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

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