9.8: The Kernel and Image of a Linear Map
( \newcommand{\kernel}{\mathrm{null}\,}\)
- Describe the kernel and image of a linear transformation.
- Use the kernel and image to determine if a linear transformation is one to one or onto.
Here we consider the case where the linear map is not necessarily an isomorphism. First here is a definition of what is meant by the image and kernel of a linear transformation.
Let V and W be vector spaces and let T:V→W be a linear transformation. Then the image of T denoted as im(T) is defined to be the set {T(→v):→v∈V} In words, it consists of all vectors in W which equal T(→v) for some →v∈V. The kernel, ker(T), consists of all →v∈V such that T(→v)=→0. That is, ker(T)={→v∈V:T(→v)=→0}
Then in fact, both im(T) and ker(T) are subspaces of W and V respectively.
Let V,W be vector spaces and let T:V→W be a linear transformation. Then ker(T)⊆V and im(T)⊆W. In fact, they are both subspaces.
- Proof
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First consider ker(T). It is necessary to show that if →v1,→v2 are vectors in ker(T) and if a,b are scalars, then a→v1+b→v2 is also in ker(T). But T(a→v1+b→v2)=aT(→v1)+bT(→v2)=a→0+b→0=→0
Thus ker(T) is a subspace of V.
Next suppose T(→v1),T(→v2) are two vectors in im(T). Then if a,b are scalars, aT(→v2)+bT(→v2)=T(a→v1+b→v2) and this last vector is in im(T) by definition.
Consider the following example.
Let T:P1→R be the linear transformation defined by T(p(x))=p(1) for all p(x)∈P1. Find the kernel and image of T.
Solution
We will first find the kernel of T. It consists of all polynomials in P1 that have 1 for a root. ker(T)={p(x)∈P1 | p(1)=0}={ax+b | a,b∈R and a+b=0}={ax−a | a∈R} Therefore a basis for ker(T) is {x−1} Notice that this is a subspace of P1.
Now consider the image. It consists of all numbers which can be obtained by evaluating all polynomials in P1 at 1. im(T)={p(1) | p(x)∈P1}={a+b | ax+b∈P1}={a+b | a,b∈R}=R Therefore a basis for im(T) is {1} Notice that this is a subspace of R, and in fact is the space R itself.
Let T:M22↦R2 be defined by T[abcd]=[a−bc+d] Then T is a linear transformation. Find a basis for ker(T) and im(T).
Solution
You can verify that T represents a linear transformation.
Now we want to find a way to describe all matrices A such that T(A)=→0, that is the matrices in ker(T). Suppose A=[abcd] is such a matrix. Then T[abcd]=[a−bc+d]=[00] The values of a,b,c,d that make this true are given by solutions to the system a−b=0c+d=0 The solution is a=s,b=s,c=t,d=−t where s,t are scalars. We can describe ker(T) as follows. ker(T)={[sst−t]}=span{[1100],[001−1]} It is clear that this set is linearly independent and therefore forms a basis for ker(T).
We now wish to find a basis for im(T). We can write the image of T as im(T)={[a−bc+d]} Notice that this can be written as span{[10],[−10],[01],[01]}
However this is clearly not linearly independent. By removing vectors from the set to create an independent set gives a basis of im(T). {[10],[01]}
Notice that these vectors have the same span as the set above but are now linearly independent.
A major result is the relation between the dimension of the kernel and dimension of the image of a linear transformation. A special case was done earlier in the context of matrices. Recall that for an m×n matrix it was the case that the dimension of the kernel of A added to the rank of A equals n.
Let T:V→W be a linear transformation where V,W are vector spaces. Suppose the dimension of V is n. Then n=dim(ker(T))+dim(im(T)).
- Proof
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From Proposition 9.8.1, im(T) is a subspace of W. By Theorem 9.4.8, there exists a basis for im(T),{T(→v1),⋯,T(→vr)}. Similarly, there is a basis for ker(T),{→u1,⋯,→us}. Then if →v∈V, there exist scalars ci such that T(→v)=r∑i=1ciT(→vi) Hence T(→v−∑ri=1ci→vi)=0. It follows that →v−∑ri=1ci→vi is in ker(T). Hence there are scalars ai such that →v−r∑i=1ci→vi=s∑j=1aj→uj Hence →v=∑ri=1ci→vi+∑sj=1aj→uj. Since →v is arbitrary, it follows that V=span{→u1,⋯,→us,→v1,⋯,→vr} If the vectors {→u1,⋯,→us,→v1,⋯,→vr} are linearly independent, then it will follow that this set is a basis. Suppose then that r∑i=1ci→vi+s∑j=1aj→uj=0 Apply T to both sides to obtain r∑i=1ciT(→vi)+s∑j=1ajT(→uj)=r∑i=1ciT(→vi)=→0 Since {T(→v1),⋯,T(→vr)} is linearly independent, it follows that each ci=0. Hence ∑sj=1aj→uj=0 and so, since the {→u1,⋯,→us} are linearly independent, it follows that each aj=0 also. It follows that {→u1,⋯,→us,→v1,⋯,→vr} is a basis for V and so n=s+r=dim(ker(T))+dim(im(T))
Consider the following definition.
Let T:V→W be a linear transformation and suppose V,W are finite dimensional vector spaces. Then the rank of T denoted as rank(T) is defined as the dimension of im(T). The nullity of T is the dimension of ker(T). Thus the above theorem says that rank(T)+dim(ker(T))=dim(V).
Recall the following important result.
Let V be a vector space of dimension n and let W be a subspace. Then W=V if and only if the dimension of W is also n.
From this theorem follows the next corollary.
Let T:V→W be a linear map where the dimension of V is n and the dimension of W is m. Then T is one to one if and only if ker(T)={→0} and T is onto if and only if rank(T)=m.
- Proof
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The statement ker(T)={→0} is equivalent to saying if T(→v)=→0, it follows that →v=→0. Thus by Lemma 9.7.1 T is one to one. If T is onto, then im(T)=W and so rank(T) which is defined as the dimension of im(T) is m. If rank(T)=m, then by Theorem 9.8.2, since im(T) is a subspace of W, it follows that im(T)=W.
Let S:P2→M22 be a linear transformation defined by S(ax2+bx+c)=[a+ba+cb−cb+c] for all ax2+bx+c∈P2. Prove that S is one to one but not onto.
Solution
You may recall this example from earlier in Example 9.7.1. Here we will determine that S is one to one, but not onto, using the method provided in Corollary 9.8.1.
By definition, ker(S)={ax2+bx+c∈P2 | a+b=0,a+c=0,b−c=0,b+c=0}.
Suppose p(x)=ax2+bx+c∈ker(S). This leads to a homogeneous system of four equations in three variables. Putting the augmented matrix in reduced row-echelon form:
[1100101001−100110]→⋯→[1000010000100000].
Since the unique solution is a=b=c=0, ker(S)={→0}, and thus S is one-to-one by Corollary 9.8.1.
Similarly, by Corollary 9.8.1, if S is onto it will have rank(S)=dim(M22)=4. The image of S is given by
im(S)={[a+ba+cb−cb+c]}=span{[1100],[1011],[01−11]}
These matrices are linearly independent which means this set forms a basis for im(S). Therefore the dimension of im(S), also called rank(S), is equal to 3. It follows that S is not onto.