A Hadamard matrix $H$ is a matrix with entries $\pm1$ and orthogonal columns.

Given that the matrix is nxn, I got that the determinant is $2^n\times4$. However, this is clearly not correct since the determinant of a $4\times4$ Hadamard Matrix is 16, but according to my answer it is 64.

This is how I derived my answer:


I got $\det(-2H^2)$ by noting that the determinant of block matrices is $\det(AD - BC)$.

Where am I going wrong?

  • $\begingroup$ If the column vectors of a square matrix is orthogonal to each other, then up to a sign, the determinant is the product of the lengths of the column vectors. $\endgroup$ – achille hui Mar 31 '18 at 10:44
  • $\begingroup$ In the statement on determinants of block matrices. It doesn't hold. $\endgroup$ – metamorphy Mar 31 '18 at 11:19

Your first equality, where you break apart the block matrix, isn't true.

A more straightforward way to find the determinant of an $n$ dimensional Hadamard matrix $H$ is to look at it this way:

Let the determinant of $H$ = $d$. If we divide a row by $\sqrt{n}$, the resulting matrix has determinant $\frac{d}{\sqrt{n}}$. Divide all rows by $n$, and the resulting matrix $H'$ is orthogonal (orthonormal columns) and thus has determinant $\frac{d}{\sqrt{n}^n} = \pm 1$ Thus, the determinant of $H$ is $\pm \sqrt{n}^n = \pm n^{n/2}$.


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