7.2 Kernal and Image of LT
Definition
T: VW is a LT. The kernal of T (ker T) and the image of
T (im T or T(V)) are defined by:
ker T = {v in V | T(v) = 0} all v such that trans is 0
(also called nullspace of T.)
im T = {T(v) | v in V} trans of every V
(also called the range of T)
(See w/ diagrams of sets V and W)
Theorem 1:
If T: VW is a LT, then:
kert T is a subspace of V and im T is a subspace of W.
Proof: 1) Need to show they contain 0 which they do since
T(0) = 0.
2) Show that closed under addition:
ker T:
T(v)=0=T(v1)
T(v+v1)=T(v)+T(v1)=0
im T:
T(v)=w, T(v1)=w1
T(v+v1)=T(v)+T(v1)=w+w1
Thm 1 (proof cont)
3) must show that closed under scalar mult.:
ker T: T(rv) = rT(v)=0r = 0
im T: T(rv) = rT(v)= rw •
Example
If T : 3  3 (T(x,y,z)=(x-y, z, y-x) find ker T, im T and
their dimesions:
kerT={(x,y,z)|(x-y,z,y-x) = (0,0,0)}
so z=0, x=y so: {(t,t,0)| t in 
dim (ker T) = 1
im T = {(x-y,z,y-x) | x,y,z in }={(s,t,-s) | s,t in }
dim (im T) = 2
Example
If TA : n  m w/ A (mxn) (TA(X)=AX)
kerTA={X|AX = 0} which is the nullspace of A
im TA ={AX|X in n} which is the range of A
Definition
Given T: VW is a LT:
dim (ker T) is called the nullity of T (nullity (T))
dim(im T) is called the rank of T (rank (T))
Example
A (mxn), show that im (TA) = col A, so rank TA = rank A.
Proof: A = [C1 … Cn]


imTA  {AX | X  n }  C1


 {x1 C1  ...  x n Cn | x i }

x1 
x  
... Cn 
...

  i


x n 


This tells us that im TA = span{C1,…,Cn} =col space of A.
Earlier, we said that rank A = dim (col A), so
rank TA = rank A
Example
Define a transformation T: Mnn Mnn by T(A) = A-AT .
Show that T is linear and that:
a) ker T consists of all symmetric matrices
b) im T consists of all skew-symmetric matrices. (ST=-S)
Proof: Relatively easy to show that it is LT.
a) T(A) = A-AT = 0 for ker T. So A = AT (ie A is sym)
b) T(A)T = (A - AT)T = AT-A = -T(A) so every mtx is skew s
Definition
T: VW is an LT:
A) T is onto if im T = W (i.e. every element maps to one
element of W and covers every element of W)
B) T is said to be one-to-one if T(v) = T(v1) implies v=v1
(i.e. two different elements in V cannot map to the same
element in W)
(Diagram)
With onto: im T is as large as possible
With one-to-one: ker T is as small as possible (no two values
in V can both map to 0)
Theorem 2
If T: VW is an LT, then T is one-to-one iff ker T = 0
Proof:
If T is one-to-one, we assign v to be any vector in
ker T. Then T(v) = 0 = T(0).
 If ker T = 0, let T(v) = T(v1).
Then T(v-v1) = T(v) - T(v1) = 0
Thus v - v1 is in ker T=0. So v - v1 = 0 so v=v1
So T is one-to-one.
Example
S: 3 2 S(x,y,z) = (x+y,x-y)
T: 2 3 T(x,y) = (x+y,x-y,x)
Show T one-to-one, not onto; S onto, not one-to-one
Proof: 1) T one-to-one (just show ker T=0)
Ker T = {(x,y)|x+y=x-y=x=0}={(0,0))=0
2) T not onto: need to find some element of 3 not in im T
ex. (1,1,0)
3) S not one-to-one since (0,0,1) is in ker S
4) S onto: we can cover every single value in 2 with
(s,t)=(x+y,x-y) -- then x=s-y=t+y so y=1/2(s-t), x=1/2(s+t)
Example
U-invertible (mxm)
T:MmnMmn T(X) = UX for all X in Mmn (show 1-1, onto)
1-1
ker T={X|UX=0} UX=0  X=0 (mult by U-1) so ker T=0
Onto let Y be any element of Mmn then T(U-1Y)=UU-1Y=Y so
we are able to cover any element of Mmn Y w/ input U-1Y. •
Theorem 3
A (mxn), TA: nm is TA(X) = AX
1. TA onto iff rank A=m (LI rows)
2. TA one-to-one iff rank A =n (LI columns)
Proof:1. Im TA = col A (shown earlier), so TA is onto iff col A
is m. Since rank A = dim (colA) (also shown earlier),
Rank A = m for col A to be m. This gives LI rows.
2. ker TA={X in n | AX=0}, so TA is 1-1 iff AX=0 implies
X=0. AX = [C1…Cn][X]=x1C1+…+xnCn (a LC of cols of A)
So if AX=0 implies X=0, we have LI cols. This is equivalent
to saying rank A = n.
•
Theorem 4-Dimension Thm
T: V W an LT and ker T and im T are finite dimensional.
Then V is finite dimensional and:
dim V = dim(ker T) + dim (im T)
or
dim V = nullity(T) + rank T
Proof: each vector in im T is of form T(v).
Let {T(e1),…,T(er)} be basis of im T (ei’s in V)
Let {f1,…,fk} be a basis of ker T (fi’s in V).
Then dim(im T)= r , and dim(ker T) = k.
Just show B={e1,…,er,f1,…,fk} is a basis of V.
1. B spans V: T(v) = t1T(e1)+…+trT(er) since ei’s form basis
T(v-t1e1-…-trer)=T(v)- t1T(e1)-…-trT(er)=0
so (v-t1e1-…-trer) is in ker T and is a linear combo of fi’s.
Theorem 4-cont
so (v-t1e1-…-trer)=s1f1+…+skfk
So v = t1e1+…+trer + s1f1+…+skfk
So v must be a LC of vectors in B. Thus B spans V.
2. B is LI: Suppose ti and sj satisfy:
t1e1+…+trer+s1f1+…+skfk = 0 (*)
If we apply T: t1T(e1)+…+trT(er) = 0 (since T(fi)=0 for all i)
Since T(ei)’s are LI, t1=…=tr = 0.
So (*) becomes: s1f1+…+skfk = 0
Since fj’s are LI, s1=…=sk=0. So B is LI!!!•
Theorem 5
T: VW is LT. Let {e1,…,er,er+1,…,en} be a basis of V such
that {er+1,…,en} is a basis of ker T. Then {T(e1),…,T(er)} is a
basis of im T, and so r = rank T.
(just putting together a few things we have seen in this
section--proof is homework problem)
Example
A (mxn) with rank r. Show that the space of all solutions of
the system AX=0 of m homogeneous equations in n variables
has dimension n-r.
Proof: since we are looking for all X such that AX=0, we just
have ker TA where TA: n m is TA(X)=AX
dim (im TA) = rank TA = rank A = r (as shown earlier)
So dim (ker TA) = dim(V) - dim(im(T))=n-r
(by dimension theorem)
Example
T: V  W is LT where V is finite dimensional.
Since dim V= dim(ker T) + dim (im T) we have:
dim(ker T) ≤ dim V and dim (im T) ≤ dim V
Example
D: Pn Pn-1 is defined by D[p(x)] = p’(x) (differentiation)
Find ker D and show that D is onto.
Proof: ker D = all p(x) that make D[p(x)]=p’(x)=0 which
only happens if p(x) is constant, so ker D = all constant fn’s.
Therefore, dim(ker D) = 1.
dim(Pn) = n+1 which then gives (by dimension thm)
dim(im D) = (n+1) - dim(kerD) = n = dim(Pn-1)
This implies that im D = Pn-1 which tells us that D is onto.
Example
Evaluation map: Ea: Pn  is Ea[p(x)] = p(a).
Show that Ea is linear and onto, and thus show that
{(x-a),(x-a)2,…,(x-a)n} is a basis of ker Ea, the subspace of all
polynomials p(x) for which p(a) = 0.
Proof: we will just go onto the second part of it:
dim(im Ea) = dim  = 1, so dim (ker Ea) = (n+1) - 1 = n
(x-a),(x-a)2,…,(x-a)3 all lie in ker Ea since if we evaluate any
of them at x=a, we get 0.
They are each LI (since all have diff’t degrees).
So we have n LI vectors all in ker Ea which has dim = n.
This tells us that we have a basis of Ea.
Example
A (mxn). Show that rank A = rank ATA = rank AAT.
Proof: Let B = ATA, and write: TA: nm, TB: nn
rank A = rank TA = dim (im TA) = n - dim (ker TA)
rank B = rank TB = dim (im TB) = n - dim (ker TB)
To show: ker (TA) = ker (TB)
AX = 0 implies that BX = ATAX = 0
This tells us that ker TA ker TB
Also, BX=0 means ATAX = 0 so: (from dot products)
||AX||2 = (AX)T(AX) = XTATAX = XT0 = 0
Which implies that AX=0, so ker TB  ker TA
Thus ker TB=ker TA
(and rank A = rank AAT follows same logic)