Linear Algebra Problem - Ph.D exam I stole this problem from a Ph.D exam from another university.

Let $V$ be a real vector space and let $T: V \to \mathbb{R}$ be a linear transformation. Suppose $(v_1, \dots, v_n)$ is a bssis for $\ker(T)$. Suppose $v \in V, v \neq 0$ is not in $\ker T$. Prove that $(v, v_1, \dots, v_n)$ is a basis for $V$.

Here is some scratch work I wrote.
$T(dv + a_1v_1 + \dots + a_nv_n) = T(dv) + 0 = dT(v)$
Now I have also observed that $\beta = (v, v_1, \dots, v_n)$ is linearly independent in $V$. I am not sure if I am allowed to assume $V$ is finite dimensional. If it is, I can use a theorem to extend $\beta$ to a basis. Otherwise, I am stuck
EDIT I just noticed the problem tells me that $\beta$ has $n + 1$ vectors, so maybe i am allowed to?
 A: Let $u \in V$.  We want to show that $u$ is in the subspace generated by $\ker T$ and $v$.  Because $Tv$ is a nonzero real number, it spans $\mathbb{R}$ and so we have $Tu = \alpha Tv$ for some real number $\alpha$.  Therefore $u - \alpha v \in \ker T$, so $u$ is the sum of $\alpha v$ and a vector in $\ker T$ as desired.
A: It is easiest to go for proving $V=\langle v\rangle\oplus \ker T$, as it will ten follow that any bases of $\langle v\rangle$ and of $\ker T$ will combine to one of$~V$ (one does not even need finite dimensionality for this). Now since $v\notin\ker T$ it is obvious that $\langle v\rangle\cap \ker T=\{0\}$. To see that $V=\langle v\rangle+\ker T$ write any $u\in V$ as $u=\alpha v+(u-\alpha v)$ where $\alpha=\frac{T(u)}{T(v)}$; check easily that this ensures $u-\alpha v\in\ker T$.
A: The most general approach for this could probably be:
Let $\;V\;$ be a linear sapce over a field $\;\Bbb F\;$ , and let $\;f\;$ be a non-zero linear functional (i.e. , $\;0\neq f\in V^*\;$) , then
== $\;f\;$ is onto
== $\;\ker f\;$ is a proper maximal subspace of $\;V\;$ , meaning that
$$\forall\,x\in V\setminus\ker f\;,\;\;V=\text{span}\,\{\ker f\;,\;x\}$$
If $\;\dim V=n<\infty\;$ the above means $\;\dim\ker f=n-1\;$ and , in any case, $\;\ker f\;$ is called a hyperplane (or hyperspace) of $\;V\;$
