Measure of an ellipsoid I'm looking to simplify the set {$xMx<1$}, where x is a vector in $R^n$ and $M$ is a positive definite matrix, in order to measure it. Can somebody explain to me a way to rewrite the linear relation (liner algebra is not my strong point) so that the resultant set is easily measurable (in terms of M and the measure of the unit ball).

Any positive definite symmetric matrix $M$ can be written as $M = A^tA$ with some invertible matrix $A$ with positive eigenvalues, and then the inequality becomes $\| Ax \| < 1$, and the set of solutions is the image of the unit ball $\| y \|<1$ under the inverse of the matrix $x = A^{-1} y$. The change of volume under linear maps is given by the determinant, so the volume of the ellipsoid is $c_n\det A^{-1} = \frac{c_n}{\det A} = \frac{c_n}{\sqrt{\det M}}$, where $c_n$ is the volume of the $n$-dimensional unit ball.

Oops, you don't assume that the matrix is symmetric. In that case, the ellipsoid is described by the equation $x^t N x$ with $N = \frac12 (M+M^t)$ symmetric and positive definite, so you get $ \frac{c_n}{\sqrt{\det N}}$