Let $E$ be a vector space and $A$ a subspace of $E$. Let $q$ be a positive integer. Then we can define a subspace $\Lambda^q A$ of the $q$-th exterior power of $E$ by

$$ \Lambda^qA=span\{\ a_1\wedge\dots\wedge a_q\ |\ a_1,\dots,a_q\in A\ \} $$

On the other hand, assume that $\varphi$ is a linear application from $E$ into $F$. We "exteriorize" $\varphi$ to obtain a linear application $\Lambda^q$ from $\Lambda^qE$ into $\Lambda^qF$ satisfying

$$ \Lambda^q\varphi(e_1\wedge\dots\wedge e_q)=\varphi(e_1)\wedge\dots\wedge\varphi(e_q) $$

for all $e_1,\dots,e_q$ in $E$.

Now, if $A=Ker(\varphi)$, then $\Lambda^qA$ is a subspace of $Ker(\Lambda^q\varphi)$ but the inclusion is in general strict, since the decomposable elements of the form $a\wedge e_2\wedge\dots\wedge e_q$ are also in $Ker(\Lambda^q\varphi)$

My questions:

  1. How can I describe $Ker(\Lambda^q\varphi)$ using $A$?
  2. Is there a simple way to define a linear application $\Phi$, related to $\varphi$, and whose kernel is $\Lambda^qA$?

That's two questions but there are closely related, so I think it is worth having them at the same place.


Answer to 1:

The kernel of $\wedge^n \phi$ is precisely the antisymmetric subspace generated by the elements you describe. In particular, we have $$ \ker \wedge^n\phi = A \wedge \left(\wedge^{n-1} E\right) $$

Answer to 2:

Perhaps you should consider the map $$ \Phi = \phi \wedge \left(\wedge^{n-1} I\right) $$ where $I$ denotes the identity mapping over $E$.

This map is sometimes denoted as $A^{[n]}$. It is notable that $A^{[n]} = \frac {d}{dt} \wedge^{n}(I + tA)$. Bhatia discusses the map briefly in section I.5 of his "Matrix Analysis", but he doesn't say much else about it.

  • $\begingroup$ Feel free to ask for clarification $\endgroup$ – Omnomnomnom Aug 10 '15 at 12:58

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