# Show sets of dual vectors span the same subspace iff wedge products are similar

The original question is as follows:

Let $$(f_1, f_2, ..., f_k)$$ and $$(g_1, g_2, ..., g_k)$$ be linearly independent sets of dual vectors in $$(\mathbb{R}^n)^*$$. Show that both sets span the same k-dimensional subspace of $$(\mathbb{R}^n)^*$$ iff $$f_1 \wedge f_2 \wedge\ ...\ \wedge f_k = c \cdot g_1 \wedge g_2 \wedge\ ...\ \wedge g_k$$, $$c \neq 0$$.

My proof is as follows, but I am not sure that it is sufficient:

$$(f_1, f_2, ..., f_k)$$ and $$(g_1, g_2, ..., g_k)$$ are linearly independent in $$(\mathbb{R}^n)^*$$, so they form a basis for a k-dimensional subspace of $$(\mathbb{R}^n)^*$$ such that, for an ordered basis $$(v_1, v_2, ..., v_k)$$ for a k-dimensional subspace of $$\mathbb{R}^n$$, $$f_i(v_i) = g_i(v_i) = \delta_{ij}$$ where $$\delta$$ is the Kronecker delta.

This is true iff $$f_i(v_i)$$ is some scalar multiple of $$g_i(v_i)$$, or $$f_i(v_i) = c_j \cdot g_i(v_i)$$, where $$c_j \neq 0$$.

From this we can deduce: $$\begin{eqnarray*} f_1 \wedge f_2 \wedge\ ...\ \wedge f_k &=& \begin{vmatrix} f_1(v_1) & f_1(v_2) & ... & f_1(v_k) \\ f_2(v_1) & f_2(v_2) & ... & f_2(v_k) \\ \vdots & \vdots & \ddots & \vdots \\ f_k(v_1) & f_k(v_2) & ... & f_k(v_k) \\ \end{vmatrix} \\ &=& \begin{vmatrix} c_1 \cdot g_1(v_1) & c_1 \cdot g_1(v_2) & ... & c_1 \cdot g_1(v_k) \\ c_2 \cdot g_2(v_1) & c_2 \cdot g_2(v_2) & ... & c_2 \cdot g_2(v_k) \\ \vdots & \vdots & \ddots & \vdots \\ c_k \cdot g_k(v_1) & c_k \cdot g_k(v_2) & ... & c_k \cdot g_k(v_k) \\ \end{vmatrix} \\ &=& c_1 c_2 ... c_k \begin{vmatrix} g_1(v_1) & g_1(v_2) & ... & g_1(v_k) \\ g_2(v_1) & g_2(v_2) & ... & g_2(v_k) \\ \vdots & \vdots & \ddots & \vdots \\ g_k(v_1) & g_k(v_2) & ... & g_k(v_k) \\ \end{vmatrix} \\ &=& c \begin{vmatrix} g_1(v_1) & g_1(v_2) & ... & g_1(v_k) \\ g_2(v_1) & g_2(v_2) & ... & g_2(v_k) \\ \vdots & \vdots & \ddots & \vdots \\ g_k(v_1) & g_k(v_2) & ... & g_k(v_k) \\ \end{vmatrix} \\ &=& c \cdot g_1 \wedge g_2 \wedge\ ...\ \wedge g_k \end{eqnarray*}$$

• $k$ linearly independent (dual) vectors are not a basis for $(\Bbb R^n)^*$ if $k < n$, so I didn't bother reading things in detail after this. But you're right in that determinants should appear, because the constant $c$ (which by the way must be required from the start to be non-zero) will be the determinant of the change-of-basis matrix. Nov 21, 2021 at 23:55
• Sorry about that, not a basis for $\mathbb{R}^n$, but rather the k-dimensional subspace in question. Nov 22, 2021 at 0:02

I've since found a better solution to the problem:

Proof: $$(\implies)$$ Assume the two sets of vectors span the same $$k$$-dimensional subspace of $$(\mathbb{R}^n)^*$$. Then we may write each of the $$f_i$$ as a linear combination of the $$g_j$$ in a unique manner, since $$\{g_1, g_2, ..., g_k\}$$ is a basis for span$$(g_1, g_2, ..., g_k)$$. Write

$$\begin{eqnarray*} f_i = \sum_j c_{ij}g_j & & \textrm{for} \ 1 \leq j \leq k. \end{eqnarray*}$$

Observe that the $$k \times k$$ matrix $$(c_{ij})$$ is a change-of-basis matrix representing the basis $$\beta_1 = \{f_1, f_2, \dots, f_k\}$$ in terms of the second basis $$\beta_2 = \{g_1, g_2, \dots, g_k\}$$ for the space span$$(g_1, g_2, \dots, g_k)$$ = span$$(f_1, f_2, \dots, f_k)$$.

We now have

$$\begin{eqnarray*} f_1 \wedge f_2 \wedge \cdots \wedge f_k &=& \left(\sum_j c_{1j}g_j\right) \wedge \left(\sum_j c_{2j}g_j\right) \wedge \cdots \wedge \left(\sum_j c_{kj}g_j\right) \\ &=& \textrm{det}(c_{ij})g_1 \wedge g_2 \wedge \cdots \wedge g_k \end{eqnarray*}$$

$$(\impliedby)$$ Assume now that $$f_1 \wedge f_2 \wedge \cdots \wedge f_k = c\ \cdot\ g_1 \wedge g_2 \wedge \cdots \wedge g_k$$. By way of contradiction, let us assume additionally that span$$(g_1, g_2, \dots, g_k) \neq$$ span$$(f_1, f_2, \dots, f_k)$$. Then we may find a one-form $$h \in$$ span$$(g_1, g_2, \dots, g_k) -$$span$$(f_1, f_2, \dots, f_k)$$ such that $$(f_1, f_2, \dots, f_k, h)$$ is a linearly independent set, and so it follows that $$f_1 \wedge f_2 \wedge \cdots \wedge f_k \wedge h \neq 0$$. This leads to a contradiction:

$$\begin{eqnarray*} (f_1 \wedge f_2 \wedge \cdots \wedge f_k) \wedge h &=& c \cdot (g_1 \wedge g_2 \wedge \cdots \wedge g_k) \wedge h \\ &=& 0 \end{eqnarray*}$$

Because $$\{g_1, g_2, \dots, g_k, h\}$$ is a linearly dependent set. $$\square$$