Fundamental matrices are non-singular for all $t$ Some definitions
Definition 4.1, An $n \times n$ matrix $\Phi$ with the property that its columns are solutions of $\dot X = AX$, and linearly independent at $t_0$, is said to be a fundamental matrix of $\dot X = AX$ at $t_0$
Definition 4.10, A fundamental matrix $\Phi (t)$ of $\dot X = AX$ at $t_0$ with the property  that $\Phi (t_0) = I$, is said to be a normalized fundamental matrix at $t_0$
Example being that, assume that $\psi$ is a fundamental matrix then $\Phi (t) = \psi(t)\psi(t_0)^{-1}$ and $\Phi (t_0) = \psi(t_0)\psi(t_0)^{-1} = I$. Thus, $\psi(t)$ is a normalized fundamental matrix at $t_0$.
Theorem 4.15, Assume that $\Phi$ and $\Psi$ are normalized fundamental matrices of $\dot X = AX$ and $\dot X = −A^TX$ at $t_0$ respectively. Then $\Psi^T(t)\Phi(t) = I$ for all $t$, i.e., $\Psi^T(t) = (\Phi(t))^{-1}$.
Now the part I don't understand
its Corollary 4.16 Fundamental matrices are non-singular for all t.
Proof. Suppose $\Psi$ is a fundamental matrix at $t_0$ as well as non–singular in $t_0$, i.e. $\Psi^{−1} (t0)$ exists. From the definition $\Psi(t)\Psi^{−1} (t_0)$ is then a normalized fundamental matrix at $t_0$, so that $\Psi(t)\Psi^{−1} (t_0)$ is non–singular for every $t$, i.e. a matrix $B$ exists so that $\Psi(t)\Psi^{−1} (t_0)B(t)$ = I for every $t$. This implies that
$\Psi(t)$ itself is non–singular for every $t$.
My Question
I don't understand the part that is bold in the Proof "so that $\Psi(t)\Psi^{−1} (t_0)$ is non–singular for every $t$", how did they jump to this conclusion?
The only way I can think of what they did is the used the theorem 4.15 meaning because $\Psi^T(t) = (\Phi(t))^{-1}$ exists then $\Psi(t)\Psi^{−1} (t_0)$ is non–singular
But the theorem says: "Assume that $\Phi$ and $\Psi$ are normalized fundamental matrices of $\dot X = AX$ and $\dot X = −A^TX$ at $t_0$" meaning that they may not exist? and I can go further as to that say that $\Phi$ and $\Psi$ could be normalized fundamental matrices but maybe only at different $t_0$?
If so then how can they make the statement in the corollary 4.16?

M. Grobbelaar, Ordinary Differential Equations, MAT3706/1/2014, University of South Africa, 2013.
 A: After many many hours of research I believe that the answer is better understood by first understanding "Liouville's formula"
So dump the proof of Corollary 4.16
First our goal is to test if $\Phi(t)$ is linearly independent for all $t$, where is $\Phi(t)$ fundamental matrix (our solution matrix)
Our Equation is $\Phi'(t) = A \Phi(t)$, note the difference between the Wikipedia one which their $A$ is time dependent, $A(t)$ while ours is constant. Example of each:
Constant: $A = \left[{\begin{array}{cc}2 & 1\\0 & 2\end{array}}\right]$, Time dependent: $A(t) = \left[{\begin{array}{cc}2t & 1\\0 & t^2\end{array}}\right]$
To fully understand the statement, we must begin with $(\det(\Phi(t)))'$ finding the derivative of the determinant.
which turns out: $(\det(\Phi(t)))' = \mbox{Tr}(A)\det(\Phi(t))$
Where $\mbox{Tr}(A)$ is the trace, once again Wikipedia is time dependent so theirs has:
$(\det(\Phi(t)))' = \mbox{Tr}(A(t))\det(\Phi(t))$
The matrix $A$ enters the equation as finding the $(\det(\Phi(t)))'$ uses $\Phi'(t)$ in its derivation which is defined a $A \Phi(t)$
Looking closely at $(\det(\Phi(t)))' = \mbox{Tr}(A)\det(\Phi(t))$ we can see this is a scalar ode, (note: $\det(\Phi)$ is a time dependent scalar, $\mbox{Tr}A$ is a constant scalar )
(This can be seen more easier by replacing $\det(\Phi(t))$ with $y$ and $\mbox{Tr}A$ with $p$ where $p$ is a constant making $y' =py$)
solving the ode:
$\frac{y'}{y} = p$
taking the definite integral with respect to $x$ on both sides with limits from $t_0$ to $t$
$\int^t_{t_0} \frac{y'}{y}dt = \int^t_{t_0}p dt$
by performing the change of variable $dy = y'dt $ for the left hand side, and converting the limits to match the change, $t, t_0$ becomes $y(t),y(t_0)$ respectively,
$\int^{y(t)}_{y(t_0)} \frac{dy}{y} = \int^t_{t_0}p dt$
$\implies \ln(y(t)) - \ln(y(t_0)) = \int^t_{t_0}p dt$
$\implies \ln(y(t)) = \int^t_{t_0}p dt +\ln(y(t_0))$
taking the $e$ of both sides
$\implies e^{\ln(y(t))} = e^{\int^t_{t_0}p dt +\ln(y(t_0))}$
$\implies y(t) = e^{\int^t_{t_0}p dt}e^{\ln(y(t_0))}$
$\implies y(t) = y(t_0)e^{\int^t_{t_0}p dt}$
now replacing the variable $y(t), p$ with $\det(\Phi(t))$ and $\mbox{Tr}(A)$ respectively,
$\implies \det(\Phi(t)) = \det(\Phi(t_0))e^{\int^t_{t_0}\mbox{Tr}(A) dt}$
Or $\implies \det(\Phi(t)) = \det(\Phi(t_0))e^{\mbox{Tr}(A)\int^t_{t_0} dt}$
as $\mbox{Tr}(A)$ is constant and NOT time dependent like in Wikipedia that why we can take it out of the integral if we want.
Now we can see that out determinant of $\Phi(t)$ depends on $\det(\Phi(t_0))$ meaning that if $\det(\Phi(t_0)) = 0$, then $\det(\Phi(t)) = 0$, note too that $e^{\mbox{Tr}(A)\int^t_{t_0} dt}$ can never be less or equal to $0$ as $e^x > 0$
Now $t_0$ does not have to be $0$ meaning it can be $1,2,3...,999$ or $-99999999$ or any number on $(-\infty,\infty)$ But remember this, fundamental matrices are matrices that are linearly independent at $t_0$ by definition, meaning that $\det(\Phi(t_0))\neq 0$ thus regardless of the $t_0$ value chosen, $\det(\Phi(t_0))\neq 0$ (meaning that there will not exist $\det(\Phi(t_0)) = 0$ if $t_0 = 0$, but if we have our interval from $t_0 = 1$, then $\det(\Phi(t_0)) \neq 0$)
NOTE this is not true for normal matrices as $\det(A(t)) \neq 0$ if $t = 1$ , but $\det(A(t)) = 0$ if $t =0$ may happen as normal matrices do not have that property that fundamental matrices have.
Example 1:
Fundamental matrix $A = \left[{\begin{array}{cc}0 & 1\\-3 & 4\end{array}}\right]$ , $\Phi = \left[{\begin{array}{cc}e^t & e^{3t}\\e^t & 3e^{3t}\end{array}}\right]$, $TrA = 4$
lets take $t_0 =0$, thus $\Phi(t_0 = 0) = \left[{\begin{array}{cc}1 & 1\\1 & 3\end{array}}\right]$, $\det(\Phi(t_0 = 0)) = 2$
thus $\det(\Phi(t)) = \det(\Phi(t_0))e^{\mbox{Tr}(A)\int^t_{t_0} dt}$
$\implies \det(\Phi(t)) = 2 e^{4t}$
which clearly never $= 0$
also you can test if it holds by subbing in different values, like looking at $t$, let $t = 10$, thus $det(\Phi(10)) = \det(\left[{\begin{array}{cc}e^{10} & e^{30}\\e^{10} & 3e^{30}\end{array}}\right]) = 2e^{40}$ which is exactly what $\det(\Phi(t_0))e^{\mbox{Tr}(A)\int^t_{t_0} dt}$ will give.
Example 2:
Fundamental matrix $A = \left[{\begin{array}{cc}0 & 1\\-3 & 4\end{array}}\right]$ , $\Phi = \left[{\begin{array}{cc}e^t & e^{3t}\\e^t & 3e^{3t}\end{array}}\right]$, $\mbox{Tr}A = 4$
lets take $t_0 =2$, thus $\Phi(t_0 = 2) = \left[{\begin{array}{cc}e^2 & e^6\\e^2 & 3e^6\end{array}}\right]$, $\det(\Phi(t_0 = 2)) = 2e^8$
thus $\det(\Phi(t)) = \det(\Phi(t_0))e^{\mbox{Tr}(A)\int^t_{t_0} dt}$
letting $t = 10$ again,
$\implies \det(\Phi(10)) = \det(\Phi(t_0))e^{\mbox{Tr}(A)\int^{10}_{2} dt}$
$\implies \det(\Phi(10)) = 2e^8\cdot e^{4(10 -2)}$
$\implies \det(\Phi(10)) = 2e^8\cdot e^{32}$
$\implies \det(\Phi(10)) = 2e^{40}$
Which is exactly the same as $\det(\Phi(10))$.example 1
Conclusion: We derive $\det(\Phi)$ by it being the solution of the ode: $(\det(\Phi(t)))' = \mbox{Tr}(A(t))\det(\Phi(t))$, solution being $\det(\Phi(t)) = \det(\Phi(t_0))e^{\mbox{Tr}(A)\int^t_{t_0} dt}$
Then thanks to fundamental matrices having the property of non-singular($\det(\Phi(t_0))\neq 0$) at $t_0$, $\det(\Phi(t)) >0$ for all $t$, meaning that it is non-singular for all t
