MATH 240

Wed. October 16th, 2019


Kernel

Definition. Let T:VWT:V\rightarrow W be a linear transformation. Then the kernel of TT is given by
ker(T)={vT(v)=0, v in V}.\text{ker}(T)=\{v\mid T(v)=0, \ v \ \text{in} \ V\}. ker(T)\text{ker}(T) is a subspace of VV; the proof of this is identical to the proof we used to show that NulA\text{Nul}A is a subspace.

Range

The range of TT is given by
ran(T)={wT(v)=w for some v in V}.\text{ran}(T)=\{w\mid T(v)=w\ \text{for some} \ v \ \text{in} \ V\}.
ran(T)\text{ran}(T) is a subspace of WW. Proof:


Ex. T:C(R)C(R), T(f)=dfdxker(T)={constant functions in C(R)}\begin{gathered} T:C^\infin(\R)\rightarrow C^\infin(\R), \ T(f)=\dfrac{df}{dx}\\ \text{ker}(T)=\{\text{constant functions in }C^\infin(\R)\} \end{gathered} The kernal space of TT here is the set of all functions whose derivatives are zero, which is just the set of constant functions in the space.


Ex.
T:C[0,1]R, T(f)=01f(x)dxran(T)=R\begin{gathered} T:C[0,1]\rightarrow \R, \ T(f)=\int_0^1f(x)dx\\ \text{ran}(T)=\R \end{gathered} Take cc to be in R\R. Consider the continuous function f(x)=cf(x)=c for all xx in [0,1][0,1]: T(f)=01f(x)dx=c.T(f)=\int_0^1f(x)dx=c. Therefore the range of this transformation is R\R.


Null Space, Kernel, and Linear Independence

The linear transformation (mapping) T:RnRmT:\R^n\rightarrow \R^m is one-to-one if and only if the columns of AA, the standard matrix of TT, are linearly independent, which is true if and only if NulA={0}\text{Nul}A=\{0\}.

Let T:VWT:V\rightarrow W be a linear transformation. TT is one-to-one if and only if ker(T)={0}\text{ker}(T)=\{0\}.

An indexed set of vectors {v1,...,vp}\{v_1,...,v_p\} is said to be linearly independent if and only if c1v1+...+cpvp=0c_1v_1+...+c_pv_p=0 is only true if c1=...=cp=0c_1=...=c_p=0. (Note that this is the same as our previous definition of linear independence, but applies to any kind of vector, not just those in Rn\R^n.)


Collections and Linear Independence

Definition. A collection of vectors uu from VV is linearly independent if and only if any (every) finite subcollection of uu is linearly independent.

Ex. u={0,1,t,t2,...,tn,...}u=\{0,1,t,t^2,...,t^n,...\}. (infinitely sized collection of vectors)
The subcollection {0,t}\{0,t\} is not linearly independent, so uu is not linearly independent either.

Ex. u={1,t,1t}u=\{1,t,1-t\}.
1(1)1(t)+1(1t)=01(1)-1(t)+1(1-t)=0; therefore uu is not linearly independent.


Basis

Definition. BB is called a basis for VV if both are true:

Ex. The basis of Rn\R^n is e1,e2,...,ene_1,e_2,...,e_n.

Every vector space has a basis, but they are sometimes hard to find and represent for infinite-dimensional vector spaces.