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.

  • $\begingroup$ 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. $\endgroup$ – Michal Adamaszek May 3 at 21:00
  • $\begingroup$ @Jan that question is specifically about positive definite matrices $\endgroup$ – 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 $\pm [\det(A)]^{1/2}$.

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