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 1 5 3 2 v = −3 ,A = −1 1 2 ? 2 2 4 5 Answer. Yes, since we can simply compute Av to show that it equals zero: 5 3 2 1 5 · 1 + 3 · (−3) + 2 · 2 5 − 9 + 4 0 −1 1 2 −3 = −1 · 1 + 1 · (−3) + 2 · 2 = −1 − 3 + 4 = 0 . 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 r r + 2s = 3t W = s : 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: r 1 2 −3 r + 2s − 3t 0 s = = . 4 0 −5 4r − 5t 0 t Hence W is a subspace of R3. It follows that W is a vector space.
1 Example 3. Is r 1 2 −3 0 W = s : r + s + t = 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 r r + 2s = 7 + 3t W = s : 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 2r + s W = 4r + 5s : 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 ∗ A = 4 ∗ 6 ∗
Then put the coefficients of s in the 2nd column of the matrix A:
2 1 A = 4 5 . 6 7
Then W = Col A as 2 1 2r + s r 4 5 = 4r + 5s s 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 = {0} b in Col A Transformations One-to-one ⇔ Onto ⇔ one-to-one and onto Nul A = {0} 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 = {x (a1 + a2x): a1, a2 in R} = 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 = {(x − 1) (a1 + a2x): a1, a2 in R} = 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 .
Next let us check that the range of T is a subspace. When T was given by matrix multiplication, we merely noted that the range of T was the span of the column of the standard matrix. For a general linear transformation this does not quite work, as V could be infinite-dimensional (see future lectures) and thus the transformation T cannot be represented by matrix multiplication.
• Zero vector: Since T is a linear transformation, 0 = T (0) in range T .
• Addition: Suppose u, v lie in the range of T ; that is, T (x) = u and T (z) = v for some x, z in V . Since T is linear,
u + v = T (x) + T (z) = T (u + v) in range T.
• Scaling: Suppose u lies in the range of T and c is a scalar. That is, T (x) = u for some x in V . Since T is linear,
c u = c T (x) = T (c x) in range T.
5 Four ways to check if a set is a subspace. Let H be a subset of a vector space V . For instance, H might be a set of matrices or of polynomials. H is a subspace if: (i) H satisfies the three properties: if u, v in H and c is a scalar then 0 in H, u + v in H, and cu in H or
(ii) H is the span of a finite set of vectors v1, v2,..., vp in V or (iii) H is the kernel of a linear transformation T : V → W , i.e. the solution set to a homogeneous linear equation, or (iv) H = {0} is the zero subspace.
Example 9. Which of the following sets of polynomials in P3 are a subspace?
(i) H = {p in P3 : p(0) = 1};
(ii) H = {p in P3 : p(1) = p(2)}; (iii) H = {a + bx2 : a, b are scalars}. Answer. (ii) and (iii). For (i), H is not a subspace since the zero polynomial is not in H. Notice that H is the solution set of a nonhomogeneous linear equation T (p) = p(0) = 1 where T (p) = p(0) is the evaluation map at 0. For (ii), H is a subspace since H is the kernel of the linear transformation T (p) = p(2) − p(1). In other words, in (ii), H is the solution set to the homogeneous linear equation 2 2 3 3 p(2) − p(1) = (1 − 1) a0 + (2 − 1) a1 + (2 − 1 ) a2 + (2 − 1 ) a3 = a1 + 3a2 + 7a3 = 0 2 3 for each p(x) = a0 + a1x + a2x + a3x . In terms of the coordinate vector of p, a0 a1 p(2) − p(1) = 0 1 3 7 = 0 a2 a3 and thus we can regard H as analogous to Nul A where A = 0 1 3 7 . For (iii), H is a subspace since H is the span of 1 and x2. Example 10. Suppose we want to solve the non-homogeneous differential equation d2y + y = 0. (1) dx2 d2 We can think of y as an element of the kernel of the differential operator T = dx2 + 1, which is a linear operator on twice continuously differentiable functions. The solution set of (1) is well-known to be y = A cos(x) + B sin(x) where A, B are arbitrary constants. Thus kernel T = Span{cos(x), sin(x)}. In fact, {cos(x), sin(x)} is a basis for kernel T .
6 Theorem 3. Let V and W be vector spaces and T : V → W be a linear transformation.
• T is one-to-one if and only if kernel T = {0}.
• T is onto if and only if the range of T equals W .
Reason. Suppose T (x) = T (y) = b for some x, y in V and b in W . Then T (x−y) = T (x)−T (y) = 0, so x−y in kernel T . T is one-to-one if and only if kernel T = {0}, which guarantees that x−y = 0 so that x = y. T is onto by definition precisely when the range of T equals W .
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