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 ) ker(T ) = T −1(0) = fv 2 V j T v = 0g: 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 ) = fT (v) 2 W j v 2 V g: 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. 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 (x) = b: Theorem 2. Let V !T W and W !U 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. 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). q.e.d. the image of T . If it is in the image, however, there 1 is at least one solution a 2 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: 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 dimension theorem. 2x3 2 0 1 1 In terms of matrices, this connection can be y = 1 1 3 4 5 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 2 0 1 1 we'll find a characterization of the image of a trans- 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 β = fb1; b2;:::; bng is a Then z can be chosen freely, and x and y deter- basis of V , then T (β) = fT (b1);T (b2);:::;T (bn)g mined from z, that is, spans the image of T . 2 3 2 1 1 3 2 1 3 2 1 3 x 2 − 2 z 2 − 2 3 5 3 5 Although T (β) spans the image, it needn't be a x = 4y5 = 4 2 − 2 z5 = 4 2 5 + 4− 2 5 z 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 2 V . But β 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 fb1;:::; bng. Therefore, T (v) is same linear combi- 2 3 2 1 3 nation of the vectors T (b1);:::;T (bn). Therefore, x − 2 5 every vector in the image of T is a linear combina- x = 4y5 = 4− 2 5 z z 1 tion of vectors in T (β). q.e.d. 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 2 3 1=2 that dimension is the sum of the rank and nullity two lines are parallel, and the vector 43=25 shifts of the transformation. 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 β = fb1;:::; bkg be a basis of the kernel equation Ax = 0. of T . This basis can be extended to a basis γ = 2 fb1;:::; bk;:::; bng 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 = fT (bk+1);:::;T (bn)g Proof. 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 (γ) spans T (V ). But all the vectors in T (β) 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 (γ) − T (β) to span T (V ). Then u − v 6= 0 does not lie in the kernel of T 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. q.e.d. ck+1T (bk+1) + ··· + cnT (bn) = 0 Since the rank plus the nullity of a transforma- tion equals the dimension of its domain, r + k = n, where the c 's are scalars. Then i we have the following corollary. T (c b + ··· + b v ) = 0 k+1 k+1 n n Corollary 9. For a transformation T whose do- main is finite dimensional, T is one-to-one if and Therefore, v = ck+1bk+1 + ··· + cnbn lies in the kernel of T . Therefore, v is a linear combination of only if that dimension equals its rank. the basis vectors β, v = c0b0 +···+ckbk: These last two equations imply that 0 is a linear combination Characterization of isomorphisms. If two of the entire basis γ of V , vector spaces V and W are isomorphic, their prop- erties with respect to addition and scalar multipli- c0b0 + ··· + ckbk − ck+1bk+1 − · · · − cnbn = 0: cation will be the same.