Let $V$ be an 3-dimensional vector space over $\mathbb{R}$. Let $\Lambda^3V^*$ denote the space of alternating trilinear forms on $V$. Note: An alternating trilinear form on $V$ is a map $\omega: V \times V \times V \to \mathbb{R}$ such that $\omega(v_1,v_2,v_3)=-\omega(v_2,v_1,v_3)=-\omega(v_3,v_2,v_1)$ and $\omega(av_1+v_4,v_2,v_3)=a\omega(v_1,v_2,v_3)+\omega(v_4,v_2,v_3)$.
(a) Show $\dim(\Lambda^3V^*)$=1.
(b) Given $L \in End(V)$ and $0 \neq \omega \in \Lambda^3V^*$ show that the function $\omega_L: V \times V \times V \to \mathbb{R}$ defined by $\omega_L(v_1,v_2,v_3):=\omega(Lv_1,Lv_2,Lv_3), \forall v_1,v_2,v_3 \in V$ is of the form $|L|\omega$ where $|L|$ is a constant which is dependent on $L$ but not on $\omega$.
(c) Show that in fact $|L|$ is the determinant of any matrix representing $L$ in terms of a basis of $V$. Is it true that $|LM|=|L||M|$ for any two determinants of endomorphisms $L,M$ of $V$?
(d) Is there way to define the trace of $L$ in a manifestly basis independent way similar to the definition of determinant in part (c)?
The following parts remain unsolved:
(e) We see the trace is a linear invariant of $L$, the determinant is a trilinear invariant of $L$. Now try to construct a bilinear invariant of $L$ along the same lines.
(f) If $L$ is diagonalisable, describe the last invariant in terms of eigenvalues of $L$.
(g) Given two linearly independent vectors $e_1$ and $e_2$ in $\mathbb{R}^3$, show that any non-zero $\omega \in \Lambda^3V^{*}$ may be used to explicitly construct a vector $e_3$ orthogonal to $e_1$ and $e_2$.
Note: For PART(e) I can't think of another invariant of $L$ , based on PART(f) this third invariant is related to eigenvalues of $L$ somehow, but what invariant besides the trace is closely related to the eigenvalues of $L$?
Best Answer
About part e), you know both the trace and the determinant are coefficients of the characteristic polynomial of $L$, which in this case has degree $3= \dim V$. There is one coefficient you have not used yet, and perhaps this is the bilinear invariant you are looking for.