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.

  • $\begingroup$ You can have $f(x) = 0$ and $x \neq 0$ at the same time. $\endgroup$ – QuantumSpace Mar 29 at 21:12
  • $\begingroup$ 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. $\endgroup$ – 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.


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