We present some variation on this MO question which is equally amusing to me.

Define the $2^{n-1}\times 2^{n-1}$ matrix $A_n$ recursively as follows: $A_1(a_1)=\begin{pmatrix} a_1\end{pmatrix}$ and $$A_n(a_1,\dots,a_n)=\begin{pmatrix} A_{n-1}(a_1,\dots,a_{n-1})& a_nJ_{n-1}\\ -a_nJ_{n-1}&A_{n-1}(a_1,\dots,a_{n-1}) \end{pmatrix}.$$ Here $J_n$ is a $2^{n-1}\times 2^{n-1}$ matrix with $1$'s on the antidiagonal and zeros elsewhere.

Example. For $n=2$ and $n=3$, we have $$A_2(a_1,a_2)=\begin{pmatrix} a_1&a_2\\-a_2&a_1\end{pmatrix} \qquad\text{and} \qquad A_3(a_1,a_2,a_3)=\begin{pmatrix} a_1&a_2&0&a_3\\-a_2&a_1&a_3&0\\0&-a_3&a_1&a_2\\-a_3&0&-a_2&a_1\end{pmatrix}.$$

Question. Is there a closed (or interesting) formula for the determinant $\det(A_n)$?

  • $\begingroup$ Looks like $(a_1^2 + \ldots a_n^2)^{2^{n-1}}$. $\endgroup$ – Robert Israel Mar 16 '17 at 15:51
  • $\begingroup$ That's what I am getting too. Perhaps we should seek for a proof then. $\endgroup$ – T. Amdeberhan Mar 16 '17 at 15:52
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    $\begingroup$ Ah! There is a known trick that works for quaternions, and that also works here. Your matrices $A_n$ are "almost orthogonal": They satisfy $A_n \left(A_n\right)^T = \left(a_1^2+a_2^2+\cdots+a_n^2\right) I_{2^{n-1}}$. (You can prove this by induction, using the fact that $A_n J_n = J_n \left(A_n\right)^T$, which you can also prove by induction.) Taking determinants, we obtain $\left(\det\left(A_n\right)\right)^2 = \left(a_1^2+a_2^2+\cdots+a_n^2\right)^{2^{n-1}}$. Now, it ... $\endgroup$ – darij grinberg Mar 16 '17 at 19:27
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    $\begingroup$ @darijgrinberg: Perhaps you like to make your comments into an answer. $\endgroup$ – T. Amdeberhan Mar 17 '17 at 12:03
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    $\begingroup$ @T.Amdeberhan: I wish I had the time to write the long-ish expository answer that this question deserves... As for your second question, see math.stackexchange.com/questions/706513/… (Kaladin's proof is more complicated than necessary). I believe this matrix is something like a multiplication matrix by the quaternion $a+bi+cj+dk$, although I have not checked. The whole thing seems to be connected with the notion of a reduced norm in a central simple algebra, but I am not sure how far this connection is worth pursuing. $\endgroup$ – darij grinberg Mar 23 '17 at 21:00

Finally, here is an argument that I came up with.

We proceed with induction on $n$. The base case $n=2$ is obvious. So assume $n\geq3$. Denote $\widehat{A_n}=A_nJ_n, \widehat{B_n}=B_nJ_n$ and let $I_n$ be the $2^{n-1}$-dimensional identity matrix. It is immediate that $$\widehat{A_n}=\begin{pmatrix} a_nI_{n-1}&\widehat{A}_{n-1}\\ \widehat{A}_{n-1}&-a_nI_{n-1} \end{pmatrix}.$$ An important difference surfaces: now the two bottom pair of block-matrices \it commute. \rm As a result, $$\begin{align} \det A_n&=\det J_n\det\widehat{A}_n=\det(-a_n^2I_{n-1}-\widehat{A}_n^2)=(-1)^{2^{n-2}}\det(a_nI_{n-1}\pm i\widehat{A}_n).\end{align}$$ The proof follows. $\square$


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