Proofs from Basis and dimension

Because $B$ spans $V$, at least one representation of $v$ exists. If

then subtracting the two expressions gives

Independence of $B$ forces every $\lambda_j-\mu_j$ to be zero, so the representation is unique.

Because $S$ spans $V$, each vector of $L$ is a linear combination $\ell_j=\sum_i a_{ij}s_i$ of the vectors in $S$. Suppose $|L|>|S|$. The coefficient matrix $A=(a_{ij})$ then has more columns than rows, so $Ac=0$ has a non-zero solution $c=(c_\ell)_{\ell\in L}$. This relation lifts to $\sum_j c_j\ell_j=\sum_i(\sum_j a_{ij}c_j)s_i=0$, that is,

which contradicts the linear independence of $L$. Hence $|S|\geq|L|$.

If $B_1$ and $B_2$ are bases, the counting theorem gives $|B_1|\geq|B_2|$ by taking $B_1$ as the spanning set and $B_2$ as the independent set. Reversing their roles gives $|B_2|\geq|B_1|$, so the cardinalities are equal.

Compare $S$ with a basis $B$ of $V$ in the preceding counting theorem. When $S$ spans, taking $S$ as the spanning set and $B$ as the independent set gives $|S|\ge|B|=\dim(V)$. If equality holds but $S$ is dependent, one vector can be removed without changing its span. That would give a spanning set with fewer than $\dim(V)$ vectors, a contradiction.

When $S$ is independent, take $B$ as the spanning set and $S$ as the independent set to get $|S|\le\dim(V)$. If equality holds but $S$ does not span, adjoining a vector outside its span gives an independent set with more than $\dim(V)$ vectors, again a contradiction.