Question regarding linear transformations from a vector space to the Reals focusing on images equaling zero [Solved]

Question

Right now I'm trying to prove a conjecture exercise (ex. 15 in section 3.2 of Lang's Linear Algebra). I am aware that the space being dealt with is the Kernel, but as I have yet to reach that section, I'm trying to prove it without delving into the math in the next section to see what insights it can provide.

$\mathbf{Statement\;of\;the\;Problem:}$ Let V be a vector space and $F:V\rightarrow\mathbf{R}$ a linear map. Let $W$ be a subset of $V$ consisting of all elements $v$ such that $F(v)=0$. Assume that $W\neq V$, and let $v_0$ be an element of $V$ which does not lie in $W$. Show that every element of $V$ can be written as a sum $w+cv_0$ with some $w$ in $W$ and some number $c$.

$\mathbf{My\;attempt\;at\;a\;solution:}$ I began by letting $v$ be an arbitrary element in $V$ and breaking the problem into the following cases:

A) $v$ is an element in $W$

B) $v$ is not in $W$ but is a multiple of $v_0$

C) $v$ is not in $W$ and is not a multiple of $v_0$.

Now cases A and B are trivial to prove (for the first, let $w=v$ and $c=0$ and for the second, let $w=0$ and you're good). Case C is tripping me up though. I initially considered taking the tactic of letting $\beta=\{v_1,v_2,...,v_n\}$ being a basis for V and the splitting the basis into sets $B=\{b_1,b_2,...,b_r\}$ and $C=\{c_1,c_2,...,c_s\}$ such that $r+s=n,C\subset W,$and $B=V\setminus C$. From there I split the linear combo of $v$ in terms of $\beta$ into the sums across $B$ and $C$. However, although I can guarantee that the sum that's the linear combination of the elements in $C$ will at least make up a part of the supposed $w$ as referenced by the claim, since $v\neq cv_0$, I'm stuck on how to prove from there that the sum of the elements in $B$ is either the sum of elements which can go into the $w$ term and those that belong in the multiple of $v_0$ or how to prove the leftover part of the sum term is entirely a part of the multiple of $v_0$. Can anyone at least point me in the right direction?

NB) Because it's rather trivial to prove W is a subspace of V, we can allow that as a given for this.

Answer 1

Here's a neat proof:

Let $v$ be an arbitrary element of $V$. Let $w_v = v - \frac{f(v)}{f(v_0)}v_0$. Note that $w_v$ is an element of $w$. Verify that $w_v \in w$. With that, we may state that $$ v = w_v + \frac{f(v)}{f(v_0)}v_0 $$ which is to say that with $c = \frac{f(v)}{f(v_0)}$, we have the desired result.

Answer 2

So I managed to figure out the case C of the proof. To not spoil it, I'll give you a hint of how I went about it.

Let $v=\sum_{i=1}^r x_i(\vec{b_i})+\sum_{i=r+1}^n x_i(\vec{c_i})$ and $v_0=\sum_{i=1}^r y_i(\vec{b_i})+\sum_{i=r+1}^n y_i(\vec{c_i})$ where $B=\{\vec{b_1}...\vec{b_r}\}$ and $C=\{\vec{c_1}...\vec{c_c}\}$ (it should be noted that $n-(r+1)=s$ as defined above). Consider $F(v-v_0)$. What happens if $v-v_0\in W$? If it's not, then clearly $F(v-v_0)\neq 0$. Combine that with the fact that the map is to the reals and $F(v_0)\in\mathbf{R}$ as well as $F(v-v_0)\in\mathbf{R}$ , and you should be good to go.

Want to give special shoutouts to @{Lord Shark the Unknown} and @Omnomnomnom. I don't think I could have gotten to this without your guys's help. Thank you for your input and being there to talk it out. :)