# Proof that block matrix has determinant $1$

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?

• This is a well-known property. Actuall, with $\sqrt{1-x^2}$ instead of $\sqrt{1+x^2}$ (without difference in the calculations), this appears in the proof of Von Neumann's inequality. See the book by Nagy, Foias, Bercovici and Kérchy. Jun 28, 2021 at 12:10
• Is this really appropriate for MathOverflow? Or is it better suited for Mathematics.SE? Jun 29, 2021 at 20:31

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.
• Thank you. This also clarifies the difficulty with generalising this construction to an $N\times N$ block matrix, unless we can do a simultaneous SVD of all the off-diagonal blocks. Jul 1, 2021 at 12:32
We have $$Af(A^TA)=f(AA^T)A$$ for any reasonable function $$f$$, including $$f(x)=\sqrt{1+x}$$. This suffices to check for $$f(x)=x^k$$ when it is obvious, then approximate your function by a polynomial.
• I see, and then use the Taylor series for $\sqrt{1+x}$? Jun 28, 2021 at 11:02
• Taylor series work only if $x$ is quite small, better to use a polynomial approximating $\sqrt{1+x}$ on a larger segment Jun 28, 2021 at 11:13
• An alternative way to conclude without using polynomial approximations: for each function $f$ and each square matrix $M$ there is a polynomial $p$ (depending on $M$) such that $f(M)=p(M)$: it is the (Hermite) interpolating polynomial of $f$ on the eigenvalues of $M$. Jun 28, 2021 at 11:42
• Alternatively, given the SVD $A=PDQ^T$ with $D$ diagonal, both $Af(A^TA)$ and $f(AA^T)A$ reduce to $P g(D) Q^T$ where $g(t) = tf(t^2)$. Jun 29, 2021 at 11:39