91Ó°ÊÓ

We are given: 1. V is a finite-dimensional vector space. 2. U is a subspace of V. 3. \(\operatorname{dim}(U) = \operatorname{dim}(V)\)

Let's denote basis (a linearly independent set that spans the vector space) for U as \(\{u_1, u_2, ... , u_n\}\) And basis for V as \(\{v_1, v_2, ... , v_n\}\) Since \(\operatorname{dim}(U) = \operatorname{dim}(V) = n\), both U and V have n basis vectors.

Now we have to show that the basis of U spans V. To prove this, we'll show that the basis vectors of U can be combined linearly to get any vector in V. Suppose \(v \in V\). Since \(\{v_1, v_2, ... , v_n\}\) is a basis for V, we can express v as a linear combination of the basis vectors of V: \(v = a_1v_1 + a_2v_2 + ... + a_nv_n\) Since U is a subspace of V, we know that every basis vector of V can be expressed as a linear combination of the basis vectors of U. So for each \(v_i\), we can say that: \(v_i = b_{1i}u_1 + b_{2i}u_2 + ... + b_{ni}u_n\) Substitute these expressions of \(v_i\) in terms of basis vectors of U into the equation of v: \(v = a_1(b_{11}u_1 + b_{21}u_2 + ... + b_{n1}u_n) + a_2(b_{12}u_1 + b_{22}u_2 + ... + b_{n2}u_n) + ... + a_n(b_{1n}u_1 + b_{2n}u_2 + ... + b_{nn}u_n)\) Rearrange this expression to get v as a linear combination of the basis vectors of U: \(v = (a_1b_{11} + a_2b_{12} + ... + a_nb_{1n})u_1 + (a_1b_{21} + a_2b_{22} + ... + a_nb_{2n})u_2 + ... + (a_1b_{n1} + a_2b_{n2} + ... + a_nb_{nn})u_n\) So we have expressed the vector v, which belongs to V, as a linear combination of the basis vectors of U. Therefore, the basis of U spans V.

Since the basis vectors of U can be linearly combined to form any vector in V and U is already a subspace of V, we can conclude that U = V.