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Let $V$ be a vector space of a finite dimension $n$ over the field $K$. Let $\phi, \psi$ be two non-zero functionals on $V$. Assume that there is no non-zero element $c \in K$ such that $\psi= c \phi$. Show that $\ker \phi \cap \ker \psi$ has dimension $n -2$.

There is a coordinate-based argument I know, which I am not writing out here, but I do not like these as they are a little hard to generalize to other situations. If you have a nicer coordinate free argument you may share it.

Here is my coordinate-free argument:

Define $f: V \rightarrow K \times K: f(v) = (\phi (v), \psi (v))$. Then $f$ is non-zero since both $\phi$ and $\psi$ are not and also the dimension of $f(V)$ cannot be $1$ since there exists no non-zero $c$ such that $\psi = c\phi$. Hence, $\dim f(V) = 2$ and $ \ker f = \ker \phi \cap \ker \psi $ so that by the Rank-Nullity Theorem we get the result.

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2 Answers 2

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Note that $\ker \phi \cup \ker \psi\subseteq V$ thus $\dim (\ker\phi \cup \ker \psi)\leq n$ Thus \begin{align} \dim (\ker\phi \cup \ker \psi)&=\dim (\ker\phi) +\dim (\ker \psi)-\dim(\ker \phi \cap \ker\psi )\\ & = n-1+n-1-\dim(\ker \phi \cap \ker\psi )\leq n \end{align} so $\color{red}{\dim(\ker \phi \cap \ker\psi )\geq n-2}$ And since $\phi\neq c\psi ~\forall c$ thus $\ker\phi\not\subseteq\ker\psi$(below link) hence $$\ker \phi \cap \ker\psi\subset \ker\phi\Rightarrow \dim(\ker \phi \cap \ker\psi )<\dim\ker\psi=n-1$$Thus $\color{red}{\dim(\ker \phi \cap \ker\psi )\leq n-2}$

Let $f,g\in V^*$ and assume $g \neq 0$. Show that $f= ag$ for some $a \in F$ iff $Ker(g) \subseteq Ker(f)$.

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I would use the theorem regarding Dimension of Sum and Intersection of Vector Spaces: for $M,N$ subspaces of $V$, you have $$\dim (M+N) +\dim (M \cap N ) = \dim (M) + \dim(N).$$ And apply it with $M= \ker \psi, N =\ker N$. You have $\dim(M+N)= n$ due to the hypothesis on $\psi,\phi$.

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