Every vector in a subspace can be represented as a "unique" linear combination of the bases vectors

• The proof in the reference is pretty convincing.

• And it’s easy to think about if we consider $\hat{i},\hat{j}$ vectors.

• Every point on the plane is a unique linear combination of those two vectors.

• Now this is true because the bases vectors have to be linearly independent, hence the components contributed by “vector A” will have to be contributed by “vector A”.

• The proof works by assuming two different coefficients for the linear combinations.

• Since the two expressions work out to be the same, we equate the two expressions.

• Now we combine the coefficients into a single bracket and the RHS = 0.

• For linear independence, the equation is true only when all the coefficients = 0.

• Since each term is $(c_n-d_n){\vec{v}}_n$, we equate $c_n=d_n$.