Spectral Theory and the Inertia Tensor

The general moment of inertia tensor Iαβ{ I }_{ \alpha \beta } is represented by a 3×33\times 3 real-symmetric matrix. Show that there exist an orthogonal matrix PP and diagonal matrix DD such that D=PTIαβP.D = { P }^{ T }{ I }_{ \alpha \beta }P.

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

Solution

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

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

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

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

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

Since real-symmetric matrices satisfy the property AT=A{A}^{T} = {A} and orthogonal matrices satisfy the property PT=P1{P}^{T} = {P}^{-1}, the proof is structurally identical to the one above.

Therefore, it follows that

PTIαβP=D{ P }^{ T }{ I }_{ \alpha \beta }P = D and Iαβ=PDPT{ I }_{ \alpha \beta } = PD{ P }^{ T }

where PP is the rotation matrix, and DD is a diagonal matrix containing the principle axes of rotation.

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

Note by Steven Zheng
5 years, 2 months ago

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