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# Spectral Theory and the Inertia Tensor

The general moment of inertia tensor $${ I }_{ \alpha \beta }$$ is represented by a $$3\times 3$$ real-symmetric matrix. Show that there exist an orthogonal matrix $$P$$ and diagonal matrix $$D$$ such that $D = { P }^{ T }{ I }_{ \alpha \beta }P.$

Furthermore, deduce that ${ I }_{ \alpha \beta } = PD{ P }^{ T }.$

Solution

We first prove a general property in spectral theory: If $$A$$ is a Hermitian matrix, there exists a unitary matrix $$P$$, the conjugate transpose of the unitary matrix $${P}^{*}$$ and a diagonal matrix $$D$$ such that $${P}^{*}AP =D$$.

We begin by finding $$P$$ such that $${P}^{*}AP = U$$, where $$U$$ is an upper triangular matrix. Since Hermitian matrices follow the property $${A}^{*} = A$$, we can deduce the following:

${\left({P}^{*}AP\right)}^{*} = {U}^{*}$ ${P}^{*}{A}^{*}P = {U}^{*}$ ${P}^{*}AP = {U}^{*}$

The conjugate transpose of an upper triangular matrix is a lower triangular matrix. Let $$L$$ be the lower triangular matrix. From the observation that $$U = L$$ implies $${P}^{*}AP = D$$.

However, we know that the inertia tensor is real-symmetric. Hence, it follows the property: If $$A$$ is a real-symmetric matrix, there exists an orthogonal matrix $$P$$, the transpose of the orthogonal matrix $${P}^{T}$$ and a diagonal matrix $$D$$ such that $${P}^{T}AP =D$$.

Since real-symmetric matrices satisfy the property $${A}^{T} = {A}$$ and orthogonal matrices satisfy the property $${P}^{T} = {P}^{-1}$$, the proof is structurally identical to the one above.

Therefore, it follows that

${ P }^{ T }{ I }_{ \alpha \beta }P = D$ and ${ I }_{ \alpha \beta } = PD{ P }^{ T }$

where $$P$$ is the rotation matrix, and $$D$$ is a diagonal matrix containing the principle axes of rotation.

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
2 years, 9 months ago