To prove this statement, we will use the following definitions and properties:
A finite dimensional vector space is a vector space that has a finite number of basis vectors.
A basis for a vector space is a set of vectors that span the space, meaning that any vector in the space can be written as a linear combination of the basis vectors.
A linear combination of a set of vectors is an expression of the form a_1v_1 + a_2v_2 + ... + a_nv_n, where v_1, v_2, ..., v_n are the basis vectors and a_1, a_2, ..., a_n are scalars from the field.
A non-zero element of a finite dimensional vector space is a vector that is not equal to the zero vector, the additive identity of the space.
Now, let's assume that V is a finite dimensional vector space over a field F, and let B be a basis for V. We will show that every non-zero element of V can be expressed as a unique linear combination of the basis vectors in B.
Let v be a non-zero element of V. Since B spans V, there exists a set of scalars a_1, a_2, ..., a_n from the field F such that v = a_1v_1 + a_2v_2 + ... + a_nv_n, where v_1, v_2, ..., v_n are the basis vectors in B. This is a valid linear combination representation of v with respect to B.
To show that this linear combination representation of v is unique, let's assume that there exists another set of scalars b_1, b_2, ..., b_n from the field F such that v = b_1v_1 + b_2v_2 + ... + b_nv_n. Then, subtracting the second equation from the first one, we get:
0 = (a_1 - b_1)v_1 + (a_2 - b_2)v_2 + ... + (a_n - b_n)v_n.
Since B is a basis for V, the only way for the left-hand side of the equation to be the zero vector is for all the scalars in the linear combination to be equal, i.e. a_1 = b_1, a_2 = b_2, ..., a_n = b_n. Thus, v can be expressed as a unique linear combination of the basis vectors in B, which completes the proof.