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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

There are no comments in this discussion.