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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?

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    $\begingroup$ 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. $\endgroup$
    – Denis Serre
    Commented Jun 28, 2021 at 12:10

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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.

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  • $\begingroup$ 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. $\endgroup$
    – eaglebrain
    Commented Jul 1, 2021 at 12:32
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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.

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  • $\begingroup$ I see, and then use the Taylor series for $\sqrt{1+x}$? $\endgroup$
    – eaglebrain
    Commented Jun 28, 2021 at 11:02
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    $\begingroup$ Taylor series work only if $x$ is quite small, better to use a polynomial approximating $\sqrt{1+x}$ on a larger segment $\endgroup$
    – Fedor Petrov
    Commented Jun 28, 2021 at 11:13
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    $\begingroup$ 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 $M$. $\endgroup$
    – Federico Poloni
    Commented Jun 28, 2021 at 11:42
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    $\begingroup$ @FedericoPoloni oh, this is better $\endgroup$
    – Fedor Petrov
    Commented Jun 28, 2021 at 12:40
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    $\begingroup$ 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 $g(t) = tf(t^2)$. $\endgroup$
    – jlewk
    Commented Jun 29, 2021 at 11:39

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