Characteristic Subspace of a k-form I have to show that two decomposable k-forms on an n-dimensional vector space $\Bbb V^n$ have the same characteristic subspace (of dimension $n-k$) if and only if one is scalar multiple of the other.
$\underline{\text{Definition}}$: Characteristic subspace of a $k$-form $\omega$ on $\Bbb V$, is the set of all vectors in $\Bbb V$ such that $\iota_u\omega=0$, where $\iota : \Bbb V^n\to\Lambda^{k-1}(\Bbb V)$ is the interior product operation.
Any advice would be very helpful. Thank you in advance.

Let $Z$ denote the characteristic subspace, and let $\{z_1,\cdots,z_{n-k}\}$ be a basis for $V$. You can extend this to a basis $\{z_1,...,z_{n-k},w_1,...,w_k\}$ of $V$. Define an inner product on $V$ by saying the dot product of a vector in the basis with itself is $1$, and is $0$ with any other vector in the basis, and extending by bilinearity. From this, it follows that $\{w_1,...,w_k\}$ is a basis for $Z^\perp = W$. 
Now $\omega$ is decomposable, so it is of the form $\alpha_1 \wedge ... \wedge \alpha_k$. In particular, for each $\alpha_i$, we can find a $w_i'$ wuch that $\alpha_i(w_i') = 1$ (this also relies on the inner product). You can see that $\{z_i,...,z_{n-k},w_1',...,w_k'\}$ is also a basis for $V$, and there is a change of basis matrix that is invertible. Since the top left of the matrix is an identity block, the bottom $k \times k$ block is also invertible. So the projections of $\{w_i'\}$ onto $W$ are also a basis for $W$. Let us call this block $K$. Then, note that $$\omega(w_1,...,w_k) = \omega(Tw_1',...,Tw_k') = (\det T) \omega(w_1,...,w_k) = \det T \neq 0$$
Last, we note that this works for any extension of the $\{z_i\}$ to a basis of $V$, so the choice is irrelevant. So $\omega_W$ is an alternating, multilinear form on $W$ whose value on a basis is non-zero and so is some multiple of $\det_W$. Since $\omega$ is essentially the same as $\omega_W$, this shows that $\omega$ is some multiple of $\det_W$. 
 This is true for any such $\omega$, so they are all multiples of each other. 
Edit: As Ted Shifrin pointed out, I meant that $\{z_i\}$ form a basis for $Z$ in my first sentence
