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 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.   Back to the Linear Algebra Home Page