Kernel, image, nullity, and rank Math 130 Linear Algebra

1 Definition 3. The dimensions of the kernel and image of a transformation T are called the trans- formation's rank and nullity, and they're denoted rank(T ) and nullity(T ), respectively. Since a ma- trix represents a transformation, a matrix also has a rank and nullity. kernel , image, nullity, and rank Math 130 Linear Algebra For the time being, we'll look at ranks and nullity D Joyce, Fall 2015 of transformations. We'll come back to these topics again when we interpret our results for matrices. Definition 1. Let T : V W be a Linear trans- The above theorem implies this corollary. formation between vector spaces. The kernel of T , T U. Corollary 4. Let V W and W X. Then also called the null space of T , is the inverse image of the zero vector, 0, of W , nullity(T ) nullity(U T ).

2 Ker(T ) = T 1 (0) = {v V | T v = 0}. and rank(U T ) rank(U ). It's sometimes denoted N (T ) for null space of T . The image of T , also called the range of T , is the Systems of Linear equations and Linear trans- set of values of T , formations. We've seen how a system of m lin- ear equations in n unknowns can be interpreted as a T (V ) = {T (v) W | v V }. single matrix equation Ax = b, where x is the n 1. column vector whose entries are the n unknowns, This image is also denoted im(T ), or R(T ) for range and b is the m 1 column vector of constants on of T . the right sides of the m equations . We can also interpret a system of Linear equations Both of these are vector spaces. ker(T ) is a sub- in terms of a Linear transformation.

3 Let the Linear space of V , and T (V ) is a subspace of W . (Why? transformation T : Rn Rm correspond to the Prove it.). matrix A, that is, T (x) = Ax. Then the matrix We can prove something about kernels and im- equation Ax = b becomes ages directly from their definition. T U. T (x) = b. Theorem 2. Let V W and W X be Linear transformations. Then Solving the equation means looking for a vector x in the inverse image T 1 (b). It will exist if and ker(T ) ker(U T ) only if b is in the image T (V ). When the system of Linear equations is homoge- and neous, then b = 0. Then the solution set is the im(U T ) im(U ). subspace of V we've called the kernel of T . Thus, kernels are solutions to homogeneous Linear equa- Proof.

4 For the first statement, just note that tions. T (v) = 0 implies U (T (v)) = 0. For the second When the system is not homogeneous , then the statement, just note that an element of the form solution set is not a subspace of V since it doesn't U (T (v) in X is automatically of the form U (w) contain 0. In fact, it will be empty when b is not in where w = T (v). the image of T . If it is in the image, however, there 1. is at least one solution a V with T (a) = b. All In general, the solution set for the nonhomoge- the rest can be found from a by adding solutions x neous equation Ax = b won't be a one-dimensional of the associated homogeneous equations , that is, line. It's dimension will be the nullity of A, and it will be parallel to the solution space of the associ- T (a + x) = b iff T (x) = 0.)

5 Ated homogeneous equation Ax = 0. Geometrically, the solution set is a translate of the kernel of T , which is a subspace of V , by the vector The dimension theorem. The rank and nullity a. of a transformation are related. Specifically, their Example 5. Consider this nonhomogeneous Linear sum is the dimension of the domain of the trans- system Ax = b: formation. That equation is sometimes called the .. x dimension theorem. 2 0 1 1 In terms of matrices, this connection can be y =. 1 1 3 2 stated as the rank of a matrix plus its nullity equals z the number of rows of the matrix. Solve it by row reducing the augmented matrix Before we prove the Dimension Theorem, first we'll find a characterization of the image of a trans.

6 2 0 1 1. 1 1 3 2 formation. to Theorem 6. The image of a transformation is 1 0 1/2 1/2. spanned by the image of the any basis of its do- 0 1 5/2 3/2. main. For T : V W , if = {b1 , b2 , .. , bn } is a Then z can be chosen freely, and x and y deter- basis of V , then T ( ) = {T (b1 ), T (b2 ), .. , T (bn )}. mined from z, that is, spans the image of T . 1 1 1 1 . x 2. 2z 2. 2. x = y = 2 2 z = 2 + 52 z 3 5 3 Although T ( ) spans the image, it needn't be a z z 0 1 basis because its vectors needn't be independent. This last equation is the parametric equation of a Proof. A vector in the image of T is of the form line in R3 , that is to say, the solution set is a line. T (v) where v V.

7 But is a basis of V , But it's not a line through the origin. There is, so v is a Linear combination of the basis vectors however, a line through the origin, namely {b1 , .. , bn }. Therefore, T (v) is same Linear combi- 1 . x 2 nation of the vectors T (b1 ), .. , T (bn ). Therefore, x = y = 52 z every vector in the image of T is a Linear combina- z 1 tion of vectors in T ( ). and that line is the solution space of the associated Theorem 7 (Dimension Theorem). If the domain homogeneous system Ax = 0. Furthermore, these of a Linear transformation is finite dimensional, then 1/2 that dimension is the sum of the rank and nullity two lines are parallel, and the vector 3/2 shifts of the transformation.

8 0. the line through the origin to the other line. In Proof. Let T : V W be a Linear transformation, summary, for this example, the solution set for the let n be the dimension of V , let r be the rank of T. nonhomogeneous equation Ax = b is a line in R3 and k the nullity of T . We'll show n = r + k. parallel to the solution space for the homogeneous Let = {b1 , .. , bk } be a basis of the kernel equation Ax = 0. of T . This basis can be extended to a basis =. 2. {b1 , .. , bk , .. , bn } of all of V . We'll show that Theorem 8. A transformation is one-to-one if and the image of the vectors we appended, only if its kernel is trivial, that is, its nullity is 0. C = {T (bk+1 ), .. , T (bn )} Proof.

9 Let T : V W . Suppose that T is one-to- one. Then since T (0) = 0, therefore T can send no is a basis of T (V ). That will show that r = n k other vector to 0. Thus, the kernel of T consists as required. of 0 alone. That's a 0-dimensional space, so the First, we need to show that the set C spans the nullity of T is 0. image T (V ). From the previous theorem, we know Conversely, suppose that the nullity of T is 0, that T ( ) spans T (V ). But all the vectors in T ( ) that is, its kernel consists only of 0. We'll show T is are 0, so they don't help in spanning T (V ). That one-to-one. Suppose that u 6= v. Then u v 6= 0. leaves C = T ( ) T ( ) to span T (V ). Then u v 6= 0 does not lie in the kernel of T.

10 Next, we need to show that the vectors in C are which means that T (u v) 6= 0. But T (u v) =. linearly independent. Suppose that 0 is a Linear T (u) T (v), therefore T (u) 6= T (v). Thus, T is combination of them, one-to-one. ck+1 T (bk+1 ) + + cn T (bn ) = 0 Since the rank plus the nullity of a transforma- tion equals the dimension of its domain, r + k = n, where the ci 's are scalars. Then we have the following corollary. T (ck+1 bk+1 + + bn vn ) = 0. Corollary 9. For a transformation T whose do- Therefore, v = ck+1 bk+1 + + cn bn lies in the main is finite dimensional, T is one-to-one if and kernel of T . Therefore, v is a Linear combination of only if that dimension equals its rank.