# $\dim (\ker f \cap \ker g) = n-2$

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.

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 $$dim(Ker f\cap Ker g)=n-2$$.

• Could you explain why the first equality is true? May 17, 2021 at 23:45
• 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 $A+B=\mathbb{R}^n$. May 17, 2021 at 23:46
• Same question as @BokaPeer, can you please detail the first sentence please "$\ker f + \ker g = \mathbb R^n$"? May 17, 2021 at 23:47
• My question was why $Ker f +Ker g = V$. May 17, 2021 at 23:51
• 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. May 17, 2021 at 23:53

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 $$h$$.

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) = $$n > 2.$$