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Mathematics LibreTexts

9.8: The Kernel and Image of a Linear Map

( \newcommand{\kernel}{\mathrm{null}\,}\)

Outcomes

  1. Describe the kernel and image of a linear transformation.
  2. 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.

Definition 9.8.1: Kernel and Image

Let V and W be vector spaces and let T:VW be a linear transformation. Then the image of T denoted as im(T) is defined to be the set {T(v):vV} In words, it consists of all vectors in W which equal T(v) for some vV. The kernel, ker(T), consists of all vV such that T(v)=0. That is, ker(T)={vV:T(v)=0}

Then in fact, both im(T) and ker(T) are subspaces of W and V respectively.

Proposition 9.8.1: Kernel and Image as Subspaces

Let V,W be vector spaces and let T:VW be a linear transformation. Then ker(T)V and im(T)W. In fact, they are both subspaces.

Proof

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 av1+bv2 is also in ker(T). But T(av1+bv2)=aT(v1)+bT(v2)=a0+b0=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(av1+bv2) and this last vector is in im(T) by definition.

Consider the following example.

Example 9.8.1: Kernel and Image of a Transformation

Let T:P1R 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,bR and a+b=0}={axa | aR} Therefore a basis for ker(T) is {x1} 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+bP1}={a+b | a,bR}=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.

Example 9.8.2: Kernel and Image of a Linear Transformation

Let T:M22R2 be defined by T[abcd]=[abc+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]=[abc+d]=[00] The values of a,b,c,d that make this true are given by solutions to the system ab=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)={[sstt]}=span{[1100],[0011]} 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)={[abc+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.

Theorem 9.8.1: Dimension of Kernel + Image

Let T:VW 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

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 vV, there exist scalars ci such that T(v)=ri=1ciT(vi) Hence T(vri=1civi)=0. It follows that vri=1civi is in ker(T). Hence there are scalars ai such that vri=1civi=sj=1ajuj Hence v=ri=1civi+sj=1ajuj. 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 ri=1civi+sj=1ajuj=0 Apply T to both sides to obtain ri=1ciT(vi)+sj=1ajT(uj)=ri=1ciT(vi)=0 Since {T(v1),,T(vr)} is linearly independent, it follows that each ci=0. Hence sj=1ajuj=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.

Definition 9.8.2: Rank of Linear Transformation

Let T:VW 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.

Theorem 9.8.2: Subspace of Same Dimension

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.

Corollary 9.8.1: One to One and Onto Characterization

Let T:VW 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

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.

Example 9.8.3: One to One Transformation

Let S:P2M22 be a linear transformation defined by S(ax2+bx+c)=[a+ba+cbcb+c] for all ax2+bx+cP2. 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+cP2 | a+b=0,a+c=0,bc=0,b+c=0}.

Suppose p(x)=ax2+bx+cker(S). This leads to a homogeneous system of four equations in three variables. Putting the augmented matrix in reduced row-echelon form:

[1100101001100110][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+cbcb+c]}=span{[1100],[1011],[0111]}

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.


This page titled 9.8: The Kernel and Image of a Linear Map is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) via source content that was edited to the style and standards of the LibreTexts platform.

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