Let $V$ be a finite-dimensional vector space over a field $F$, and let $\varphi_1,\ldots,\varphi_k \in V^*$ be linearly independent. Then $$ \dim(V) = k + \dim \left(\bigcap_{i=1}^k \ker(\varphi_i) \right). $$
I have a proof which I provide below. I am interested to see different approaches.
Proof:
Define $T:V \to F^k$ by $$ T(x)=\big(\varphi_1(x),\ldots,\varphi_k(x)\big). $$ Then $\ker T=\bigcap_{i=1}^k \ker(\varphi_i)$.
By the Rank–nullity theorem, $$ \dim (\ker T)+\dim\big( \operatorname{Image}(T)\big)=\dim V, $$ thus it suffices to prove that $\dim \operatorname{Image}(T)=k$.
In order to shows this we shall use the following lemma:
Claim: There exist $x_1,\ldots,x_k \in V$ such that $\varphi_i(x_j) = \delta_{ij}$.
Then we have $T(x_j)=e_j$ form the standard basis of $F^k$. Thus $T$ is surjective as required.
Question: Can we prove $T$ is surjective without using the lemma of the $x_j$? Or maybe is there a different approach to proving the claim without defining an auxiliary map $T$?
Note that $\ker\varphi_i=\langle\varphi_i\rangle^{\perp}$, that is, the orthogonal complement of the $1$-dimensional subspace spanned by $\varphi_i$ under the scalar product between $V^*$ and $V$ given by $\langle\varphi,x\rangle=\varphi(x)$. Now $$\bigcap_{i=1}^k\langle\varphi_i\rangle^{\perp}=\bigl(\sum_{i=1}^k\langle\varphi_i\rangle\bigr)^{\perp}$$ Since the $\varphi_i$ are linearly independent, we know the sum is direct and has dimension $k$, therefore the dimension of its complement is $\dim V-k$.