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*}

  • $\begingroup$ $k$ 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. $\endgroup$
    – Ivo Terek
    Nov 21, 2021 at 23:55
  • $\begingroup$ Sorry about that, not a basis for $\mathbb{R}^n$, but rather the k-dimensional subspace in question. $\endgroup$ Nov 22, 2021 at 0:02

1 Answer 1


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. $\square$


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