15
$\begingroup$

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?

$\endgroup$
2
  • 2
    $\begingroup$ 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. $\endgroup$ Jun 28 at 12:10
  • 1
    $\begingroup$ Is this really appropriate for MathOverflow? Or is it better suited for Mathematics.SE? $\endgroup$ Jun 29 at 20:31
24
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
    Jul 1 at 12:32
21
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ I see, and then use the Taylor series for $\sqrt{1+x}$? $\endgroup$
    – eaglebrain
    Jun 28 at 11:02
  • 2
    $\begingroup$ Taylor series work only if $x$ is quite small, better to use a polynomial approximating $\sqrt{1+x}$ on a larger segment $\endgroup$ Jun 28 at 11:13
  • 13
    $\begingroup$ 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 $M$. $\endgroup$ Jun 28 at 11:42
  • 3
    $\begingroup$ @FedericoPoloni oh, this is better $\endgroup$ Jun 28 at 12:40
  • 2
    $\begingroup$ 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
    Jun 29 at 11:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.