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The principal moments of inertia are the eigenvalues of the symmetric inertia tensor. In units of $\mathrm{kg}{\mathrm{m}}^2$, its characteristic determinant isTherefore the requested algebraic answer isAn orthonormal basis of principal axes is $(1,1,0)/\sqrt{2}$, $(1,-1,0)/\sqrt{2}$, and $(0,0,1)$; the last two span the degenerate eigenspace and may be replaced by any orthonormal basis of it.

There is a physical inconsistency in calling the supplied matrix a rigid body's inertia tensor. The moment-of-inertia triangle inequalities require each principal moment to be no greater than the sum of the other two. Here $3>1+1$. In principal coordinates, $I_2+I_3-I_1=2\int x_1^{2}dm$ cannot be negative for a nonnegative mass distribution. Thus the eigenvalue calculation is well defined, but no ordinary rigid mass distribution realizes this tensor. The positive second-moment matrix test for an inertia tensor makes the same obstruction explicit.

Choose three mutually perpendicular axes through the same point, and label coordinates $x,y,z$. The moments of inertia of any rigid body areSubtracting givesFor a thin planar lamina lying in $z=0$, the right-hand side vanishes, proving the perpendicular axis theorem:The two in-plane axes and the perpendicular axis must intersect at one point; this need not be the centre of mass. For an extended three-dimensional mass distribution, the equality generally fails. The general moment-of-inertia triangle inequalities arewith equality in the first precisely when all mass lies in the plane perpendicular to its $z$ axis, and cyclically for the others. None of these identities requires principal axes.

For a nonnegative mass distribution relative to an origin, the second-moment matrix is $\mathsf{S}=\int \mathbf{r}{\mathbf{r}}^{\mathsf{T}}dm$ and the inertia tensor is $\mathsf{I}=\mathrm{tr}(\mathsf{S})1-\mathsf{S}$. Taking the trace gives the displayed inverse formula, which must be positive semidefinite. Conversely, any such $\mathsf{S}$ can be realized by pairs of point masses along its orthogonal principal directions. Thus this is also a realizability test when total mass and dimensions are unrestricted. In principal coordinates it is equivalent to the moment-of-inertia triangle inequalities: $I_2+I_3-I_1=2S_{11}\ge 0$, with cyclic counterparts. A symmetric positive definite matrix alone need not be a physical inertia tensor.