# Clarification for a question in functional analysis

Why the nontrivial nullspace of a linear has codimension 1?

The answer that was top voted I understood for the most part, but at some point the author says that $$f(y-\frac{f(y)}{f(x_0)}x_0)=0$$. I understand that $$f(0)=0$$, I just am not sure why the author is able to assume that $$y=\frac{f(y)}{f(x_0)}x_0$$.

Edit: It has been pointed out that $$y=\frac{f(y)}{f(x_0)}x_0$$ is not necessarily true, and that $$y-\frac{f(y)}{f(x_0)}x_0\in\ker f$$. I guess this would be the better question for me to ask here, since I am not sure why this is able to be assumed.

• You can have $f(x) = 0$ and $x \neq 0$ at the same time. – QuantumSpace Mar 29 at 21:12
• That's true. But $f(0)=0$, and I was unsure why someone is able to assume that $y-\frac{f(y)}{f(x_0)}x_0\in\ker f$, unless they were implying that to be true. – NewbieMather Mar 29 at 21:15

Since $$f$$ is linear, we have
$$f\left(y- \frac{f(y)}{f(x_0)}x_0\right) = f(y) - \frac{f(y)}{f(x_0)}f(x_0) = f(y)-f(y) = 0$$ hence $$y- \frac{f(y)}{f(x_0)}\in \ker f.$$
Let $$V$$ be a vector space and $$f$$ be a nonzero linear functional on $$V$$ over a field $$\mathbb{F}$$. Clearly, $$f$$ is onto. By the first isomorphism theorem, you have $${V\over \ker(f)} \simeq \mathbb{F}$$ via the mapping $$x + \ker(f) \mapsto f(x).$$ We conclude that the codimension of $$\ker(f)$$ is 1.