Let $V$ be a $\mathbb{K}-$vector space finite dimensional ($\mathbb{K}$ is a field). Let $v\in V$ be a vector.
Is there a non-zero linear functional $\varphi\colon V\to\mathbb{K}$ such that $\varphi(v)=0$?
I would like to know if this is true and from what result it can be derived.
Thanks!
In the case that you meant $\varphi(v)=0$ should be true for all $v\in V$ then $\varphi\neq 0$ is wrong by definition.
If this should only be true for a particular $0\neq v\in V$, you need to consider whether $\dim V=1$ or $\dim V>1$. In the case that $v=0$, you can choose $\varphi=\mathrm{Id}$ and you are done.
$\dim V=1$: Suppose $\varphi(v)=0, \; \varphi\neq 0$ is a linear functional. Because $v\neq 0$, every other Element in $V$ is a linear combination of $v$. It follows, that for every $w\in V: \varphi(w)=\varphi(\alpha v)=\alpha\varphi(v)=\alpha\cdot 0=0$, $\alpha \in \mathbb{K}$, which contradicts $\varphi\neq 0$.
$\dim V=n\geq 2$: Let $\mathcal{B}=\{b_1,\dots, b_n\} $ be a basis of $V$ such that $b_1=v$ and let $w\in V$. Then $w=\sum_{i=1}^{n} \alpha_i b_i,\; \alpha_i\in\mathbb{K}$. Choose $\varphi: V\to\mathbb{K}$ where $$\varphi(w)=\varphi(\sum_{i=1}^{n} \alpha_i b_i)=\sum_{i=2}^{n} \alpha_i $$ Because every $b_i\neq 0$ you have that in particular $b_1=v\neq 0$ and $\varphi(v)=0$ but $\varphi(b_i)\neq 0\; \forall i\geq 2$ implies $\varphi\neq 0$. I will leave proving $\varphi$ is linear up to you.
$\hspace{12pt}$
Edit to adress your concerns whether $v\in \mathcal{B}$ is always possible: Either you accept that $v\in\mathcal{B}$ w.l.o.g. to simplify the proof or you have to define $\varphi$ a little differently:
$v\in V$ and $\mathcal{B}$ are already chosen. $v$ is a linear combination of at most $n$ basis vectors, so $v=\sum_{i=1}^{m}\alpha_{k_i} b_{k_i}$ where $m\leq n$. Also let $w\in V$ then $w=\sum_{i=1}^{n} \beta_i b_i$.
Case $m=n$: Let $\mathcal{B'}=\{v,b_2,\dots,b_n\}$, then its $n$ elements are still linearly independent, so $\mathcal{B'}$ is still a basis. Define $\varphi: V\to\mathbb{K}$ with $$\varphi(w)=\varphi(\beta_1' v+\sum_{i=2}^{n} \beta_i' b_i)=\sum_{i=2}^{n} \beta_i' $$ then all the required conditions are met.
Case $m<n$: Reorganize $\mathcal{B'}=\{b_{k_1},\dots,b_{k_m},b_{k_{m+1}},\dots,b_{k_n}\}$. Then define $\varphi: V\to\mathbb{K}$ with $$\varphi(w)=\varphi(\sum_{i=1}^{n} \beta_{k_i} b_{k_i})=\sum_{i=m+1}^{n} \beta_{k_i} $$ and you are done!
