Prove that any two bases of a finite dimensional vector space \( V(F) \) have the same number of elements.

To show that any two bases of a finite-dimensional vector space \( V \) over a field \( F \) have the same number of elements, we can proceed as follows: Assume \( B = \{ v_1, v_2, \ldots, v_n \} \) is a basis of \( V \), which means that \( B \) is a linearly independent set that spans \( V \) with \( n \) elements. Now, let \( B' = \{ u_1, u_2, \ldots, u_m \} \) be another basis of \( V \) with \( m \) elements. Since \( B' \) is a basis, it is also linearly independent and spans \( V \). To establish the relationship between \( n \) and \( m \), we can show two inequalities. First, since the set \( B' \) spans \( V \), each vector \( v_i \) in \( B \) can be expressed as a linear combination of the vectors in \( B' \). Because \( B' \) is linearly independent, this implies that the number of vectors \( m \) in \( B' \) must be at least \( n \): \[ m \geq n. \] Next, since \( B \) forms a basis, \( B' \) must also be expressible in terms of the vectors from \( B \), which gives us that each vector \( u_j \) in \( B' \) can be written as a linear combination of the vectors in \( B \). This suggests that the number of vectors \( n \) in \( B \) must also be sufficient to express the vectors in \( B' \), hence: \[ n \geq m. \] Combining these inequalities, we arrive at: \[ m \leq n \quad \text{and} \quad n \leq m. \] Thus, it follows that: \[ m = n. \] As a result, any two bases of a finite-dimensional vector space have the same number of elements, thereby completing the proof!