Invertibility of inertia matrices The inertia matrix is defined, for a discrete body composed of material points $P_i$ of mass $m_i$ whose position vector $\overrightarrow{CP_i}$ with respect to the centre of mass $C$ has Cartesian coordinates $(x_i,y_i,z_i)$, as the matrix $$I:=\left( \begin{array}{ccc} \sum_im_i(x_i^2+z_i^2) & -\sum_im_i x_iy_i & -\sum_im_i x_iz_i \\ -\sum_im_i x_iy_i & \sum_im_i(y_i^2+z_i^2) & -\sum_im_i y_iz_i \\ -\sum_im_i x_iz_i & -\sum_im_i y_iz_i & \sum_im_i(x_i^2+y_i^2) \end{array} \right).$$
For a continuous body extending on a domain$V\subset\mathbb{R}^3$, with density $\rho:V\to\mathbb{R}$ and centre of mass having coordinates $(x_C,y_C,z_C)$, the inertia matrix is expressed -I do not write the argument $(x,y,z)$ of $\rho$ and synthesise $dxdydz$ as $dV$ for formatting reasons- as
$$\left( \begin{array}{ccc}
\int_V\rho[(y-y_C)^2+(z-z_C)^2]dV & -\int_V\rho\cdot(x-x_C)(y-y_C)dV & -\int_V\rho\cdot(x-x_C)(z-z_C)dV \\
-\int_V\rho\cdot(x-x_C)(y-y_C)dV & \int_V\rho[(x-x_C)^2+(z-z_C)^2]dV & -\int_V\rho\cdot(y-y_C)(z-z_C)dV \\
-\int_V\rho\cdot(x-x_C)(z-z_C)dV & -\int_V\rho\cdot(y-y_C)(z-z_C)dV & \int_V\rho[(x-x_C)^2+(y-y_C)^2]dV \end{array} \right).$$
I read that a solid body, i.e. a body not contained in a plane, has an invertible inertia matrix.
I have tried to show that $\det(I)\ne 0$ by writing $$I=\left( \begin{array}{ccc} I_{xx} & I_{xy} & I_{xz} \\I_{xy} & I_{yy} & I_{yz} \\I_{xz} & I_{yz} & I_{zz} \end{array} \right)$$ and calculated $\det(I)=I_{xx}I_{yy}I_{zz}+2I_{xy}I_{xz}I_{yz}-(I_{xx}I_{yz}^2+I_{xy}^2I_{zz}+I_{xz}^2I_{yy})$, but I cannot prove its non-nullity.
How can we see that $I$ is invertible? I am asking here because I am interested in a mathematical proof. Thank you very much for any answer!
 A: Don't know if this is the best way, but here goes: Your inertia matrix is real and symmetric. Therefore its eigenvalues must be real, and the determinant $\Delta = \omega_1 \omega_2 \omega_3$, the product of your eigenvalues. 
So find the eigenvalues of your matrix and you will find that none of them is 0, thus $\Delta$ will also be non-zero. 
A: We're going to show positive definiteness of the inertia tensor.
Moment of inertia with respect to an axis through the origin along the unit normal $n$ is 
$$I_{n}=\int d^{3}x\,\rho(x)r^{2}(x)$$
where $r^{2}(x)=\Vert x\Vert^{2}-\langle x,n\rangle^{2}$ is the distance
between the mass at $x$ and the axis. We see that $$\forall n:\;I_{n}>0$$
and we're going to show that $$I_{n}=\langle n,I(n)\rangle$$ where $I$
is the inertia tensor thus proving the positive definiteness of its
matrix.
The inertia tensor for a continuous body is given by
$$I=\int_{V}\rho(x)\left(\langle x,x\rangle^{2}E-x\otimes x\right)\, dV,$$ 
where E is the identity tensor,
$E=e_{1}\otimes e_{1}+e_{2}\otimes e_{2}+e_{3}\otimes e_{3}.$
The 2-tensors act on vectors as $(a\otimes b)(v)=a(b\cdot v)$, and the
matrix representation of $a\otimes b$ in coordinates is $ab^{T}$ which acts on
column vectors. And so we have
$$I(n)=\int d^{3}x\,\rho(x)[\langle x,x\rangle^{2}n-x\langle x,n\rangle]$$
$$\langle n,In\rangle=\int d^{3}x\,\rho(x)[\langle x,x\rangle^{2}-\langle x,n\rangle^{2}]=I_{n}.$$
