The following real $2 \times 2$ matrix has determinant $1$:


$$\begin{pmatrix}
\sqrt{1+a^2} & a \\
a & \sqrt{1+a^2}
\end{pmatrix}$$


The natural generalisation of this to a real $2 \times 2$ block matrix would appear to be the following, where $A$ is an $n \times m$ matrix:


$$\begin{pmatrix}
\sqrt{I\_n+AA^T} & A \\
A^T & \sqrt{I\_m+A^TA}
\end{pmatrix}$$


Both $I\_n+AA^T$ and $I\_m+A^TA$ are positive-definite so the positive-definite square roots are well-defined and unique.


Numerically, the determinant of the above matrix appears to be $1$, for any $A$, but I am struggling to find a proof. Using the Schur complement, it would suffice to prove the following (which almost looks like a commutativity relation):


$$A\sqrt{I\_m + A^TA} = \sqrt{I\_n + AA^T}A$$


Clearly, $A(I\_m + A^TA) = (I\_n + AA^T)A$. But I'm not sure how to generalise this to the square root. How can we prove the above?



Proof that block matrix has determinant $1$
Write the SVD of $A$, say $A=PDQ^T$ with $D$ diagonal with non-negative entries and $P\in O(n),Q\in O(m)$. Then $\sqrt{I\_n + AA^T} = P\sqrt{1+D^2}P^T$
and $\sqrt{I\_m+ A^TA} = Q\sqrt{1+D^2}Q^T$. This gives
$$
\begin{pmatrix}
\sqrt{I\_n + AA^T} & A \\ A^T& \sqrt{I\_m+A^TA}
\end{pmatrix}
=
\begin{pmatrix}
P & 0 \\
0 & Q
\end{pmatrix}
\begin{pmatrix}
\sqrt{I\_n + D^2} & D \\ D & \sqrt{I\_m+D^2}
\end{pmatrix}
\begin{pmatrix}
P^T & 0 \\
0 & Q^T
\end{pmatrix}.
$$
Up to permutation, the matrix in the middle is diagonal by block with $n$ blocks given by 2x2 matrices of the same form as in the question.


