Prove that the generalized eigenvalue equation in normal mode analysis has positive eigenvalues Context
I am studying normal modes oscillations and normal modes [1,2].
I am trying to complete a proof to show that the eigenvalues (i.e., the eigenvalues are positive). There is a proof in [2], but I find it verbose and likely incorrect. There is a short proof of this in [4]. Nonetheless, I proceed with my questions on the topic.
Questions
Question 1
In my understanding an eigenvalue equation [3], I write the following definition.

Definition
If $T$ is a linear transformation from a finite-dimensional vector space $\mathbb{C}^n$ over the field of complex number into itself and $\mathbf{v}$ is a  nonzero  vector in $\mathbb{C}^n$, then $\mathbf{v}$ is an eigenvector of $T$ if $T\,\mathbf{v}$ is a scalar multiple of $\mathbf{v}$. This can be written as the eigenvalue equation
$$T(\mathbf{v}) = \lambda \mathbf{v},$$
where $\lambda$ is a scalar in $\mathbb{C}$, known as the eigenvalue associated with $\mathbf{v}$.

However, in the analysis of normal mode of dynamical systems, I have $n$-dimensional real symmetric matrices $\mathbf{V}$ and $\mathbf{T} $, and the "eigenvalue equation"
$$\mathbf{V}\,\mathbf{a} = \lambda\,\mathbf{T}\,\mathbf{a}.\tag{1}$$
In my mind, if $T$ is invertible, I can make an actual eigenvalue equation
$$\left( \mathbf{V}\,\mathbf{T}^{-1}\right)\left(\mathbf{T}\,\mathbf{a}\right) = \lambda\,\left(\mathbf{T}\,\mathbf{a}\right).\tag{2}$$
Now, I would have that  $\mathbf{T}\mathbf{a}$ is an eigenvector of $\mathbf{V}\mathbf{T}^{-1}$ since $\mathbf{V} \mathbf{a}$ is a scalar multiple of $\mathbf{T}\mathbf{a}$.
So my first series of questions:
(1) Can we agree that, as written, Equation 1 is not actually an eigenvalue equation?
(2) Does this type of equation have a name? If so what is it?
Question 2
In [2] there is a claim that $\lambda$ is always finite and positive. It comes down to faith. I really like [2], but when it comes to such matters as the proofs of mathematical claim, I just have no faith in [2]. For example, there is even an example later in the book when one of the eigenvalues is zero valued.
(3) How can we prove or disprove the following propositions? [Please note that both propositions cannot be proven to be true.  An amended proposition is given and proven in OP's answer below.]

Proposition. Given two $n$ dimensional real symmetric matrices $\mathbf{T}$ and $\mathbf{V}$ show that any solution of the equation
$$\mathbf{V}\,\mathbf{a} = \lambda\,\mathbf{T}\,\mathbf{a}$$
has a scalar lambda that is positive.


Proposition. Given two $n$ dimensional real symmetric matrices $\mathbf{T}$ and $\mathbf{V}$ show that any solution of the equation
$$\mathbf{V}\,\mathbf{a} = \lambda\,\mathbf{T}\,\mathbf{a}$$
has a scalar lambda that is non-negative.

These two propositions are closely related. Thus, I assume that some other predicates need to be set in order to distinguish the proofs.
Bibliography
[1] https://en.wikipedia.org/wiki/Normal_mode
[2] Goldstein, "Classical Mechanics," 3rd edition, page 242-3.
[3] https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors#Formal_definition
[4] Prove that the eigenvalues for $A\vec{w} = \lambda B\vec{w}$ are real and positive
 A: I believe it's called a generalized eigenvalue equation.
If $T$ is positive definite, you can write the equation as
$$T^{-1/2} V T^{-1/2} w = \lambda w $$
where $w = T^{1/2} v$.  The advantage of this over the form $T^{-1} V v = \lambda v$ is that $T^{-1/2} V T^{-1/2}$ is symmetric (and positive definite if $V$ is).
Your propositions are not true.  Try $V = I$ (the identity matrix) and $T = -I$, or vice versa.
A: 
Proposition. Given two $n$-dimensional, real, symmetric, and positive definite  matrices $\mathbf{T}$ and $\mathbf{V}$ show that any solution of the generalized eigenvalue equation
$$\mathbf{V}\,\mathbf{a} = \lambda\,\mathbf{T}\,\mathbf{a}$$
has a scalar $\lambda$ that is positive.

Proof:
From [1] I know that  a positive semidefinite matrix (real or complex). Then there is exactly one positive semidefinite matrix $B$ such that $A = B^\dagger \, B $. Further, from [1] I understand that "this unique matrix is called the principal, non-negative, or positive square root (the latter in the case of positive definite matrices)." Lastly, from [3], I understand that "The principal square root of a positive definite matrix is positive definite; more generally, the rank of the principal square root of A is the same as the rank of A." This is important to my proof since I need to know if the principle positive square root of a positive definite matrix is invertible, and I also need to know that the principle positive square is full rank.
Since $\mathbf{T}$ is a positive definite matrix (real or complex), it has a principle positive square root, which is invertible. Hence, as suggested by Robert Israel, by $\mathbf{w}$ I denote the vector defined by the equation
$$\mathbf{w}= \mathbf{T}^{1/2}\,\mathbf{v} .$$
As Israel suggests, I then rewrite the generlized eigenvalue equation as
$$ \left(\mathbf{T}^{1/2}\right)^{-1}\mathbf{V}
\,\left(\mathbf{T}^{1/2}\right)^{-1}\,\mathbf{w}  = \lambda \,\mathbf{w}.\tag{1}$$
Now, 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.
Further, following the logic of Batman in [3]  I proceed as follows. I note that
\begin{align}
\left[\mathbf{x}^\top\,\left(\mathbf{T}^{1/2}\right)^{-1}\right]\,\mathbf{V}\,\left[\left(\mathbf{T}^{1/2}\right)^{-1}\,\mathbf{x}\right]
&=
\left[{\left(\mathbf{T}^{1/2}\right)^{-1}}^\top\mathbf{x} \right]^\top\,\mathbf{V}\,\left[\left(\mathbf{T}^{1/2}\right)^{-1}\,\mathbf{x}\right]
\\
&=
\left[{\left(\mathbf{T}^{1/2}\right)^{-1}}\mathbf{x} \right]^\top\,\mathbf{V}\,\left[\left(\mathbf{T}^{1/2}\right)^{-1}\,\mathbf{x}\right].
\end{align}
Further, for $\mathbf{x}\neq \mathbf{0}$, since $\left(\mathbf{T}^{1/2}\right)^{-1}$ is full rank,
$$\left(\mathbf{T}^{1/2}\right)^{-1} \,\mathbf{x} \neq 0;$$
and since $\mathbf{V}$ is a real, symmetric, and positive definite matrix
$$\left[{\left(\mathbf{T}^{1/2}\right)^{-1}}\mathbf{x} \right]^\top\,\mathbf{V}\,\left[\left(\mathbf{T}^{1/2}\right)^{-1}\,\mathbf{x}\right] \geq 0.$$
Thus, by definition,
$$\left(\mathbf{T}^{1/2}\right)^{-1}\mathbf{V}
\,\left(\mathbf{T}^{1/2}\right)^{-1} $$ is positive definite.  Finally, from [4] I know that all the eigenvalues of an $n$-dimensional,  real, symmetric, and positive definite matrix are positive. Returning to Equation 1, we see that all the eigenvalues of the generalized eigenvalue equation are positive. Q.E.D.
Bibliography
[1] https://en.wikipedia.org/wiki/Square_root_of_a_matrix
[2] Is the inverse of a symmetric matrix also symmetric?
[3] Product of any two arbitrary positive definite matrices is positive definite or NOT?
[4] https://en.wikipedia.org/wiki/Definite_matrix
