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Math429
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Math429 L39

Lecture 39

Chapter IX Multilinear Algebra and Determinants

Exterior Powers ?A

Definitions ?.1

Let VV be a vector space, the n-th exterior power of VV denoted mV\wedge^m V is a vector space formed by finite linear combination of expression of the form v1v2vmv_1\wedge v_2\wedge\dots \wedge v_m. subject to relations:

  1. c(v1v2vm)=(cv1)v2vmc(v_1\wedge v_2\wedge\dots \wedge v_m)=(cv_1)\wedge v_2\wedge\dots \wedge v_m
  2. (v1+w1)v2vm=(v1v2vm)+(w1v2vm)(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 (v1v2vmv_1\wedge v_2\wedge\dots \wedge v_m) gives a negative sign.

Example:

2R3\wedge^2\mathbb{R}^3

(1,0,0)(0,1,0)+(1,0,1)(1,1,1)2R3=(1,0,0)(0,1,0)+((1,0,0)+(0,0,1))(1,1,1)=(1,0,0)(0,1,0)+(1,0,0)(1,1,1)+(0,0,1)(1,1,1)=(1,0,0)(1,2,1)+(0,0,1)(1,1,1)\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

0v1vm=00\wedge v_1\wedge\dots\wedge v_m=0

Proof:

0v2vm=(00)v2vm=0(0v2vm)=0\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

v1v1vm=0v_1\wedge v_1\wedge\dots\wedge v_m=0

Proof:

swap v1v_1 and v1v_1.

v1v1v2vm=(v1v1v2vm)v1v1v2vm=0\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

v1v2vm0v_1\wedge v_2\wedge\dots\wedge v_m\neq 0 if and only if v1,,vmv_1,\dots ,v_m are linearly independent.

Proof:

We first prove forward direction,

Suppose v1,,vmv_1,\dots, v_m are linearly dependent then let a1v1++anvm=0a_1v_1+\dots +a_nv_m=0 be a linear dependence. Without loss of generality. a0a\neq 0 then consider

0=0v2vm=(a1,v1+...+amvm)v2vm=a1(v1vm)+a2(v2v2vm)+am(vmv2vm)=a1(v1vm)\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 v1,vnv_1,\dots v_n forms a basis for VV, then expressions of the form vi1vimv_{i_1}\wedge\dots \wedge v_{i_m} for 1i1imn1\leq i_1\leq i_m\leq n forms a basis of mV\wedge^m V

Proof:

Spanning: Let u1ummVu_1\wedge\dots \wedge u_m\in \wedge^m V where u1=a1,1v1++a1,nvn,um=am,1v1++am,nvnu_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 ±c(vi1vim)\plusmn c(v_{i_1}\wedge \dots \wedge v_{i_m}). Let A=(ai,j)A=(a_{i,j}) , cc is the m×mm\times m minor for the columns i1,..,imi_1,..,i_m.

Corollary ?.6

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

Note dim mV=(nm)dim\ \wedge^m V=\begin{pmatrix} n\\m \end{pmatrix}

Proof: Chose a basis v1,...,vnv_1,...,v_n of VV then v1vnv_1\wedge \dots \wedge v_n generates nv\wedge^n v.

Definition ?.7

Let TL(V)T\in\mathscr{L}(V), n=dim Vn=dim\ V define det Tdet\ T to be the unique number such that for v1vnnVv_1\wedge\dots\wedge v_n\in \wedge^n V. (Tv1Tvn)=(det T)(v1vn)(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. TT is invertible if and only if det T0det\ T\neq 0
  3. det(ST)=det(S)det(T)det(ST)=det(S)det(T)
  4. det(cT)=cndet(T)det(cT)=c^n det(T)