subspace, dimension and spans How may I go about and construct this proof? Thank you.

Let $V$ be a vector space with dimension $d:= \dim(V)$. Consider a $d$-dimensional subspace $U ≤ V$. Show that $U = V$.  Hint: Let $\vec{a}_1, \ldots, \vec{a}_d$ be a basis in $U$. Assume (for the sake of contradiction) that $U ≠ V$ i.e., there exists $\vec{v} \in V$ such that $\vec{c} \not\in \text{span} \vec{a}_1, \ldots, \vec{a}_d$. Show that $\vec{a}_1, \ldots, \vec{a}_d, \vec{v}$ must be independent ... and consider the dimension of $V$...

 A: Lets follow the hint. Let $\{a_1,\dots, a_d\}$ be a basis for $U$. Then by definition $U=\text{span}(a_1,\dots,a_d)$ Assume (for the sake of contradiction) that $U\not=V$. Then there exists $v\in V\backslash U$. Hence $v\notin \text{span}(a_1,\dots,a_d)$. But this means that $a_1,\dots,a_d,v$ is independent. Indeed if they were dependent there would exist $c_1,\dots,c_{d+1}\in\mathbb{R}$, not all zero, such that $c_1a_1+\dots c_da_d+c_{d+1}v=0$. Notice that in particular $c_{d+1}\not=0$ for otherwise $a_1,\dots,a_d$ would be linearly dependent. This equation implies $v=\frac{-c_1}{c_{d+1}}a_1+\dots+\frac{-c_d}{c_{d+1}}a_d$ meaning $v\in \text{span}(a_1,\dots,a_d)$, a contradiction. Hence $U=V$.
A: To show a(1), a(2), ..., a(d) , v are independent. Suppose not, then there are scalars not all zero r(1), r(2), ..., r(d), r(d+1) such that: r(1)a(1) + ...+ r(d)a(d) + r(d+1)v = 0. If r(d+1) = 0, then r(1)a(1) + ... + r(d)a(d) = 0 ===> r(1) = r(2) =...= r(d) = 0 since a(1), a(2), ..., a(d) form a basis hence independent. Thus r(1), r(2),..., r(d), r(d+1) are independent. If r(d+1) is not zero, then v = -r(1)/r(d+1)*a(1) - r(2)/r(d+1)*a(2) - ... - r(d)/r(d+1)*a(d) and this means v is in S = span{a(1), a(2), ..., a(d)} while it isn't in S to begin with. Since a(1), a(2), ..., a(d) are also in V, V has at least d + 1 independent vectors and so dim(V) > d + 1 or dim(V) = d + 1 and either of these contradicts dim(V) = d.
