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 $(a_1^2 + \ldots a_n^2)^{2^{n-1}}$. $\endgroup$ Commented Mar 16, 2017 at 15:51
  • $\begingroup$ That's what I am getting too. Perhaps we should seek for a proof then. $\endgroup$ Commented Mar 16, 2017 at 15:52
  • 2
    $\begingroup$ 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 ... $\endgroup$ Commented Mar 16, 2017 at 19:27
  • 3
    $\begingroup$ @darijgrinberg: Perhaps you like to make your comments into an answer. $\endgroup$ Commented Mar 17, 2017 at 12:03
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Mar 23, 2017 at 21:00

1 Answer 1


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. $\square$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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