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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 $L$?

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    For (a), you are on a good way. Now use linearity, and think about what is $\omega(v,v,w)$. For (b), show that $\omega_L$ is also a skew-symmetric trilinear form, and then think about what it means that the dimension is 1.2012-08-04
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    (a)-(d): yes. I recommend any introductory textbook on the multilinear algebra.2012-08-04
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    **Hint:** For (d): define $\omega_L(v_1,v_2,v_3)=Lv_1\wedge v_2\wedge v_3+v_1\wedge Lv_2\wedge v_3+v_1\wedge v_2\wedge Lv_3$. For (g): define $T_\omega:\mathbb{R}^2\to\mathbb{R}^\ast$ by $T_\omega(v_1,v_2)=\omega(v_1,v_2,\cdot)$, that is $T_\omega:\mathbb{R}^2\to\mathbb{R}^\ast$ by $T_\omega(v_1,v_2)v=\omega(v_1,v_2,v)$.2012-08-04
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    @user31899: oopss... sorry!!! for (d): $\omega_L(v_1,v_2,v_3)=\omega(Lv_1,v_2,v_3)+\omega(v_1,Lv_2,v_3)+\omega(v_1,v_2,Lv_3)$... no idea why I wrote $v_1\wedge v_2\wedge v_3$... sorry!!! =p Try to write $\omega_l(v_1,v_2,v_3)=k\omega(v_1,v_2,v_3)$, for some $k$.2012-08-04
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    @user31899: $T_\omega:(\mathbb{R}^3)^2\to(\mathbb{R}^3)^\ast$ (sorry, typo) is for PART(g), how can you canonically "move" from $(\mathbb{R}^3)^\ast$ to $\mathbb{R}^3$? For (e): you can do anything analogous you did in (b) ($\omega_L(v_1,v_2,v_3):=\omega(Lv_1,Lv_2,Lv_3)$), or in (c) ($\omega_L(v_1,v_2,v_3):=\omega(Lv_1,v_2,v_3)+\omega(v_1,Lv_2,v_3)+\omega(v_1,v_2,Lv_3)$) (try to "put $L$" in "two entries", and sum all the possibilities). For (f), (solve (e)) and note that "determinant" is the product of the eigenvalues and "trace" is its sum.2012-08-05
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    BTW,I don't quite understand what you mean by Canonically "move" from V* to V, as I know there is no general structural relationship between V and V*, however there is one between V and V** which is not relevant in this case?!2012-08-06

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