Diagonalization of inertia tensor 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?
 A: 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.
