Context
I am studying normal modes oscillations and normal modes [1,2]. In an earlier post [3], I asked for a proof that the generalized eigenvalue equation in normal-mode analysis has positive eigenvalues. This was proven. At this point, I am seeking a related proof regarding the eigenvectors found in normal-mode analysis. There appears to be a related proof in [5] that shows the same thing that I have shown here.
Question
How can the following proposition be proved?
Proposition. Given two real, symmetric, and positive definite matrices $\mathbf{T}$ and $\mathbf{V}$ on $\mathbb{R}^3$, show that there exists an orthonormal basis of $\mathbb{R}^3$ consisting of eigenvectors $\mathbf{a}_1$, $\mathbf{a}_2$, and $\mathbf{a}_3,$ of the generalized eigenvalue equation $$\mathbf{V}\,\mathbf{a} = \lambda\,\mathbf{T}\,\mathbf{a}.$$ Here each eigenvalue represented by $\lambda$ is positive [3].
My attempt of a proof
Just like in [3], I can rewrite the generalized eigenvalue equation $$\mathbf{V}\,\mathbf{a} = \lambda\,\mathbf{T}\,\mathbf{a}.\tag{1}$$ as an actual eigenvalue equation. First, by $\mathbf{w}$ I denote the vector defined by the equation $$\mathbf{w}= \mathbf{T}^{1/2}\,\mathbf{a} ,$$ and next I rewrite [1] as $$ \left(\mathbf{T}^{1/2}\right)^{-1}\mathbf{V} \,\left(\mathbf{T}^{1/2}\right)^{-1}\,\mathbf{w} = \lambda \,\mathbf{w}.\tag{2}$$ As in [3], I note that since $$\left(\mathbf{T}^{1/2}\right)^{-1}\mathbf{V} \,\left(\mathbf{T}^{1/2}\right)^{-1} = \left(\left(\mathbf{T}^{1/2}\right)^{-1}\mathbf{V} \,\left(\mathbf{T}^{1/2}\right)^{-1} \right)^T,$$ therefore, $$\left(\mathbf{T}^{1/2}\right)^{-1}\mathbf{V} \,\left(\mathbf{T}^{1/2}\right)^{-1} $$ is symmetric.
I now apply the spectral theorem. Since $\left(\mathbf{T}^{1/2}\right)^{-1}\mathbf{V} \,\left(\mathbf{T}^{1/2}\right)^{-1} $ is Hermitian (and more over positive definite) on $\mathbb{R}^3$, then there exists an orthonormal basis of $\mathbb{R}^3$ consisting of eigenvectors of $\left(\mathbf{T}^{1/2}\right)^{-1}\mathbf{V} \,\left(\mathbf{T}^{1/2}\right)^{-1} $. Each eigenvalue is real (and moreover positive). I can, therefore, construct a matrix $\mathbf{W}$ defined by the equation $$ \mathbf{W}= \begin{bmatrix}\mathbf{w}_1&\mathbf{w}_2&\mathbf{w}_3\end{bmatrix}.$$ I know that $\mathbf{W}$ is an orthogonal matrix.
Next, I define a new matrix $\mathbf{A}$ by the equation $$\mathbf{A} = {\left(\mathbf{T}^{1/2}\right)}^{-1}\mathbf{W} .$$ Next, I try exploit the properties of orthogonal matrices and real symmetric positive definite matrices to show that since $\mathbf{W}$ is an orthogonal matrix and ${\left(\mathbf{T}^{1/2}\right)}^{-1}$ real symmetric positive definite matrix that therefore $\mathbf{A}$ is an orthogonal matrix.
From [4], there are three characteristics or orthogonal matrices which I might exploit:
(1) $ \mathbf{W}^\top = \mathbf{W}^{-1}$,
(2) $\mathbf{W}^{-1}= \mathbf{W}^{\dagger} $, where $\dagger$ is conjugate transpose, and
(3) $ \mathbf{W}^{\dagger}\,\mathbf{W} = \mathbf{W}\,\mathbf{W}^{\dagger} $
From the first characteristic, I find that \begin{align} \mathbf{W}^\top &= \mathbf{W}^{-1} \\ \left({\left(\mathbf{T}^{1/2}\right)}\, \mathbf{A} \right)^\top &= \left({\left(\mathbf{T}^{1/2}\right)}\, \mathbf{A} \right)^{-1} \\ \mathbf{A}^\top\,\left({\left(\mathbf{T}^{1/2}\right)}\, \right)^\top &= \left( \mathbf{A} \right)^{-1} \left(\mathbf{T}^{1/2} \right)^{-1} \\ \mathbf{A}^\top\, \left(\mathbf{T}^{1/2}\right) &= \left( \mathbf{A} \right)^{-1} \left(\mathbf{T}^{1/2} \right)^{-1} \\ \mathbf{A}^\top\, &= \left( \mathbf{A} \right)^{-1} \left(\mathbf{T}^{1/2} \right)^{-1} \left(\mathbf{T}^{1/2} \right)^{-1} \\ \mathbf{A}^\top\, &= \mathbf{A} ^{-1} \mathbf{T} ^{-1}\tag{3} \end{align} Nope, the first characteristic alone does not yield the desired result.
From the second characteristic, I find that \begin{align} \mathbf{W}^\dagger &= \mathbf{W}^{-1} \\ \left({\left(\mathbf{T}^{1/2}\right)}\, \mathbf{A} \right)^\dagger &= \left({\left(\mathbf{T}^{1/2}\right)}\, \mathbf{A} \right)^{-1} \\ \mathbf{A}^\dagger\,\left({\left(\mathbf{T}^{1/2}\right)}\, \right)^\dagger &= \left( \mathbf{A} \right)^{-1} \left(\mathbf{T}^{1/2} \right)^{-1} \\ \mathbf{A}^\dagger\,\left({\left(\mathbf{T}^{1/2}\right)}\, \right) &= \left( \mathbf{A} \right)^{-1} \left(\mathbf{T}^{1/2} \right)^{-1} \\ \mathbf{A}^\dagger &= \mathbf{A}^{-1}\mathbf{T} ^{-1}\tag{4} \end{align} Nope, the second characteristic alone does not yield the desired result.
From the third characteristic I find that \begin{align} \mathbf{W}^{\dagger}\,\mathbf{W} &= \mathbf{W}\,\mathbf{W}^{\dagger} \\ \left({\left(\mathbf{T}^{1/2}\right)}\, \mathbf{A} \right)^{\dagger} \, \left({\left(\mathbf{T}^{1/2}\right)}\, \mathbf{A} \right) &= \left({\left(\mathbf{T}^{1/2}\right)}\, \mathbf{A} \right) \, \left({\left(\mathbf{T}^{1/2}\right)}\, \mathbf{A} \right)^{\dagger} %%%%%%%%%%%%% %%%%%%%%%%%%% %%%%%%%%%%%%% \\ \mathbf{A} ^{\dagger} \left({\left(\mathbf{T}^{1/2}\right)} \right)^{\dagger} \, \left({\left(\mathbf{T}^{1/2}\right)}\, \mathbf{A} \right) &= \left({\left(\mathbf{T}^{1/2}\right)}\, \mathbf{A} \right) \, \mathbf{A} ^{\dagger} \, \left({\left(\mathbf{T}^{1/2}\right)} \right)^{\dagger} %%%%%%%%%%%%% %%%%%%%%%%%%% %%%%%%%%%%%%% \\ \mathbf{A} ^{\dagger} \, \mathbf{T} \, \mathbf{A} &= \mathbf{T}^{1/2} \, \mathbf{A} \, \mathbf{A} ^{\dagger} \, \mathbf{T}^{1/2} \tag{5} \end{align} Nope, the third characteristic alone does not yield the desired result.
Let's if the characteristics combine in some useful way. Equations (4) and (5) together give that $$ \mathbf{T}^{-1} = \mathbf{A} \, \mathbf{A} ^{\dagger} \qquad \text{or that}\qquad \mathbf{A} ^{\dagger} \, \mathbf{T} \, \mathbf{A} = \mathbf{I} \,.\tag{6} $$ Next, substituting Equation (3) into Equation (6), we find that $$ \mathbf{A}^\top = \mathbf{A} ^{\dagger} \qquad \text{or that}\qquad \mathbf{A} ^{\dagger} \, \mathbf{T} \, \mathbf{A} = \mathbf{I} \,.\tag{7} $$ Finally, combining the results in Equation (7) we have that $$ \boxed{ \mathbf{A} ^{\top} \, \mathbf{T} \, \mathbf{A} = \mathbf{I} \,.} $$ In conclusion, I find that, in general, given two real, symmetric, and positive definite matrices $\mathbf{T}$ and $\mathbf{V}$ on $\mathbb{R}^3$ with the standard inner product $\left<y,x\right> = y^\dagger\,x$, there does not, in fact appear to, exists an orthonormal basis of $\mathbb{R}^3$ consisting of eigenvectors $\mathbf{a}_1$, $\mathbf{a}_2$, and $\mathbf{a}_3,$ of the generalized eigenvalue equation $$\mathbf{V}\,\mathbf{a} = \lambda\,\mathbf{T}\,\mathbf{a}.$$
Discussion
Perhaps if the inner product is defined otherwise, then we might be able to construct an affirmative proof. For example, perhaps if I incorporated, "the general form of an inner product on $\mathbb{C}^3$, [which] is known as the Hermitian form and is given by $$\left<x, y\right> = y^\dagger\,\mathbf{T}\,x= \overline{x^\dagger\,\mathbf{T}\,y},\,[6]"$$ then I would be able to prove existence of orthonormal basis consisting of eigenvectors of generalized eigenvalue equation. I leave this question open and hope that one more knowledgeable than I might construct an affirmative proof.
Bibliography
[1] https://en.wikipedia.org/wiki/Normal_mode
[2] Goldstein, "Classical Mechanics," 3rd edition, page 242-3.
[3] Prove that the generalized eigenvalue equation in normal mode analysis has positive eigenvalues
[4] https://en.wikipedia.org/wiki/Orthogonal_matrix#Matrix_properties
