Range and kernel Brian Krummel October 9, 2019 Recall from Chapter 2 that given an m × n matrix A: • Nul A is the solution set of the homogeneous equation Ax = 0. • Col A is the span of the columns of A. Let's review a few things about null spaces and column spaces. Example 1. Is v in Nul A, where 2 1 3 2 5 3 2 3 v = 4 −3 5 ;A = 4 −1 1 2 5 ? 2 2 4 5 Answer. Yes, since we can simply compute Av to show that it equals zero: 2 5 3 2 3 2 1 3 2 5 · 1 + 3 · (−3) + 2 · 2 3 2 5 − 9 + 4 3 2 0 3 4 −1 1 2 5 4 −3 5 = 4 −1 · 1 + 1 · (−3) + 2 · 2 5 = 4 −1 − 3 + 4 5 = 4 0 5 : 2 4 5 2 2 · 1 + 4 · (−3) + 5 · 2 2 − 12 + 10 0 Note that simply computing Av is much quicker than using row reduction to compute Ax = 0. Moreover, if we solved Ax = 0, then we would have to write v as a linear combination of the basis vectors for Nul A. Note that checking if v is in Col A is when we row reduce the augmented matrix A v . Example 2. Is 82 r 3 9 < r + 2s = 3t = W = s : 4 5 4r − 5t = 0 : t ; as vector space? Answer. Yes. Let's rewrite the first equation as r +2s−3t = 0 and recall that the second equation is 4r − 5t = 0. Put the coefficients of r; s; t for each equation as the rows of a 2 × 3 matrix: 1 2 −3 A = 4 0 −5 Then W = Nul A as W is the solution set to Ax = 0: 2 r 3 1 2 −3 r + 2s − 3t 0 s = = : 4 0 −5 4 5 4r − 5t 0 t Hence W is a subspace of R3. It follows that W is a vector space. 1 Example 3. Is 82 r 3 9 < 1 2 −3 0 = W = s : r + s + t = 4 5 4 0 −5 0 : t ; as vector space? Answer. Yes, this is the exact same solution set W as in the previous example, except we rewrote the linear system as a vector equation. Example 4. Is 82 r 3 9 < r + 2s = 7 + 3t = W = s : 4 5 4r − 5t = 0 : t ; as vector space? Answer. No, W is a solution set to a non-homogeneous linear equation and in particular W does not contain the zero vector. Example 5. Is 82 3 9 < 2r + s = W = 4 4r + 5s 5 : r; s in R : 6r + 7s ; as vector space? Answer. Yes. Put the coefficients of r in the 1st column of a matrix A: 2 2 ∗ 3 A = 4 4 ∗ 5 6 ∗ Then put the coefficients of s in the 2nd column of the matrix A: 2 2 1 3 A = 4 4 5 5 : 6 7 Then W = Col A as 2 2 1 3 2 2r + s 3 r 4 5 = 4r + 5s 4 5 s 4 5 6 7 6r + 7s is a general vector in W . Hence W is a subspace of R3. It follows that W is a vector space. 2 Compare and contrast column space and null space. Let A be an m × n matrix. Nul A Col A Subspace of Rn Rm Definition Implicitly defined: Explicitly defined: Solution set to Ax = 0. Span of columns of A. Find vector v Hard: Easy: v is a linear in the space Solve Av = 0. combination of columns of A. Checking if v Easy: Hard: Check if is in the space Check Av = 0. Ax = v is consistent. Basis for space Solve Ax = 0 Pivot columns of A. in vector parametric form. Existence and Uniqueness , Existence , uniqueness for Ax = b Nul A = f0g b in Col A Transformations One-to-one , Onto , one-to-one and onto Nul A = f0g Col A = Rn Now let's generalize the notion of null space and column space to abstract vector spaces. Definition 1. Let V and W be vector spaces. Recall that a transformation T : V ! W is a rule which assigns each x in V a unique vector T (x) in W . We call V the domain of T and W the codomain of T . A transformation T : V ! W is linear if (i) T (u + v) = T (u) + T (v) for every u; v in V and (ii) T (cu) = c T (u) for every scalar c and every u in V . Theorem 1. Let V and W be vector spaces and T : V ! W be a linear transformation. Then (i) T (0) = 0; (ii) T (c1v1 + c2v2 + ··· + cpvp) = c1T (v1) + c2T (v2) + ··· + cpT (vp) for all scalars c1; c2; : : : ; cp and vectors v1; v2;:::; vp. Reason. The argument is exactly the same as in Section 1.8. Now we define the subspaces associated with a linear transformation T , the kernel and range. Definition 2. Let V and W be vector spaces and T : V ! W be a linear transformation. • The kernel of T is the set of all x in V such that T (x) = 0. • The range of T is the set of all images T (x) in W over all x in V . Example 6 (Matrix transformation). Let T : Rn ! Rm be a linear transformation. In Section 1.9 we showed that every such linear transformation T takes the form T (x) = Ax for all x in Rn, where A is an m × n matrix. Then kernel T = Nul A and range T = Col A. 3 Example 7 (Evaluation map). Recall that P2 is the space of all polynomials of degree at most 2. Take a point in R, say 0, and define the evaluation map T : P2 ! R by T (p) = p(0) 2 for each polynomial p(x) = a0 + a1x + a2x in P2. Then T is a linear transformation. To check this, given polynomials p; q in P2 and a scalar c: T (p + q) = (p + q)(0) = p(0) + q(0) = T (p) + T (q); T (cp) = (cp)(0) = c p(0) = c T (p); using the definitions of polynomial addition and scalar multiplication. To find the kernel and range of T we can write the evaluation map as 2 2 T (a0 + a1x + a2x ) = a0 + a1 · 0 + a2 · 0 = a0 so that T maps a polynomial to its constant coefficient a0. One can readily see that the range of T is the set of all possible constant coefficients a0, i.e. range T = R. In particular, T maps onto R. The kernel of T is 2 2 ker T = a1x + a2x : a1; a2 in R = fx (a1 + a2x): a1; a2 in Rg = Span x; x : Of course, we could also consider the evaluation map at another point. For instance, if we defined T by T (p) = p(1) for each polynomial p in P2, then T is onto and 2 ker T = f(x − 1) (a1 + a2x): a1; a2 in Rg = Span x − 1; x − x : Example 8 (Derivative). Let C1([0; 1]) be the space of continuously differentiable real-valued functions f : [0; 1] ! R, i.e. f takes values f(x) at each 0 ≤ x ≤ 1. Let T : C1([0; 1]) ! C1([0; 1]) be the derivative df (T f)(x) = (x) for each 0 ≤ x ≤ 1 dx and for each continuously differentiable function f : [0; 1] ! R. By the standard properties df of derivatives, T is linear. As a consequence of the Mean Value Theorem, dx (x) = 0 for each 0 ≤ x ≤ 1 if and only if f is a constant function, i.e. kernel T is the set of all constant functions. df Clearly for each continuously differential function f, its derivative T f = dx is continuous and thus the range of T is the space of continuous functions. In fact, by the Fundamental Theorem of Calculus, for each continuous function f : [0; 1] ! R there is an anti-derivative F : [0; 1] ! R given by Z x F (x) = f(t) dt 0 such that dF (x) = f(x) dx for each x in [0; 1]. The general anti-derivative of f is given by Z f(x) dx = F (x) + C for all x in [0; 1], where C is a constant and represents a function in the kernel of the derivative operator. This is why one always wrote \+C" when integrating functions. 4 Theorem 2. Let V and W be vector spaces and T : V ! W be a transformation. Then the kernel of T is a subspace of V and the range of T is a subspace of W . Reason. To check that the kernel of T is a subspace, we need to check the three properties of a subspace. This proceeds exactly as it did in the case that T was given by matrix multiplication using the linearity of T . • Zero vector: Since T is a linear transformation, T (0) = 0 and thus 0 is in the kernel of T . • Addition: Suppose u; v lie in the kernel of T ; that is, T (u) = 0 and T (v) = 0. Since T is linear, T (u + v) = T (u) + T (v) = 0 + 0 = 0 so u + v is in the kernel of T . • Scaling: Suppose u lies in the kernel of T and c is a scalar. That is, T (u) = 0. Since T is linear, T (c u) = c T (u) = c 0 = 0 so cu is in the kernel of T .