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I'm computing the principal axes and moments of a triangle at its centroid. The triangle vertices are at the points $(12,0,0)$, $(0,24,0)$, $(0,0,36)$. According to my calculations, the relevant characteristic equation is $\lambda^3 - 112\lambda^2 + 2352\lambda = 0$. This has roots 84, 28, 0. It's the zero eigenvalue that's bothering me. The values 84 and 28 are moments of inertia about axes lying in the plane of the triangle. So, according to the perpendicular axis theorem, the moment of inertia around the axis perpendicular to this plane ought to be $84 + 28 = 112$. So, why is $112$ not a root of the characteristic polynomial ?? Puzzled.

If it helps -- the characteristic polynomial is derived from the matrix (the so-called "inertia tensor") whose rows are $[104,8,12]$, $[8,80,24]$, $[12,24,40]$. It's symmetric, obviously.

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    Well, not a lot of time was wasted! I just put your matrix into Mathematica and found the characteristic polynomial $\det(\lambda I - A) = \lambda^3 - 224\lambda^2 + 14896\lambda - 263424$. This has roots $28$, $84$, and $112$, like you expect.2012-05-20

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The characteristic polynomial of $\mathbf A = \begin{bmatrix}104 & 8 & 12 \\ 8 & 80 & 24 \\ 12 & 24 & 40\end{bmatrix}$ is $\det(\lambda \mathbf 1_{n\times n} - \mathbf A)$, which according to Mathematica is equal to $\lambda^3 - 224\lambda^2 + 14896\lambda - 263424.$ This has roots $28$, $84$, and $112$ like you expect. Perhaps you mixed up two matrices when you were working it out.

Conceptually, the moment of inertia tensor is "morally" continuous in some physically intuitive sense, so there shouldn't be any reason to worry that it won't work for planar objects -- you can think of them as the limit of thinner and thinner solid bodies. On physical grounds, the only way you can get a zero moment about some axis is if the body is an infinitesimally thin rod lying along that axis. Any other object requires energy to spin it about the axis, and rotational kinetic energy is $\frac12 \mathbf\omega^T\mathbf I\mathbf\omega$ for angular velocity $\mathbf\omega$, so $\mathbf I$ cannot have a zero eigenvector there.

In general, though, I find it easier to think about the tensor $\mathbf \Sigma = \int \mathbf r\mathbf r^T \mathrm dV$, which a statistician might call the covariance matrix, because its eigenvalues correspond to how "stretched out" the body is in a given direction. The moment of inertia can be easily derived from this as $\mathbf I = \mathbf 1_{3\times3}\operatorname{tr} \mathbf \Sigma - \mathbf\Sigma$, but $\mathbf\Sigma$ corresponds more closely with my intuition: it scales naturally with the body under nonuniform scaling, while the moment of inertia tensor does not.

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    I mixed up the inertia tensor and the "covariance" tensor. The characteristic polynomial I cited actually belongs to the latter. Bone-head mistake. Sorry.2012-05-20