Let our equations be represented by $\displaystyle \varphi_1(v)=0,\cdots,\varphi_m(v)=0$ where $\displaystyle \varphi_k\in\text{Hom}\left(\mathbb{R}^n,\mathbb{R }\right)$

(the dual space). Note then that since $\displaystyle m<n$ we have that

$\displaystyle \dim \text{Ann}\left(\text{span}\{\varphi_1,\cdots,\var phi_m\}\right)=n-m>0$

(where $\displaystyle \text{Ann}\left(\text{span}\{\varphi_1,\cdots,\var phi_m\}\right)\subseteq\left(\text{Hom}\left(\math bb{R}^n,\mathbb{R}\right)\right)^{\ast}$ is the set of all linear functionals on

$\displaystyle \text{Hom}\left(\mathbb{R}^n,\mathbb{R}\right)$ which map $\displaystyle \text{span }\{\varphi_1,\cdots,\varphi_m\}$ to $\displaystyle \{0\}$)and so there exists some non-zero

$\displaystyle \Psi\in\text{Ann}\left(\text{span}{\varphi_1,\cdot s,\varphi_m\}\right)$. Thus, $\displaystyle \Psi(\varphi_1)=\cdots=\Psi(\varphi_m)=0$. Recall though that the natural identification

$\displaystyle f:\mathbb{R}^n\to\left(\text{Hom}\left(\mathbb{R}^ n,\mathbb{R}\right)\right)^{\ast}:v\mapsto \Psi_v$

(where $\displaystyle \Psi_v$ is the evaluation functional $\displaystyle \Psi_v(\varphi)=\varphi(v)$) is an isomorphism and so in particular

$\displaystyle \Psi=\Psi_{v_0}$ for some $\displaystyle v_0\in\mathbb{R}^n$. In particular we have that $\displaystyle \varphi_1(v_0)=\cdots=\varphi_m(v_0)=0$ as

required.