NoteNextra-origin/pages/Math429/Math429_L39.md

Lecture 39

Chapter IX Multilinear Algebra and Determinants

Exterior Powers ?A

Definitions ?.1

Let V be a vector space, the n-th exterior power of V denoted \wedge^m V is a vector space formed by finite linear combination of expression of the form v_1\wedge v_2\wedge\dots \wedge v_m. subject to relations:

1. c(v_1\wedge v_2\wedge\dots \wedge v_m)=(cv_1)\wedge v_2\wedge\dots \wedge v_m
2. (v_1+w_1)\wedge v_2\wedge\dots \wedge v_m=(v_1\wedge v_2\wedge\dots \wedge v_m)+(w_1\wedge v_2\wedge\dots \wedge v_m)
3. Swapping two entires in (v_1\wedge v_2\wedge\dots \wedge v_m) gives a negative sign.

Example:

\wedge^2\mathbb{R}^3

\begin{aligned}
   &(1,0,0)\wedge(0,1,0)+(1,0,1)\wedge(1,1,1)\in \wedge^2\mathbb{R}^3\\
   &=(1,0,0)\wedge(0,1,0)+((1,0,0)+(0,0,1))\wedge(1,1,1)\\
   &=(1,0,0)\wedge(0,1,0)+(1,0,0)\wedge(1,1,1)+(0,0,1)\wedge(1,1,1)\\
   &=(1,0,0)\wedge(1,2,1)+(0,0,1)\wedge(1,1,1)
\end{aligned}

Theorem ?.2

0\wedge v_1\wedge\dots\wedge v_m=0

Proof:

\begin{aligned}
\vec{0}\wedge v_2\wedge\dots \wedge v_m &=(0\cdot \vec{0})\wedge v_2\wedge \dots\wedge v_m\\
&=0(\vec{0}\wedge v_2\wedge \dots\wedge v_m)\\
&=0
\end{aligned}

Theorem ?.3

v_1\wedge v_1\wedge\dots\wedge v_m=0

Proof:

swap v_1 and v_1.

\begin{aligned}
v_1\wedge v_1 \wedge v_2\wedge\dots \wedge v_m &=-(v_1\wedge v_1 \wedge v_2\wedge\dots \wedge v_m) \\
v_1\wedge v_1 \wedge v_2\wedge\dots \wedge v_m&=0
\end{aligned}

Theorem ?.4

v_1\wedge v_2\wedge\dots\wedge v_m\neq 0 if and only if v_1,\dots ,v_m are linearly independent.

Proof:

We first prove forward direction,

Suppose v_1,\dots, v_m are linearly dependent then let a_1v_1+\dots +a_nv_m=0 be a linear dependence. Without loss of generality. a\neq 0 then consider

\begin{aligned}
0&=0\wedge v_2\wedge\dots\wedge v_m\\
&=(a_1,v_1+...+a_m v_m)\wedge v_2\wedge \dots \wedge v_m\\
&=a_1(v_1\wedge \dots v_m)+a_2(v_2\wedge v_2\wedge \dots \wedge v_m)+a_m(v_m\wedge v_2\wedge\dots\wedge v_m)\\
&=a_1(v_1\wedge \dots v_m)
\end{aligned}

reverse is the similar.

Theorem ?.5

If v_1,\dots v_n forms a basis for V, then expressions of the form v_{i_1}\wedge\dots \wedge v_{i_m} for 1\leq i_1\leq i_m\leq n forms a basis of \wedge^m V

Proof:

Spanning: Let u_1\wedge\dots \wedge u_m\in \wedge^m V where u_1=a_{1,1}v_1+\dots+a_{1,n}v_n,u_m=a_{m,1}v_1+\dots+a_{m,n}v_n

Expand: then we set expressions of the form \plusmn c(v_{i_1}\wedge \dots \wedge v_{i_m}). Let A=(a_{i,j}) , c is the m\times m minor for the columns i_1,..,i_m.

Corollary ?.6

Let n=dim\ V then dim\ \wedge^n v=1

Note $dim\ \wedge^m V=\begin{pmatrix} n\m \end{pmatrix}$

Proof: Chose a basis v_1,...,v_n of V then v_1\wedge \dots \wedge v_n generates \wedge^n v.

Definition ?.7

Let T\in\mathscr{L}(V), n=dim\ V define det\ T to be the unique number such that for v_1\wedge\dots\wedge v_n\in \wedge^n V. (Tv_1\wedge\dots\wedge Tv_n)=(det\ T)(v_1\wedge \dots \wedge v_n)

Theorem ?.8

1. Swapping columns negates the determinants
2. T is invertible if and only if det\ T\neq 0
3. det(ST)=det(S)det(T)
4. det(cT)=c^n det(T)