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 

        0  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 M^2x2 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 M^2x2.  

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 {e_i} 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, ..., v_k} 

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

        T  =  {v₁, ..., v_k, v_k+1, ..., v_n} 

We need to show that 

        U  =  {L(v_k+1k+1), ..., L(v_n)}