# Linear algebra - Property of an exterior form [closed]

It is true the following statement?

Let $E$ be a real vector space with $\dim E=n$. If $\{e_1,\ldots,e_n\}$ is a basis of $E$ and $\{e^1,\ldots,e^n\}$ is the dual basis of $E^\star$, then for every $\omega\in \Lambda^rE^\star$ (i.e. $\omega$ anti-symmetric covariant tensor) there exists $a\in\mathbb R$ such that: $$\omega=a\, e^1\wedge\ldots\wedge e^r,$$ where $\wedge$ is the exterior product of $E$.

## closed as off-topic by spaceisdarkgreen, Peter, kimchi lover, José Carlos Santos, user99914 Jan 13 '18 at 22:56

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I think you should firstly state what have you tried to solve the problem. Nevertheless, such a statement is true only if $r = n$. Counterexample for $r < n$: let $\omega$ be given by $$\omega = e^1 \wedge e^2 \wedge \dots \wedge e^r + e^2 \wedge e^3 \wedge \dots \wedge e^{r+1}.$$ Then $\omega$ is an $r$-form, i.e. $\omega \in \Lambda^r E^{\star}$ but cannot be written as $$\omega = a e^1 \wedge e^2 \wedge \dots \wedge e^r$$ for some $a \in \mathbb{R}$ since $e^1 \wedge e^2 \wedge \dots \wedge e^r$ and $e^2 \wedge e^3 \wedge \dots \wedge e^{r+1}$ are linearly independent elements of the basis of $\Lambda^r E^{\star}$
• In fact even $e^2 \wedge \dots \wedge e^{r+1}$ is a counter-example. – Nicolas Hemelsoet Jan 13 '18 at 19:14