If $f,g$ are linear forms of a finite dimensional vector $V$ that are not proportional then $ker (F) \cap ker (G)$ has dimension $(n-2)$

If $f,g$ are linear forms of a finite dimensional vector $V$ of dimension $n$ that are not proportional then $ker (f) \cap ker (g)$ has dimension $(n-2)$. I showed in previous exercises that

1. Two linear forms are proportional $\iff$ they have the same linear subspace (say N) as their kernel.

2. That linear forms of finite vector spaces have kernel (n-1) dimensional by Rank-Nullity.

Not sure how to proceed. It seems to me that f,g most have different kernels since they are not proportional. Any hints on how to solve this problem appreciated.

• What do you mean with $(2)$ ? Did you mean that the kernel of any non-zero linear functional on a $\;n\,-$ dimensional space is a subspace of dimension $\;n-1\;$ ? Also, what are proportional linear forms? – DonAntonio Sep 1 '16 at 20:29
• @DonAntonio yes sorry edited. – IntegrateThis Sep 1 '16 at 20:51

Hint: Use Grassmann's formula for linear subspaces: $$\dim U + \dim W = \dim(U \cap W) + \dim(U + W)$$