Square root of determinant equals determinant of square root?

Question

Is it true that for a real-valued positive definite matrix $X$, $\sqrt{\det(X)} = \det(X^{1/2})$?

I know that this is indeed true for the $2 \times 2$ case but I haven't been able to find the answer or prove it myself for larger $X$.

Thanks a lot for any help.

Answer 1

Call $Y = x^{\frac{1}{2}}$. Then $\det X = \det (Y^2) = (\det Y)^2$, so $\sqrt{\det X} = |\det Y|$. If $Y$ is positive definite, then $|\det Y| = \det Y$, and so we are done.

Answer 2

If you know that $\det(AB)=\det(A)\det(B)$, then this is immediate, taking $A=B=X^{1/2}$. If you don't know that, there are several proofs suggested here.

Answer 3

I assume you mean by $X^{1/2}$ a matrix $Y$ such that $Y^2=X$, in which case $$\begin{align} \det Y^2=\det X & \implies (\det Y)^2=X\\ & \implies \det Y=\pm\sqrt{\det X)} \end{align}$$