The Power of Bilinearity: To prove something is a Lie Algebra , it's not enough to show that the Lie Bracket is antisymmetric , formally `[x \ , y] = -[y \ , x]`. You must also show that it plays nicely with the vector addition we have been practising. The Lie Bracket [x , y] must be Bilinear. This means it respects the "Vector Space" rules (addition and scalar multiplication) in both slots.

Exercise: Prove that the Lie Bracket is linear in the first slot. Specifically , given three `2 times 2` matrices A , B , and C. Show that : `[A + B \ , C] = [A \ , C] + [B \ , C]`. Use the following matrices:

`A = ([1 , 0] , [0 , 0]) \ , B = ([0 , 1] , [0 , 0]) \ , C = ([0 , 0] , [1 , 0])`

General Strategy: We will calculate the Left-Hand Side (LHS) by adding A and B first , then taking the bracket with C. Then we will calculate the Right-Hand Side (RHS) by calculating two seperate brackets and adding them. If LHS = RHS , we have demonstrated bilinearity for these elements.

Specific Theory Used:
- Commutator Definition: `[X \ , Y] = XY - YX`.
- Matrix Addition: Component-wise addition , `(A + B)_{ij} = A_{ij} + B_{ij}`.
- Bilinearity Axiom: The requirement that `[ax + by \ , z] = a[x \ , z] + b[y \ , z]`.

Solution:
Step A. Calculate the LHS [A + B , C]:

First find A + B: `A + B = ([1 , 0] , [0 , 0]) + ([0 , 1] , [0 , 0]) = ([1 , 1] , [0 , 0]) `

Now , calculate `[(A + B) , C]` using `(A + B)C - C(A + B)`:

`(A + B)C = ([1 , 1] , [0 , 0])([0 , 0] , [1 , 0]) = ([1 , 0] , [0 , 0])`

`C(A + B) = ([0 , 0] , [1 , 0])([1 , 1] , [0 , 0]) = ([0 , 0] , [1 , 1])`

LHS result: `([1 , 0] , [0 , 0]) - ([0 , 0] , [1 , 1]) = ([1 , 0] , [-1 , -1])`

Step B. Calculate the RHS [A , C] + [B , C]:

Calculate `[A , C] : AC = ([0 , 0] , [0 , 0]) \ , CA = ([0 , 0] , [1 , 0]) implies [A , C] = ([0 , 0] , [-1 , 0])`

Calculate `[B , C] : BC = ([1 , 0] , [0 , 0]) \ , CB = ([0 , 0] , [0 , 1]) implies [B , C] = ([1 , 0] , [0 , -1])`

RHS result: `[A , C] + [B , C] = ([1 , 0] , [-1 , -1])`.

Conclusion: `LHS = RHS`. The operation is linear. QED