A little problem on Lie Brackets relating it with the commutator of matrices 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.
 A: 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.
$\square$
