115 lines
3.1 KiB
Markdown
115 lines
3.1 KiB
Markdown
# Lecture 9
|
|
|
|
## Chapter III Linear maps
|
|
|
|
**Assumption: $U,V,W$ are vector spaces (over $\mathbb{F}$)**
|
|
|
|
### Vector Space of Linear Maps 3A
|
|
|
|
Review
|
|
|
|
$\mathscr{L}(V,W) =$ space of linear maps form $V$ to $W$.
|
|
|
|
$\mathscr{L}(V)=\mathscr{L}(V,V)$
|
|
|
|
Key facts:
|
|
|
|
* $\mathscr{L}(V,W)$ is a vector space
|
|
* given $T\in\mathscr{L}(U,V),S\in \mathscr{L}(V,W)$, we have $TS\in \mathscr{L}(V,W)$
|
|
* not commutative
|
|
|
|
### Null spaces and Range 3B
|
|
|
|
#### Definition 3.11
|
|
|
|
Null space and injectivity
|
|
|
|
For $T\in \mathscr{L}(V,W)$, the **null space** of $T$, denoted as $null(T)$ (sometime also noted as $ker\ T$), is a subset of $V$ given by
|
|
|
|
$$
|
|
ker\ T=null(T)=\{v\in V \vert Tv=0\}
|
|
$$
|
|
|
|
Examples:
|
|
|
|
* $0\in \mathscr{L}(V,W)$, then $null\ 0=V$, $null(I)=\{0\}$
|
|
* $T\in \mathscr{L}(\mathbb{R}^3,\mathbb{R}^2),T(x,y,z)=(x+y,y+z)$, to find the null space, we set $T(x,y,z)=0$, then $x+y=0,y+z=0$, $x=-y,x=z$. So $null(T)=\{(x,-x,x)\in \mathbb{R}^2\vert x\in \mathbb{R}\}$
|
|
* Let $D\in \mathscr{L}(\mathscr{P}(\mathbb{R})),D(f)=f'$, $null (D)=$ the set of constant functions. (because the derivatives of them are zero.)
|
|
|
|
#### Theorem 3.13
|
|
|
|
Given $T\in \mathscr{L}(V,W)$, $null(T)$ is a subspace of $V$.
|
|
|
|
Proof:
|
|
|
|
We check the conditions for the subspace.
|
|
* $T0=0$, so $0\in null(T)$
|
|
* $u,v\in null(T)$, then consider $T(u+v)=Tu+Tv=0+0=0$, so $u+v\in null(T)$
|
|
* Let $v\in null (T),\lambda \in \mathbb{F}$, then $T(\lambda v)=\lambda (Tv)=\lambda 0=0$, so $\lambda v \in null (T)$
|
|
|
|
So $null(T)$ is a subspace.
|
|
|
|
#### Definition 3.14
|
|
|
|
A function $f:V\to W$ is **injective** (also called one-to-one, 1-1) if for all $u,v\in V$, if $Tv=Tu$, then $T=U$.
|
|
|
|
#### Lemma 3.15
|
|
|
|
Let $T\in \mathscr{L}(V,W)$ then $T$ is injective if and only if $null(T)=\{0\}$
|
|
|
|
Proof:
|
|
|
|
$\Rightarrow$
|
|
|
|
Let $T\in \mathscr{L}(V,W)$ be injective, and let $v\in null (T)$. Then $Tv=0=T0$ so because $T$ is injective $v=0\implies null (T)=\{0\}$
|
|
|
|
$\Leftarrow$
|
|
|
|
Suppose $T\in \mathscr{L}(V,W)$ with $null (T)=\{0\}$. Let $u,v\in V$ with $Tu=Tv$, $Tu-Tv=0,T(u-v)=0,u-v=0,u=v$, so $T$ is injective
|
|
|
|
#### Definition 3.16
|
|
|
|
Range and surjectivity
|
|
|
|
For $T\in \mathscr{L}(V,W)$ the range of $T$ denoted $range(T)$, is given by
|
|
|
|
$$
|
|
range(T)=\{Tv\vert v\in V\}
|
|
$$
|
|
|
|
Example:
|
|
|
|
* $0\in \mathscr{L}(V,W)$, $range(0)=\{0\}$
|
|
* $I\in \mathscr{L}(V,W)$, $range(I)=V$
|
|
* Let $T:\mathbb{R}\to \mathbb{R}^2$ given by $T(x)=(x,2x)$, $range(T)=\{(x,2x)\vert x\in \mathbb{R}\}$
|
|
|
|
#### Theorem 3.18
|
|
|
|
Given $T\in \mathscr{L}(V,W)$, $range(T)$ is a subspace of $W$
|
|
|
|
Proof:
|
|
|
|
Exercise, not interesting.
|
|
|
|
#### Definition 3.19
|
|
|
|
A function $T:V\to W$ is surjective (also called onto) if $range(T)=W$
|
|
|
|
#### Theorem 3.21 (The Fundamental Theorem of Linear Maps, Rank-nullity Theorem)
|
|
|
|
Suppose $V$ is finite dimensional, and $T\in \mathscr{L}(V,W)$, then $range(T)$ is finite dimensional ($W$ don't need to be finite dimensional). and
|
|
|
|
$$
|
|
dim(V)=dim(null (T))+dim(range(T))
|
|
$$
|
|
|
|
#### Theorem 3.22
|
|
|
|
Let $T\in \mathscr{L}(V,W)$ and suppose $dim(V)>dim(W)$. Then $T$ is not injective.
|
|
|
|
Proof:
|
|
|
|
By **Theorem 3.21**, $dim(V)=dim(null (T))+dim(range(T))$, $dim(V)=dim(null (T))+dim(W)$, $0<dim (V)-dim(W)\leq dim(null(T))\implies dim(null(T))>0\implies null (T)\neq \{0\}$
|
|
|
|
So $T$ is not injective.
|