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.

How are you defining $X^{1/2}$? In general, $\det(AB)=\det(A)\det(B)$. So if $X^{1/2}\cdot X^{1/2}=X$ then $(\det(X^{1/2}))^2=\det(X)$. — Thomas Andrews, Feb 19 '15 at 22:11

score 8 · Accepted Answer · answered Feb 19 '15 at 22:13

8

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.

answered Feb 19 '15 at 22:13

Crostul

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I appreciate all of the answers but this one was first. Thanks everyone for the help. – jld Feb 20 '15 at 15:49

score 4 · Answer 2 · answered Feb 19 '15 at 22:18

4

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

answered Feb 19 '15 at 22:18

Tim Raczkowski

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score 3 · Answer 3 · edited Apr 13 '17 at 12:20

3

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.

edited Apr 13 '17 at 12:20

Community

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answered Feb 19 '15 at 22:14

Kevin Carlson

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Square root of determinant equals determinant of square root?

3 Answers3