I am trying to solve this problem:
There is a hint saying that I can use the fact that for $$X=a_i\frac{\partial }{\partial x_i}\text{ with } a_i\in C^\infty(U)$$ and $$Y=b_i\frac{\partial }{\partial x_i}\text{ with } b_i\in C^\infty(U)$$ we have $$[X,Y]=\left(a_{i}\frac{\partial b_{j}}{\partial x_{i}}-b_{i}\frac{\partial a_{j}}{\partial x_{i}}\right)\frac{\partial}{\partial x_{j}} $$
But I did not manage to use this hint in any useful way. Also, the obvious procedure seems to be the following: $$\Psi([X,Y])(P)=P[X,Y]=P(XY-YX)=PXY-PYX$$ and $$[\Psi(X),\Psi(Y)](P)=\left(\Psi(X)\Psi(Y)-\Psi(Y)\Psi(X)\right)(P)\\=\Psi(X)\left(\Psi(Y)(P)\right)-\Psi(Y)\left(\Psi(X)(P)\right)=\Psi(X)\left(PY\right)-\Psi(Y)\left(PX\right)=PYX-PXY$$
And this is clearly wrong. So:
1) What did I miss in my attempt?
2) Any hints on solving this using the given hint?
EDIT: I believe I can solve this using the hint if $[\tilde{X},\tilde{Y}](P)=[\tilde{X}(P),\tilde{Y}(P)]$. However, I do not know why/if this is true, and question 1) still holds.
The first line : ('canonical identification') means we say that an element of $T_I(GL(n,\mathbb{R}))$ can be thought of as an element of $M_n(\mathbb{R})$ and vice versa by means of vectorization.
Here an example of this canonical identification for $n=2$ :
Let : $Q \in T_I(GL(2,\mathbb{R})) $ , $Q= \left( \begin{array}{c} q_1 \\ q_2 \\ q_3 \\ q_4 \\ \end{array} \right)$ ,
the corresponding matrix $\in M_2(\mathbb{R})$ becomes : $ \left( \begin{array}{c , c} q_1 & q_3 \\ q_2 & q_4 \\ \end{array} \right)$.
With the canonical identification we can also define the map : $\Psi : T_I(GL(n,\mathbb{R})) \mapsto \mathfrak{X}(GL(n,\mathbb{R}))$ as a product of matrices because a vector field maps an element of $GL(n,\mathbb{R})$ to an element of $T_P(GL(n,\mathbb{R}))$ at each point $P$ :
To make a vector field from a vector at $T_I$ we have to assign a vector to every element of $GL(n,\mathbb{R})$.
We choose to do so by 'left-multiplying' with said element of $GL(n,\mathbb{R})$ which we have called $P$ .
So (for $n=2$ ) in coordinates, $P$ is in fact just : $\left( \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{array} \right)$ or $\left( \begin{array}{c , c} x_1 & x_3 \\ x_2 & x_4 \\ \end{array} \right)$.
This way $\Psi \left( X \right) = PX $ becomes :
$ \left( \begin{array}{c , c} x_1 & x_3 \\ x_2 & x_4 \\ \end{array} \right) \left( \begin{array}{c , c} X_1 & X_3 \\ X_2 & X_4 \\ \end{array} \right) = \left( \begin{array}{c , c} x_1X_1 + x_3X_2 & x_1X_3 + x_3X_4 \\ x_2X_1 + x_4X_2 & x_2X_3 + x_4X_4 \\ \end{array} \right)$
Or :
$\Psi \left( X \right) =\left( \begin{array}{c , c} x_1X_1 + x_3X_2 \\ x_2X_1 + x_4X_2 \\ x_1X_3 + x_3X_4 \\ x_2X_3 + x_4X_4 \\ \end{array} \right)$ and $\Psi \left( Y \right) =\left( \begin{array}{c , c} x_1Y_1 + x_3Y_2 \\ x_2Y_1 + x_4Y_2 \\ x_1Y_3 + x_3Y_4 \\ x_2Y_3 + x_4Y_4 \\ \end{array} \right)$
Now according to the hint we have :
$[\Psi \left( X \right) ,\Psi \left( Y \right) ]= \left(\Psi \left( X \right)_{i}\frac{\partial \Psi \left( Y \right)_{j}}{\partial x_{i}}-\Psi \left( Y \right)_{i}\frac{\partial \Psi \left( X \right)_{j}}{\partial x_{i}}\right)\frac{\partial}{\partial x_{j}} $ (with Einstein summation convention ).
We work out the first component :
$[\Psi \left( X \right) ,\Psi \left( Y \right) ]_1= \left(\Psi \left( X \right)_{i}\frac{\partial \Psi \left( Y \right)_{1}}{\partial x_{i}}-\Psi \left( Y \right)_{i}\frac{\partial \Psi \left( X \right)_{1}}{\partial x_{i}}\right) = \left( \Psi \left( X \right)_{1}\frac{\partial \Psi \left( Y \right)_{1}}{\partial x_{1}} + \Psi \left( X \right)_{3}\frac{\partial \Psi \left( Y \right)_{1}}{\partial x_{3}} -\Psi \left( Y \right)_{1}\frac{\partial \Psi \left( X \right)_{1}}{\partial x_{1}} -\Psi \left( Y \right)_{3}\frac{\partial \Psi \left( X \right)_{1}}{\partial x_{3}}\right) = \left( \Psi \left( X \right)_{1}Y_1 + \Psi \left( X \right)_{3}Y_2 -\Psi \left( Y \right)_{1}X_1 -\Psi \left( Y \right)_{3}X_2 \right) = \left( (x_1X_1 +x_3X_2)Y_1 + (x_1X_3 +x_3X_4)Y_2 -(x_1Y_1 +x_3Y_2)X_1 -(x_1Y_3 +x_3Y_4)X_2 \right) = x_1(X_3Y_2-X_2Y_3) + x_3(X_2Y_1 -X_1Y_2) + x_3(X_4Y_2-X_2Y_4) $
Now calculate the first component of $\Psi \left( [X,Y] \right)$ :
$ \Psi \left( [X,Y] \right) = \left( \begin{array}{c , c} x_1 & x_3 \\ x_2 & x_4 \\ \end{array} \right) \left( \begin{array}{c , c} X_1Y_1 + X_3Y_2 -Y_1X_1 - Y_3X_2 & X_1Y_3 + X_3Y_4 -Y_1X_3 - Y_3X_4 \\ X_2Y_1 + X_4Y_2 -Y_2X_1 - Y_4X_2 & X_2Y_3 + X_4Y_4 -Y_2X_3 - Y_4X_4 \\ \end{array} \right) $
We see the first components (and also the other components) are equal, and the generalization to $n > 2$ is trivial.
$\square$