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Chapter 22. Subspaces, linear maps and the Kernel-Image theorem A subset U of a vector space V is called a subspace of V if the following two conditions hold: S1) If u, u0 are any two vectors in U then u + u0 ∈ U. S2) If u ∈ U and c ∈ R, then cu ∈ U. In brief, U is a subspace if it is closed under addition and scalar multiplication. Examples: • A hyperplane in Rn defined by an equation a1 x1 + · · · + an xn = 0 (where the ai are given constants) is a subspace of Rn . An intersection of such hyperplanes is also a subspace of Rn . • The set of polynomials f ∈ Pn such that f (1) = 0 is a subspace of Pn . • The set of solutions to the differential equation f 00 + af 0 + bf = 0 (where a and b are given constants) is a subspace of C ∞ (R). A subspace is itself a vector space. Proposition 1 If dim V = n, then any subspace U ⊂ V is also finite-dimensional and dim U ≤ n. And if dim U = n, then U = V. Proof: Since dim V = n, no subset of V with more than n elements can be linearly independent, by the Goldilocks Principle in chap. 21. Now suppose U is a subspace of V that is not finite dimensional. That means U has no finite spanning set. Let u1 be a nonzero vector in U. Since {u1 } cannot span U, there is another vector u2 ∈ U which is not a scalar multiple of u1 . Now {u1 , u2 } cannot span U so there is u3 ∈ U which is not in the span of {u1 , u2 }. Then there is u4 ∈ U not in the span of {u1 , u2 , u3 }, etc. Continuing like this we arrive at a subset {u1 , . . . , un+1 } of U such that no uj is in the span of {u1 , . . . , uj−1 }. This set of vectors is linearly independent. For suppose c1 , . . . , cn+1 are scalars such that c1 u1 + · · · + cn+1 un+1 = 0. Since un+1 is not in the span of {u1 , . . . , un }, we must have cn+1 = 0. Now c1 u1 + · · · + cn un = 0. Since un is not in the span of {u1 , . . . , un−1 }, we must have cn = 0, etc. So we find that all ci = 0. We now have a linearly independent subset of V with more than n elements, which is a contradiction. Hence U is finite-dimensional, as claimed. Any basis of U would be a linearly independent set in V, so cannot have more than n elements. Therefore dim U ≤ n. Finally, suppose dim U = n. By Prop. 2 in chapter 21, any basis B of U is contained in a basis B 0 of V. But since dim U = n = dim V, we must have B = B 0 , so B is also a basis of V. Now if v is any 1 element of V, we can write v as a linear combination of the vectors in B, which lies in U. Therefore V ⊂ U, so U = V. As a corollary, we find that the various subspaces of V are stratified according to their dimension. More precisely: Proposition 0.1 Let U ⊂ U0 be finite dimensional subspaces of a vector space V. Then dim U = dim U0 if and only if U = U0 Proof: If U = U0 it is obvious that dim U = dim U0 . If dim U = dim U0 , then applying Prop. 1 to the pair U ⊂ U0 , we get U = U0 . Subspaces arise from linear functions between vector spaces. Suppose we have two vector spaces V and W. A function L : V −→ W is called linear if the following two conditions hold: L1) If v, v0 are any two vectors in V then L(v + v0 ) = L(v) + L(v0 ). L2) If v ∈ V and c ∈ R, then L(cv) = cL(u). In brief, L is a linear function if it preserves the addition and scalar multiplication in V and W. There are two subspaces associated with a linear function L : V → W, namely the kernel of L: ker L = {v ∈ V : L(v) = 0}, and the image of L: im L = {L(v) : v ∈ V} = {w ∈ W : w = L(v) for some v ∈ V}. Conditions S1) and S2) combined with L1) and L2) imply that ker L is subspace of V and im L is a subspace of W. (See the exercises.) Examples: • The function L : Rn → R given by L(x1 , . . . , xn ) = x1 + · · · + xn is linear. So ker L is the hyperplane in Rn defined by an equation x1 +· · ·+xn = 0 and im L = R. • The function L : Pn → Pn given by df dx is linear. So ker D consists of the constant polynomials and im D = Pn−1 . D(f ) = 2 • The function L : C ∞ (R) −→ C ∞ (R) given by L(f ) = f 00 + af 0 + bf is linear (where a, b are given constants). So ker L is the set of solutions to the differential equation f 00 + af 0 + bf = 0. It is a fact from differential equations that for every g ∈ C ∞ (R) there exists f ∈ C ∞ (R) such that g = f 00 + af 0 + bf . This means that im L = C ∞ (R). We can also describe ker L and im L in terms of solutions of equations: Proposition 2: Let L : V → W be a linear function. 1. A vector w ∈ W lies in im L iff the equation L(v) = w has a solution for some v ∈ V. 2. The difference between any two solutions of L(v) = w lies in ker L. Proof: Part 1 is immediate from the definition. The proof of part 1 is in the exercises. Example: Let us find all solutions of the differential equation f ” + f = x2 . Here the linear map is L : C ∞ (R) → C ∞ (R), given by L(f ) = f ” + f . We have seen that ker L has basis {cos x, sin x}. One solution of L(f ) = x2 is f = x2 − 2. From Prop. 2 part 2 it follows that every solution of L(f ) = x2 is of the form f (x) = c1 cos x + c2 sin x + x2 − 2, for some constants c1 , c2 . The kernel and image of a linear function are related by the 1 Kernel-Image Theorem: Let L : V → W be a linear function between two vector spaces V and W. Assume that V is finite-dimensional. Then ker L and im L are both finite dimensional and we have dim(ker L) + dim(im L) = dim V. Proof: By Prop. 1, ker L is finite dimensional because it is a subspace of the finite-dimensional vector space V. To prove that im L is finite dimensional we must find a finite spanning set for im L. Let {y1 , . . . , yn } be a basis of V. Every vector in im L is of the form L(v) for some v ∈ V. We can write v as a linear combination: v = c1 y1 + · · · + cn yn and L(v) = c1 L(y1 ) + · · · + cn L(yn ). Thus, {L(y1 ), . . . , L(yn )} is a finite spanning set for im L, as desired. So we have proved that ker L and im L are both finite dimensional. Let k = dim(ker L), 1 m = dim(im L) In chap. 19 we discussed the Kernel-Image Theorem informally, in terms of “conservation of information”. 3 and let {v1 , . . . , vk } be a basis of ker L and let {w1 , . . . , wm } be a basis of im L. By definition each wi = L(ui ) for some (possibly many) ui ∈ V. These ui are not unique; we choose them arbitrarily. Let B = {v1 , . . . , vk , u1 , . . . , um } be the set of vectors obtained, however arbitrarily, in this way. I claim that B is a basis of V. 2 To show spanning, let v ∈ V. Then L(v) ∈ im L, which is spanned by the wi ’s, so there are scalars ck+1 , . . . , ck+m such that L(v) = ck+1 w1 + · · · + ck+m wm . Since each wi = L(ui ), we have L(v) = ck+1 L(u1 ) + · · · + ck+m L(u)m = L(ck+1 u1 + · · · + ck+m um ). It follows that L(v − ck+1 u1 − · · · − ck+m um ) = 0, which means that v − ck+1 u1 − · · · − ck+m um ∈ ker L. Since ker L is spanned by the vi0 s, there are scalars c1 , . . . , ck such that v − ck+1 u1 − · · · − ck+m um = c1 v1 + · · · + ck vk . Therefore, v = c1 v1 + · · · + ck vk + ck+1 u1 + · · · + ck+m um , so we have proved that B spans V. To show linear independence, suppose c1 , . . . , ck+m are scalars such that c1 v1 + · · · + ck vk + ck+1 u1 + · · · + ck+m um = 0. Then 0 = L(0) = L(c1 v1 + · · · + ck vk + ck+1 u1 + · · · + ck+m um ) = c1 L(v1 ) + · · · + ck L(vk ) + ck+1 L(u1 ) + · · · + ck+m L(um ) = ck+1 w1 + · · · + ck+m wm , since each L(vi ) = 0 and each L(ui ) = wi . Since the wi ’s are linearly independent, we must have ck+1 = · · · = ck+m = 0. But now c1 v1 + · · · + ck vk = 0, and since the vi ’s are linearly independent we must have c1 = · · · = ck = 0. We have proved that B is linearly independent, so is a basis of V, Since dim V is the number of elements in any basis of V we have dim V = k + m = dim(ker L) + dim(im L), and the theorem is proved. 2 In general, this basis will not be the same as the basis {y1 , . . . , yn } used in the first part of the proof. 4 Exercise 22.1 Suppose L : V → W is a linear function. Prove that ker L is a subspace of V, and that im L is a subspace of W. Exercise 22.2 Let L : Pn → R be the function given by Z 1 L(f ) = f (x) dx. 0 (a) Show that L is a linear function. (b) Determine the kernel and image of L. (c) Determine the dimension of the subspace {f ∈ Pn : R1 0 f (x) dx = 0}. Exercise 22.3 Let D : Pn → Pn be the function given by D(f ) = df + f. dx (a) Show that D is a linear function. (b) Show that im D = Pn . (c) Use the Kernel-Image Theorem to compute ker D. (d) The function f = e−x satisfies the equation f 0 + f = 0. Does this contradict your result in (c)? Exercise 22.4 Compute the kernel and image of the matrix that you used to project the hypercube in to R2 . Exercise 22.5 Use the Kernel-Image Theorem to compute the dimension of the vector space V = {f ∈ Pn : f (0) = f (1) = 0}. [Hint: Find a linear map L : Pn → R2 such that V = ker L.] Exercise 22.6 A function L : V → W is called one-to-one if L(v) = L(v0 ) implies v = v0 . Assume that L is linear. Prove that L is one-to-one if and only if ker L = {0}. Exercise 22.7 (Part 2 of Prop. 2) Let L : V → W be a linear function, suppose w ∈ im L, and that L(v) = w. Suppose also that L(v0 ) = w. Prove that v0 = v + u for some vector u ∈ ker L. 5