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Imagine a part with an inertia matrix $I$. This matrix is symmetric and thus can be diagonalized. This means that a base $B$ exists where the $I$ matrix is diagonal (i.e $I=BDB^{-1}$), which is actually eigen vectors or $I$

This also means that a rotation matrix $R$ exists such that the part get its principal axes aligned with $x$, $y$ and $z$

Are $B$ and $R$ actually the same thing?

If yes, why?

If no, how can $R$ be found?

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  • $\begingroup$ Yes. When we "diagonalize" a matrix, it is precisely the same thing as rotating our basis to line up with the eigenvectors. $\endgroup$
    – knzhou
    Aug 18 '16 at 21:46
  • $\begingroup$ However, this point can sometimes get obscured in linear algebra courses that portray matrices as blocks of numbers. $\endgroup$
    – knzhou
    Aug 18 '16 at 21:46
  • $\begingroup$ Ok, but rotating the inertia tensor is not the same as rotating the part and computing its inertia tensor, right? $\endgroup$
    – Gregwar
    Aug 18 '16 at 21:58
  • $\begingroup$ No, that's exactly my point. "Rotating the inertia tensor" by taking $I \to RIR^{-1}$ is exactly the same as rotating your basis by $R$ and computing the components of $I$ in that new basis, which is the same as rotating your physical object by $R^{-1}$. (I might have dropped some inverses in there, though.) $\endgroup$
    – knzhou
    Aug 18 '16 at 21:59
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    $\begingroup$ @Kashmiri If $B$ works but isn't proper, then $-B$ will also work, and it is proper. $\endgroup$
    – knzhou
    Dec 5 '21 at 17:36
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The answer is yes.

To understand why, recall that the inertia matrix is the matrix of the linear function that maps the angular velocity vector to the angular momentum vector:

$$\vec{L} = I\,\vec{\omega}\tag{1}$$

Now rotate the co-ordinate basis, so that the components of $\vec{L}$ and $\vec{\omega}$ transform like $\vec{L}^\prime = R\,\vec{L}$ and $\vec{\omega}^\prime = R\,\vec{\omega}$. Now plug these equations (in the form $\vec{L} = R^{-1}\,\vec{L}^\prime$, $\vec{\omega} = R^{-1}\,\vec{\omega}^\prime$) into (1) to show that, in these co-ordinates, the inertia matrix must have elements given by:

$$I^\prime = R\,I\,R^{-1}\tag{2}$$

Next, we witness that $I$ is a symmetric, real matrix; therefore its eigenvalues are all real, therefore its eigenvectors are all real and:

Exercise Given that $I\,\vec{x}=\lambda_x\,\vec{x}$ and $I\,\vec{y}=\lambda_y\,\vec{y}$ for two eigenvectors $\vec{x}$ and $\vec{y}$, show that the symmetry of $I$ means that $\langle\vec{x},\,\vec{y}\rangle=\vec{x}^T\,\vec{y} = 0$ whenever $\lambda_x,\,\lambda_y$ are different.

Exercise Given the above result, show that an orthonormal transformation $R$ diagonalizes $I$. Therefore, we can find a co-ordinate rotation in (2) that indeed diagonalizes $I$.


You should be able to see that much of this reasoning applies to the matrix of any homogeneous, linear mapping between vectors, i.e. the transformation law (2) holds generally; there's nothing special about the inertia matrix. You need the matrix to be symmetric to show that you can diagonalize the matrix through a co-ordinate rotation, though.

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