3B. Null Spaces and Ranges

Definition

Null space of a linear map is defined as: $\text{null}T=\{v\in V\mid Tv=0\}$ Basically the kernel, or the part of the function that maps to 0.

Null space is a subspace

Suppose $T\in \mathcal{L}(V,W)$, then $\text{null }T$ is a subspace of $V$

Proof:

1. Identity: As every linear map sends $0$ to $0$, we know $T(0)=0$ thus 0 is in the null space
2. Additive closure: Let some $a,b\in \text{null }T$. That means $Ta=0,Tb=0$. Then: $T(a+b)=Ta+Tb=0+0=0$
3. Scaler Multiplicative Closure Let some $\lambda \in F$ and $a\in \text{null }T$. then $T(\lambda a)=\lambda Ta=\lambda 0=0$

Straight from kernels in abstract algebra

Injective or one-to-one

A function $T$ is called injective if $Tv=Tw⟹v=w$

This also means that $v\ne w⟹Tv\ne Tw$

This also just means that distinct inputs map to distinct outputs. A function only needs each input to map to one output, but allows for multiple inputs to map to the same output. So if you read the definition, if two inputs are not the same then their outputs are also not the same.

Check for Injective:

A function $T$ is injective $⟺$ $\text{null}T=\{0\}$

Proof:

1. $\Rightarrow$
   1. Assume a function $T$ is injective. That means $Ta=Tb⟹a=b$.
   2. By the same logic, we know $0$ has to map to $0$ for any linear map. Thus, only $0$ can map to $0$ as the function is injective, as if any other $Ta=T0$ then $a=0$
2. $\Leftarrow$
   1. Assume $\text{null}T=\{0\}$
   2. We have to show that $Ta=Tb⟹a=b$.
   3. If $Ta=Tb$ then we know $Ta-Tb=0$
   4. Thus $T(a-b)=0$
   5. As the only thing in the null space is $0$, we know $a-b=0$ or $a=b$.

Range

Definition

For a function $T\in \mathcal{L}(V,W)$, the range is a subset of $W$ equal to $Tv$ for all $v\in V$ $\text{Range}T=\{Tv\mid v\in V\}$

This is just a collection of the outputs of the function

Range is a subspace

For $T\in \mathcal{L}(V,W)$, we have it so $\text{range}T$ is a subspace of $W$

Proof:

1. 0
   1. $T0=0$ and $0\in V$
2. Additive Closure
   1. Let $a,b\in \text{range}T$. this means there exists $u,v\in Vs.t.T(u)=a,T(v)=b$
   2. $T(u+v)=T(u)+T(v)=a+b$
3. Scaler Multiplicative closure
   1. Let $a\in \text{range}T,\lambda \in F$. Then $\exists u\in V\mid T(u)=a$
   2. $T(\lambda u)=\lambda Tu=\lambda a$

Surjective or onto

A function $T:V\to W$ is surjective if $\text{range}T=W$

todo look at example

Fundamental Theorem of linear maps

Suppose $V$ is a finite-dimensional and $T\in \mathcal{L}(V,W)$. Then range $T$ is finite dimensional and $\text{dim}V=\text{dim}\text{null}T+\text{dim}\text{range}T$

todo take a look at this proof

By the proof above we get some important results, that you should rewrite and prove:

Proof:

1. We just need to show that the null space of a function $T\in \mathcal{L}(V,W)$ is not $\{0\}$
2. Dim V = Dim null T + Dim range T
3. Dim null T = Dim V - Dim range T
4. Dim null T >= Dim V - Dim range W (as range is a subspace which always has a lower dimension)
5. Dim null T > 0

similar proof with iniqualities.