The Kernel and the Range of a Linear Transformation

 

One to One Linear Transformations

Recall that a function is 1-1 if 

        f(x)  =  f(y) 

implies that 

        x  =  y

Since a linear transformation is defined as a function, the definition of 1-1 carries over to linear transformations.  That is 

 

Definition

        A linear transformation L is 1-1 if for all vectors u and v

                L(u)  =  L(v

        implies that

                u  =  v


Example

Let L be the linear transformation from R2 to P2 defined by

        L((x,y))  =  xt2 + yt

We can verify that L is indeed a linear transformation.  We now check that L is 1-1.  Let

        L(x1,y1)  =  L(x2,y2)  

then

        x1t2 + y1t  =  x2t2 + y2t

If two polynomials are equal to each other, then their coefficients are all equal.  In particular,

        x1  =  x2          and        y1  =  y2 

We can conclude that L is a 1-1 linear transformation.


Example

Let L be the linear transformation from P1 to R1 defined by

        L(f(t))  =  f(0)

Then L is not a 1-1 linear transformation since 

        L(0)  =  L(t)

and 

       1


The Kernel

Related to 1-1 linear transformations is the idea of the kernel of a linear transformation.

Definition

The kernel of a linear transformation L is the set of all vectors v such that 

        L(v)  =  0 

 

Example

Let L be the linear transformation from M2x2 to P1 defined by

       

Then to find the kernel of L, we set

        (a + d) + (b + c)t  =  0

        d  =  -a        c  =  -b

so that the kernel of L is the set of all matrices of the form

       

Notice that this set is a subspace of M2x2.  


Theorem

The kernel of a linear transformation from a vector space V to a vector space W is a subspace of V.

 

Proof

Suppose that u and v are vectors in the kernel of L.  Then 

     L(u)  =  L(v)  =  0

We have 

        L(u + v)  =  L(u) + (v)  =  0 + 0  =  0 

and

        L(cu)  =  cL(u)  =  c 0  =  0

Hence u + v and cu are in the kernel of L.  We can conclude that the kernel of L is a subspace of V.


In light of the above theorem, it makes sense to ask for a basis for the kernel of a linear transformation.  In the previous example, a basis for the kernel is given by

       

Next we show the relationship between 1-1 linear transformations and the kernel.

 

Theorem

A linear transformation L is 1-1 if and only if Ker(L)  =  0.

 

Proof

Let L be 1-1 and let v be in Ker(L).  We need to show that v is the zero vector.  We have both

        L(v)  =  0        and        L(0)  =  0

Since L is 1-1

        v  =  0 

Now let Ker(L)  =  0.  Then

        L(u) =  L(v)  

implies that

        0  =  L(v) - L(u)  =  L(v - u)

Hence v - u is in Ker(L), so that

        v - u  =  0         or        u  =  v

and L is 1-1.


Range

We have seen that a linear transformation from V to W defines a special subspace of V called the kernel of L.  Now we turn to a special subspace of W.

 

Definition

Let L be a linear transformation from a vector space V to a vector space W.  Then the range of L is the set of all vectors w in W such that there is a v in V with 

        L(v)  =  w 


Theorem

The range of a linear transformation L from V to W is a subspace of W.

 

Proof

Let w1 and w2 vectors in the range of W.  Then there are vectors v1 and v2 with

        L(v1)  =  w1         and        L(v2)  =  w2 

We must show closure under addition and scalar multiplication.  We have

        L(v1 + v2)  =  L(v1) + L(v2)  =  w1 + w2 

and

        L(cv1)  =  cL(v1)  =  cw1 

hence w1 + w2 and cw1 are in the range of L.  Hence the range of L is a subspace of W.


We say that a linear transformation is onto W if the range of L is equal to W.

 

Example

Let L be the linear transformation from R2 to R3 defined by

        L(v)  =  Av

with

       

A.  Find a basis for Ker(L).

B.  Determine of L is 1-1.

C.  Find a basis for the range of L.

D.  Determine if L is onto.

 

Solution

The Ker(L) is the same as the null space of the matrix A.  We have

       

Hence a basis for Ker(L) is 

        {(3,-1)}

L is not 1-1 since the Ker(L) is not the zero subspace.

Now for the range.  If we let {ei} be the standard basis for R2, then 

        {L(e1), L(e2)} 

will span the range of L.  These two vectors are just the columns of A.  In general 

The columns of A span the range of L.

A basis for the column space is L is given by the first column of A (the only corner of rref(A)).  That is a basis is

        {(1,2,3)}

Since the dimension of the range of A is 1 and the dimension of R3 is 3, L is not onto.


A Few Theorems

In the last example the dimension of R2 is 2, which is the sum of the dimensions of Ker(L) and the range of L.  This will be true in general.  

Theorem

Let L be a linear transformation from V to W.  Then

        dim(Ker(L))  + dim(range(L))  =  dim(V)

 

Proof

Let 

        S  =  {v1, ..., vk

be a basis for Ker(L). Then extend this basis to a full basis for V.

        T  =  {v1, ..., vk, vk+1, ..., vn

We need to show that 

        U  =  {L(vk+1k+1), ..., L(vn)} 

is a basis for range L.  If w is in the range of L then there is a v in V with L(v)  =  w.  Since T spans V, we can write

        v  =  c1v1 + ... + ckvk + ck+1vk+1 + ... +  cnvn 

so that

        w  =  L(v)  =  L(c1v1 + ... + ckvk + ck+1vk+1 + ... +  cnvn)

        =  c1L(v1) + ... + ckL(vk) + ck+1L(vk+1) + ... +  cnL(vn)

        =  c10 + ... + ck0 + ck+1L(vk+1) + ... +  cnL(vn)

        =  ck+1L(vk+1) + ... +  cnL(vn)

hence U spans the range of L.  Now we need to show that U is a linearly independent set of vectors.  If 

        ck+1L(vk+1) + ... +  cnL(vn)  =  0

then

        0  =  L(ck+1vk+1 + ... +  cnvn)

hence 

        v  =  ck+1vk+1 + ... +  cnvn

is in Ker(L).  But then v can be written as a linear combination of vectors in S.  That is

        ck+1vk+1 + ... +  cnvn  =  c1v1 + ... + ckvk 

Which means that all of the constants are zero since these are linearly independent.  


We call the dimension of Ker(L) the nullity of L and the dimension of the rang of L the rank of L.

We end this discussion with a corollary that follows immediately from the above theorem.

Theorem

Let L be a linear transformation from a vector space V to a vector space W with dim V  =  dim W, then the following are equivalent:

        1.  L is 1-1.

        2.  L is onto.

 



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