Definition. Let T:V→W be a linear transformation. Then the kernel of T is given by
ker(T)={v∣T(v)=0, v in V}. ker(T) is a subspace of V; the proof of this is identical to the proof we used to show that NulA is a subspace.
The range of T is given by
ran(T)={w∣T(v)=w for some v in V}.
ran(T) is a subspace of W. Proof:
- T(0V)=0W
- Thus 0W is in ran(T).
- Take u,v in ran(T).
- Now there are u′,v′ in V such that T(u′)=u and T(v′)=v.
- u+v=T(u′)+T(v′)
- u+v=T(u′+v′)
- Thus u+v is also in ran(T).
- Let c be any scalar.
- cu=cT(u′)
- cu=T(cu′)
- Thus cu is also in ran(T).
- Thus ran(T) is a subspace of W.
Ex. T:C∞(R)→C∞(R), T(f)=dxdfker(T)={constant functions in C∞(R)} The kernal space of T 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 Take c to be in R. Consider the continuous function f(x)=c for all x in [0,1]: T(f)=∫01f(x)dx=c. Therefore the range of this transformation is R.
Null Space, Kernel, and Linear Independence
The linear transformation (mapping) T:Rn→Rm is one-to-one if and only if the columns of A, the standard matrix of T, are linearly independent, which is true if and only if NulA={0}.
Let T:V→W be a linear transformation. T is one-to-one if and only if ker(T)={0}.
An indexed set of vectors {v1,...,vp} is said to be linearly independent if and only if c1v1+...+cpvp=0 is only true if c1=...=cp=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.)
Collections and Linear Independence
Definition. A collection of vectors u from V is linearly independent if and only if any (every) finite subcollection of u is linearly independent.
Ex. u={0,1,t,t2,...,tn,...}. (infinitely sized collection of vectors)
The subcollection {0,t} is not linearly independent, so u is not linearly independent either.
Ex. u={1,t,1−t}.
1(1)−1(t)+1(1−t)=0; therefore u is not linearly independent.
Definition. B is called a basis for V if both are true:
- B is linearly independent.
- spanB=V.
Ex. The basis of Rn is e1,e2,...,en.
Every vector space has a basis, but they are sometimes hard to find and represent for infinite-dimensional vector spaces.