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Let $V$ a vector subspace of dimension $n$ on $\mathbb R$ and $f,g \in V^* \backslash \{0\}$ two linearly independent linear forms. I want to show that $\dim (\ker f \cap \ker g) = n-2$.

Since $f$ and $g$ are linear forms, I know that dim ker $f =n-1$ and dim ker $g =n-1$. I think I should use the fact that the two forms are linearly independent to $\dim (\ker f \cap \ker g) = n-2$ but I don't really see how...

I saw a proof with scalar product but I would like to see an alternative proof or maybe an explanation of the scalar product's proof.

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3 Answers 3

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If $f,g$ are linearly independent, $Ker f +Ker g=\mathbb{R}^n$.

$dim(Ker f+Kerg) =dim Ker f +dim Ker g-dim(Ker f\cap Ker g)$ implies that

$n = n-1+n-1-dim(Ker f\cap Ker g)$, and $dim(Ker f\cap Ker g)=n-2$.

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  • $\begingroup$ Could you explain why the first equality is true? $\endgroup$
    – Boka Peer
    May 17, 2021 at 23:45
  • $\begingroup$ This is a general result on vector subspaces, $dim(A+B)=dimA+dimB-dim A\cap B$. Here $A =Ker f, B=Ker g$ and $A+B=\mathbb{R}^n$. $\endgroup$ May 17, 2021 at 23:46
  • $\begingroup$ Same question as @BokaPeer, can you please detail the first sentence please "$\ker f + \ker g = \mathbb R^n$"? $\endgroup$
    – Kilkik
    May 17, 2021 at 23:47
  • $\begingroup$ My question was why $Ker f +Ker g = V$. $\endgroup$
    – Boka Peer
    May 17, 2021 at 23:51
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    $\begingroup$ If $Ker f=Ker g$, consider $x$ which is not in $Ker f$, and such that $f(x)=1, g=g(x)f$. Contradiction,there exists $y\in Ker g$ not in $Ker f$ implies that $Ker f+Ker g=\mathbb{R}^n$ because of the dimension. $\endgroup$ May 17, 2021 at 23:53
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Another way:

Consider the linear maps $h : V \to \mathbb R^2$ and $\tilde h : \mathbb R^2 \to V^*$ given by $h(v) = (f(v),g(v))$ for all $v \in V$ and $\tilde h(a,b) = af+bg$ for all $(a,b) \in \mathbb R^2$. Prove that $h$ is surjective if $\tilde h$ is injective (hint: find an isomorphism $\psi : (\mathbb R^2)^* \to \mathbb R^2$ such that the tranpose of $h$ can be written as $\tilde h \circ \psi$). Thus, the linear independence of $f$ and $g$ shows that $h$ is surjective, and clearly $\ker h = \ker f \cap \ker g$. Finally, use the rank-nullity theorem on $h$.

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Comments:

The following post contains the key part of the proof of your problem. Tsemo has mentioned it in his comment.

Proving that linear functionals are linearly independent if and only if their kernels are different

A bit more explanation/belaboring:

From this above post, you conclude that $kerf \cap ker g$ strictly contained in ker $f$ and ker $g.$ Then we have the following:

$$ dim ( ker f + ker g) = n-1 + (dim ker f - dim (ker f \cap ker g).$$ Since $ (ker f - dim (ker f \cap ker g )>0$, we have dim$ (ker f + ker g)$ is strictly greater than $n-1.$ This shows that $ ker f + ker g = V.$ by the way, I assumed that dim (V) = $n > 2.$

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