ are square matrices with same dimensions.

  • 1
    depends if $ABA^T$ is invertible – 2017-02-05
  • 1
    @A.Molendijk by writing and discussing $(ABA^T)^{-1}$ in the first place assumes that it is invertible. What needs to be worried about though would be the invertibility of $A$ and $B$ since $A$ need not necessarily be square. – 2017-02-05
  • 0
    So if I had: $A^T (A B A^T)^{-1} A B^{-1}$ this would give me $I$? $B$ is a diagonal matrix and $A$ is also not singular. All matrix are square. – 2017-02-05

3 Answers 3

1

This is how inverse is defined.

If you have $(AB)^{-1}$, then that is equivalent to $B^{-1}A^{-1}$.

I suppose you need a proof?

If $A^{-1}$ and $B^{-1}$ exist, then $AA^{-1} = I$ and $BB^{-1} = I$

So, $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A = AIA^{-1} = AA^{-1} = I$

and you can do it the other way to show that $(B^{-1}A^{-1}(AB)) = I$ aswell.

So you are correct.

0

$ (ABA^T)^{-1} = ((AB)A^T)^{-1} = A^{-T}(AB)^{-1} = A^{-T}B^{-1}A^{-1}$

  • 0
    So if I had: $A^T (A B A^T)^{-1} A B^{-1}$ this would give me $I$? – 2017-02-05
  • 0
    What do you denote $A^{\color{red}{-}T}$? – 2017-02-05
  • 0
    @Bernard its the inverse of transpose. – 2017-02-05
  • 0
    @今天春天 yes it is ( where your last term is $(AB)^{-1}$) – 2017-02-05
  • 0
    what do you mean with my last term is $(AB)^{-1}$ @ChirantanChowdhury? – 2017-02-05
  • 0
    The notation $-T$ looks strange, as $T$ is not a number… – 2017-02-05
0

In the case that $A$ and $B$ are both square matrices, yes it can, but just because $ABA^T$ is invertible (and thus square) does not imply that $A$ is even square. In the case that $A$ is not square then it cannot possibly be invertible.

$\underbrace{\begin{bmatrix}1&0\end{bmatrix}}_A\underbrace{\begin{bmatrix}1\end{bmatrix}}_B\underbrace{\begin{bmatrix}1\\0\end{bmatrix}}_{A^T}=\begin{bmatrix}1\end{bmatrix}$ is invertible however $A^{-1}$ does not exist.

In the case that all are square, one can use that $(ABA^{T})^{-1}~\text{exists}\iff \det(ABA^T)=\det(A)^2\det(B)\neq 0$

$\iff \det(A)\neq 0\wedge \det(B)\neq 0\iff A^{-1}\text{ and}~ B^{-1}~\text{exist}$

which will imply $(ABA^T)^{-1}=(A^{-1})^TB^{-1}A^{-1}$