Finding a Basis for a Bivector in $\Lambda^2 (V)$

[smile1]

Hello everyone

Can anyone help me to solve the following problem

Prove that for any bivector $v\in \Lambda^2 (V)$ there is a basis $\{e_1,e_2,...,e_n\}$ of $V$ such that $v=e_1\Lambda e_2 +e_3\Lambda e_4 +...+e_{k-1}\Lambda e_k$

I did it in this way, since the out product $e_i\Lambda e_j$ forms a bivector in $\Lambda^2 (V)$ then, it forms a basis for $v$. However, I am not sure about the dimension of the bivector $v$ how many basis do we need to form the bivector $v$?

Thanks

 

[jvicens]

for your question! Solving this problem involves understanding the properties and structure of bivectors in a vector space.

First, let's define a bivector as an element of the exterior power $\Lambda^2 (V)$ of a vector space $V$. This means that a bivector is a two-dimensional subspace of $V$.

Now, to prove that for any bivector $v\in \Lambda^2 (V)$ there is a basis $\{e_1,e_2,...,e_n\}$ of $V$ such that $v=e_1\Lambda e_2 +e_3\Lambda e_4 +...+e_{k-1}\Lambda e_k$, we can use the following steps:

1. Since $v$ is a bivector, it can be represented as a linear combination of wedge products of basis vectors in $V$. Let's say $v=\sum_{i=1}^{k}a_i e_i\Lambda e_j$, where $a_i$ are scalars and $e_i, e_j$ are basis vectors.

2. Now, we need to show that we can choose a basis $\{e_1,e_2,...,e_n\}$ of $V$ such that the wedge products $e_i\Lambda e_j$ form a basis for $v$. To do this, we will use the Gram-Schmidt process.

3. Start by choosing any two linearly independent basis vectors $e_1, e_2$ for $v$. Then, we can write $v$ as $v=e_1\Lambda e_2 + v'$, where $v'$ is a bivector in the subspace orthogonal to $e_1\Lambda e_2$. This means that $v'$ can be represented as a linear combination of wedge products of basis vectors from this subspace.

4. Next, choose a third basis vector $e_3$ that is linearly independent from $e_1$ and $e_2$. Then, we can write $v'$ as $v'=e_3\Lambda e_4 + v''$, where $v''$ is a bivector in the subspace orthogonal to $e_3\Lambda e_4$.

5. Continue this process until we have $k$ linearly independent basis vectors $e_1, e_2,...,