If I have $A(AA)^{-1} A$ where $A$ is an $n \times n$ matrix, will I get the Identity matrix?

Question

If I have $A(AA)^{-1}A$ where $A$ is an $n \times n$ matrix, will I get the Identity matrix?

Also, is it possible to get a representation of the eigenvector of $A$ without actually knowing what $A$ is?

Answer 1

For any two invertible square matrices of the same size,

$$(BC)^{-1}=C^{-1}B^{-1}\implies A(AA)^{-1}A=AA^{-1}A^{-1}A\stackrel{\text{asociativity}}=II=I$$

I can't see how you can know anything about a matrix if you don't know what the matrix is, so your last question is a little strange to me.

Answer 2

Invertable matrices form a group under matrix multiplication. Whenever we have a group we have what is called the socks and shoes property, which states $(AB)^{-1}=B^{-1}A^{-1}$ for any two elements $A,B$ in the group. So yes, as long as $A$ is invertable then you will get the identity. But it is important to have invertability.

I do believe that for invertible matrices, it is implied that all of the eigenvalues are non-zero. I cant infer any more than that, unfortunately.

Answer 3

If $A^2$ is invertible, then also $A$ is: indeed, if $B=(A^2)^{-1}$, then $$ I=A^2B=A(AB) $$ and $I=BA^2=(BA)A$, so $A$ has both a left and a right inverse and therefore it is invertible and the left and right inverses coincide (standard exercise). Actually, for a square matrix existence of a left or a right inverse suffices.

Now $A(A^2)^{-1}A=ABA=(AB)A=I$, because $AB$ is a right inverse of $A$, hence also a left inverse.