# 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. $det(cT)=c^n det(T)$