Symmetric and invertible matrix

[Yankel]

Hello all,

If A and B are both squared invertible matrices and A is also symmetric and:

\[AB^{-1}AA^{T}=I\]

Can I say that

\[B=A^{3}\] ?

In every iteration of the solution, I have multiplied both sides by a different matrix. At first by the inverse of A, then the inverse of the transpose, etc...Is this the correct approach to solve this ? Thank you in advance !

Another question. If A in both symmetric and invertible, it doesn't mean that the inverse of A is equal to A, right ?

 

[I like Serena]

Hello all,

If A and B are both squared invertible matrices and A is also symmetric and:

\[AB^{-1}AA^{T}=I\]

Can I say that

\[B=A^{3}\] ?

In every iteration of the solution, I have multiplied both sides by a different matrix. At first by the inverse of A, then the inverse of the transpose, etc...Is this the correct approach to solve this ? Thank you in advance !

Sure.
With $A$ symmetric and invertible, we have indeed:
$$
AB^{-1}AA^{T}=I
\quad\Rightarrow\quad AB^{-1}AA=I
\quad\Rightarrow\quad A^{-1}AB^{-1}AA=A^{-1}I
\quad\Rightarrow\quad B^{-1}AA=A^{-1} \\
\quad\Rightarrow\quad BB^{-1}AAA=BA^{-1}A
\quad\Rightarrow\quad AAA=B
\quad\Rightarrow\quad B=A^3
$$

Another question. If A in both symmetric and invertible, it doesn't mean that the inverse of A is equal to A, right ?

Nope.
That will only be the case if $A$ is either identity or a reflection (only eigenvalues $\pm 1$).

 

[Yankel]

Thanks !