# Determinant of a Hadamard Matrix as a function of n?

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:

derivation

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

Where am I going wrong?

• 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. – achille hui Mar 31 '18 at 10:44
• In the statement on determinants of block matrices. It doesn't hold. – metamorphy Mar 31 '18 at 11:19

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 $$\pm \sqrt{n}^n = \pm n^{n/2}$$.