# Determinant of Square Root of Positive Define Matrix

Suppose the matrix $$A \in \mathbb{R}^{n\times n}$$ is positive definite symmetric. To begin, I want to investigate if the following equality holds

$$|\det A^{1/2}| = |\det A|^{1/2}.$$

Since $$A$$ is positive definite symmetric, then we can diagonalize it as $$A = V\Lambda V$$, where $$V = V^\intercal$$ and $$VV^\intercal = I$$, i.e. $$V$$ is a symmetric, orthogonal matrix, and $$\Lambda = \text{diag}(\lambda_1,\ldots,\lambda_n)$$ is a diagonal matrix. Then computing the left hand side gives

$$|\det A^{1/2}| = |\det(V\Lambda^{1/2}V)| = |\det(V^2)\det(\Lambda^{1/2})| = |\det{\Lambda}|^{1/2},$$ where $$\det(\Lambda^{1/2}) = \det(\Lambda)^{1/2}$$ comes from the fact that $$\Lambda$$ is diagonal, and you can pull the fraction outside the absolute value since all eigenvalues are positive. Computing the right hand side gives

$$|\det A|^{1/2} = |\det (V\Lambda V)|^{1/2} = |\det\Lambda|^{1/2},$$

so the two are the same. My question is if this is a general result even for non positive definite matrices. I know that for integer powers, the equality holds because of the property $$\det(AB) = \det(A)\det(B)$$, but I'm not sure if it will hold for fractional powers.

• Are you not overcomplicating? If $A^{1/2}$ is any matrix with the property $(A^{1/2})^2=A$ then applying det to both sides you get what you want. – Michal Adamaszek May 3 at 21:00
• @Jan that question is specifically about positive definite matrices – Ben Grossmann May 3 at 21:05

Whenever $$B$$ is a matrix such that $$B^2 = A$$, we must have $$\det(B)^2 = \det(B \cdot B) = \det(B^2).$$ Whenever the expression $$A^{1/2}$$ makes sense, we must have $$(A^{1/2})^2 = A$$. It follows from the above then that $$\det(A^{1/2})^2 = \det(A)$$, so that $$\det(A^{1/2})$$ is one of the two values $$\pm [\det(A)]^{1/2}$$.