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?

  • $\begingroup$ 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, 2016 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, 2016 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, 2016 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, 2016 at 21:59
  • 1
    $\begingroup$ @Kashmiri If $B$ works but isn't proper, then $-B$ will also work, and it is proper. $\endgroup$
    – knzhou
    Dec 5, 2021 at 17:36

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.