107 lines
3.2 KiB
Markdown
107 lines
3.2 KiB
Markdown
# Lecture 39
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## Chapter IX Multilinear Algebra and Determinants
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### Exterior Powers ?A
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#### Definitions ?.1
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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:
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1. $c(v_1\wedge v_2\wedge\dots \wedge v_m)=(cv_1)\wedge v_2\wedge\dots \wedge v_m$
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2. $(v_1+w_1)\wedge v_2\wedge\dots \wedge v_m=(v_1\wedge v_2\wedge\dots \wedge v_m)+(w_1\wedge v_2\wedge\dots \wedge v_m)$
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3. Swapping two entires in ($v_1\wedge v_2\wedge\dots \wedge v_m$) gives a negative sign.
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Example:
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$\wedge^2\mathbb{R}^3$
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$$
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\begin{aligned}
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&(1,0,0)\wedge(0,1,0)+(1,0,1)\wedge(1,1,1)\in \wedge^2\mathbb{R}^3\\
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&=(1,0,0)\wedge(0,1,0)+((1,0,0)+(0,0,1))\wedge(1,1,1)\\
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&=(1,0,0)\wedge(0,1,0)+(1,0,0)\wedge(1,1,1)+(0,0,1)\wedge(1,1,1)\\
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&=(1,0,0)\wedge(1,2,1)+(0,0,1)\wedge(1,1,1)
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\end{aligned}
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$$
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#### Theorem ?.2
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$0\wedge v_1\wedge\dots\wedge v_m=0$
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Proof:
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$$
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\begin{aligned}
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\vec{0}\wedge v_2\wedge\dots \wedge v_m &=(0\cdot \vec{0})\wedge v_2\wedge \dots\wedge v_m\\
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&=0(\vec{0}\wedge v_2\wedge \dots\wedge v_m)\\
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&=0
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\end{aligned}
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$$
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#### Theorem ?.3
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$v_1\wedge v_1\wedge\dots\wedge v_m=0$
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Proof:
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swap $v_1$ and $v_1$.
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$$
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\begin{aligned}
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v_1\wedge v_1 \wedge v_2\wedge\dots \wedge v_m &=-(v_1\wedge v_1 \wedge v_2\wedge\dots \wedge v_m) \\
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v_1\wedge v_1 \wedge v_2\wedge\dots \wedge v_m&=0
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\end{aligned}
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$$
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#### Theorem ?.4
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$v_1\wedge v_2\wedge\dots\wedge v_m\neq 0$ if and only if $v_1,\dots ,v_m$ are linearly independent.
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Proof:
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We first prove forward direction,
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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
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$$
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\begin{aligned}
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0&=0\wedge v_2\wedge\dots\wedge v_m\\
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&=(a_1,v_1+...+a_m v_m)\wedge v_2\wedge \dots \wedge v_m\\
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&=a_1(v_1\wedge \dots v_m)+a_2(v_2\wedge v_2\wedge \dots \wedge v_m)+a_m(v_m\wedge v_2\wedge\dots\wedge v_m)\\
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&=a_1(v_1\wedge \dots v_m)
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\end{aligned}
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$$
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reverse is the similar.
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#### Theorem ?.5
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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 $\wedge^m V$
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Proof:
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Spanning: Let $u_1\wedge\dots \wedge u_m\in \wedge^m V$ where $u_1=a_{1,1}v_1+\dots+a_{1,n}v_n,u_m=a_{m,1}v_1+\dots+a_{m,n}v_n$
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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 $i_1,..,i_m$.
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#### Corollary ?.6
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Let $n=dim\ V$ then $dim\ \wedge^n v=1$
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Note $dim\ \wedge^m V=\begin{pmatrix}
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n\\m
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\end{pmatrix}$
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Proof: Chose a basis $v_1,...,v_n$ of $V$ then $v_1\wedge \dots \wedge v_n$ generates $\wedge^n v$.
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#### Definition ?.7
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Let $T\in\mathscr{L}(V)$, $n=dim\ V$ define $det\ T$ to be the unique number such that for $v_1\wedge\dots\wedge v_n\in \wedge^n V$. $(Tv_1\wedge\dots\wedge Tv_n)=(det\ T)(v_1\wedge \dots \wedge v_n)$
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#### Theorem ?.8
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1. Swapping columns negates the determinants
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2. $T$ is invertible if and only if $det\ T\neq 0$
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3. $det(ST)=det(S)det(T)$
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4. $det(cT)=c^n det(T)$ |