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If you have an inertia matrix of the form
$$\begin{pmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{pmatrix}=I$$
If the matrix is symmetric ($I_{xy} = I_{yx}$, $I_{xz} = I_{zx}$, $I_{yz} = I_{zy}$) and positive definite (all its eigenvalues are positive)
How can you show the Triangle Inequalities $I_{xx} + I_{yy} \geq I_{zz}$, $I_{xx} + I_{zz} \geq I_{yy}$, and $I_{yy} + I_{zz} \geq I_{yy}$?

I don't seem to understand how these relationships work or even how one would start a derivation to show them.

Qmechanic
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Caido Phell
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1 Answers1

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The three inequalities follow from

$$I_{xx}=\sum_k m_k(y_k^2+z_k^2),$$

$$I_{yy}=\sum_k m_k(x_k^2+z_k^2),$$

and

$$I_{zz}=\sum_k m_k(x_k^2+y_k^2).$$

They do not follow from the symmetry and positive-definiteness conditions that you mention. For example, the diagonal matrix with diagonal elements 1, 3, and 5 is symmetric and has positive-definite eigenvalues but does not obey one of the triangle inequalities.

G. Smith
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  • Well if the inequalities follow from those three equations, I still don't understand how you can relate them to each other. It might be something obvious but I just can't seem to piece it together. How can you relate Ixx, Iyy, and Izz together? Is there a certain theory or other relationship that connects them? – Caido Phell Mar 09 '20 at 01:58
  • Just take an inequality, say $I_{xx}+I_{yy}\ge I_{zz}$, substitute the expressions on the right side of my equations, and cancel the common terms that appear on both sides. What’s left is an obviously true inequality. – G. Smith Mar 09 '20 at 03:47