The Kernel and the Range of a Linear Transformation
One to One Linear Transformations Recall that a function is 11 if f(x) = f(y) implies that x = y Since a linear transformation is defined as a function, the definition of 11 carries over to linear transformations. That is
A linear transformation L is 11 if for all vectors u and v, L(u) = L(v) implies that u = v Example Let L be the linear transformation from R^{2} to P_{2} defined by L((x,y)) = xt^{2} + yt We can verify that L is indeed a linear transformation. We now check that L is 11. Let L(x_{1},y_{1}) = L(x_{2},y_{2}) then x_{1}t^{2} + y_{1}t = x_{2}t^{2} + y_{2}t If two polynomials are equal to each other, then their coefficients are all equal. In particular, x_{1} = x_{2} and y_{1} = y_{2} We can conclude that L is a 11 linear transformation. Example Let L be the linear transformation from P_{1} to R^{1} defined by L(f(t)) = f(0) Then L is not a 11 linear transformation since L(0) = L(t) and 0 1 The Kernel Related to 11 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 P^{1} 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 11 linear transformations and the kernel.
Theorem A linear transformation L is 11 if and only if Ker(L) = 0.
Proof Let L be 11 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 11, 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 11. 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.
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 w_{1} and w_{2} vectors in the range of W. Then there are vectors v_{1} and v_{2} with L(v_{1}) = w_{1} and L(v_{2}) = w_{2} We must show closure under addition and scalar multiplication. We have L(v_{1} + v_{2}) = L(v_{1}) + L(v_{2}) = w_{1} + w_{2} and L(cv_{1}) = cL(v_{1}) = cw_{1} hence w_{1} + w_{2} and cw_{1} 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 R^{2} to R^{3} defined by L(v) = Av with
A.
Find a basis for Ker(L). B.
Determine of L is 11. 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 11 since the Ker(L) is not the zero subspace. Now for the range. If we let {e_{i}} be the standard basis for R^{2}, then {L(e_{1}), L(e_{2})} 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 R^{3} is 3, L is not onto. A Few Theorems In the last example the dimension of R^{2} 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 = {v_{1}, ..., v_{k}} be a basis for Ker(L). Then extend this basis to a full basis for V. T = {v_{1}, ..., v_{k}, v_{k+1}, ..., v_{n}} We need to show that U = {L(v_{k+1}k+1), ..., L(v_{n})} 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 = c_{1}v_{1} + ... + c_{k}v_{k} + c_{k+1}v_{k+1} + ... + c_{n}v_{n} so that w = L(v) = L(c_{1}v_{1} + ... + c_{k}v_{k} + c_{k+1}v_{k+1} + ... + c_{n}v_{n}) = c_{1}L(v_{1}) + ... + c_{k}L(v_{k}) + c_{k+1}L(v_{k+1}) + ... + c_{n}L(v_{n}) = c_{1}0 + ... + c_{k}0 + c_{k+1}L(v_{k+1}) + ... + c_{n}L(v_{n}) = c_{k+1}L(v_{k+1}) + ... + c_{n}L(v_{n}) hence U spans the range of L. Now we need to show that U is a linearly independent set of vectors. If c_{k+1}L(v_{k+1}) + ... + c_{n}L(v_{n}) = 0 then 0 = L(c_{k+1}v_{k+1} + ... + c_{n}v_{n}) hence v = c_{k+1}v_{k+1} + ... + c_{n}v_{n} is in Ker(L). But then v can be written as a linear combination of vectors in S. That is c_{k+1}v_{k+1} + ... + c_{n}v_{n} = c_{1}v_{1} + ... + c_{k}v_{k} 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 11. 2. L is onto.
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