Hyperplanes question in Tu's book on Manifolds

Question

I am trying an exercise from Tu's book "An introduction to Manifolds". Specifically, Subchapter 1.3, exercise 2 (b).

I would like, before continuing with my question, to discuss something about the way the second sub-question is written: How is ker$f$=ker$g$ implying $g=cf$ for some constant $c\in\mathbb{R}$, and for some $f,g:V\rightarrow\mathbb{R}$ nonzero functionals on $V$ mean that a nonzero linear functional on a vector space $V$ is determined up to a multiplicative constant by its kernel?

...Continuing with the solution (which, after some time trying to think, I found online). A solution I found starts by stating that $$V=\text{ker}f\oplus W=\text{ker}g\oplus W\ \text{for some}\ W\subset V$$ I realise that if $V$ can be decomposed as a direct sum of the kernel of $f$ and another vector space $W$, and it can be decomposed as well as a direct sum of the kernel of $g$ and another vector space say $\tilde{W}$, then $\tilde{W}=W$, as the kernel of $f$ is the same as the kernel of $g$. I also realise that if $\text{dim} V=\text{dim(ker}f)+\text{dim}W$ and if $\text{Im}f=\mathbb{R}$, then $\dim(\text{ker}f)=\dim(\text{ker}g)=\dim V-1$. However, why am I allowed to write a generic vector space as a direct sum of its kernel with another vector space? Is this widely known? Does it require proof, or is it derived from logic?

Then, the author of the solution I found, deduces that $\dim(W)=1$, which only makes sense, and proceeds to state that if $w$ is an element of $\ker{f}$ or of $\ker{g}$, then clearly $f=g=0$ and hence $f=cg$ for some constant $c\in\mathbb{R}$. If, however, $w\ne0\in W$ (rather than in $\ker{f}$ or in $\ker{g}$, then $f(w)\in\mathbb{R}$ and $g(w)\in\mathbb{R}$ (by definition of $f$ and of $g$). Then, there exists $c\in\mathbb{R}$, such that $f(w)=cg(w)$. Why doesn't the proof stop here? Instead of stopping, they proceed as follows: If $w'$ is any other element in $W$, then $w'=\beta w$ for some $\beta\in\mathbb{R}$, so $g(w')=g(\beta w)=\beta g(w)=\beta cf(w)=c f(\beta w)=c f(w')$ and hence (I guess) now it is proven that $g=cf$.

1. Why is the last step necessary?

2. Why does the $\ker{f}=\ker{g}$ implying $g=cf$ means that a nonzero functional is determined up to a multiplicative constant by its kernel (for me this statement would be somethin of the form $f=c\ker{f}$, which I do not think holds, as I equate a function with its kernel)?

Thanks

Answer 1

Regarding your first question:

The "English" statement is

• If $f : V \to \mathbb R$ then $f$ is determined by its kernel up to a multiplicative constant.

Your confusion is the determined by part. "Determined by" does not mean "is equal to". It is saying that "all you need to know is $\ker f$", but the words "determined by" do not tell you how to use $\ker f$ once you know it. In this case you are supposed to recall the theorem and then know that $f$ is "determined by" $\ker f$ because, once we have this information, $\ker f = \ker g$ implies $f = cg$ for some scalar $c$ any other functional $g$.

Another way we could state this is:

• A nonzero functional $\color{blue}{\text{has one very important piece of information}}$: its $\color{green}{\text{kernel}}$. $\color{blue}{\text{This is because if two}}$ nonzero functionals $\color{blue}{\text{have the same}}$ $\color{green}{\text{kernel}}$ then they $\color{red}{\text{have to be scalar multiples of each other}}$.

This is what "$\color{blue}{\text{determined by}}\text{ its }\color{green}{\text{kernel }}\color{red}{\text{up to a scalar multiple}}$" means.

Regarding your second question:

For every subspace $S \subseteq V$ there is a subspace $T \subseteq V$ such that $V = S\oplus T$. You should think hard about why this is true.

Regarding your third question:

The proof does not stop there because the author is saying that the fact that $f(w)$ and $g(w)$ are two parallel vectors means there is a scalar $c$ (potentially dependent on $w$) such that $f(w) = cg(w)$. They then proceed to show that $c$ is not dependent on $w$ by showing that the same $c$ also suffices for any other $w' \in W$.