Linear functions.

Definition. Suppose V and W are vector spaces and

L : V → W.

(Remember: This means that L is a function; the domain of L equals V ; and the range of L is a subset of W .)

We say L is linear if L(cv) = cL(v) whenever c ∈ R and v ∈ V and L(v1 + v2) = L(v1) + L(v2) whenever v1, v2 ∈ V . We let ker L = {v ∈ V : L(v) = 0} and call ker L the kernel of L and we let

rng L = {L(v): v ∈ V } so rng L is the range of L. We let L(V,W ) be the set of L such that L : V → W and L is linear.

Example. Suppose m and n are positive integers and L ∈ L(Rn, Rm). For each i = 1, . . . , m and each j = 1, . . . , n we let i lj

be the i-th component of L(ej). Thus

Xm 1 m i L(ej) = (lj , . . . , lj ) = ljei whenever j = 1, . . . , n. i=1

n Suppose x = (x1, . . . , xn) ∈ R . Then Xn x = xjej j=1 so à !   Xn Xn Xn Xn Xm Xm Xn i  i  L(x) = L( xjej) = L(xjej) = xjL(ej) = xj ljei = xjlj ei. j=1 j=1 j=1 j=1 i=1 i=1 j=1 We call  1 1 1  l1 l2 ··· ln    2 1 2   l1 22 ··· ln     . . .. .   . . . .  m m m l1 l2 ··· ln the standard matrix of L. Given the standard matrix of L one can compute L by the above formula.

Theorem. Suppose V and W are vector spaces and L ∈ L(V,W ). Then ker L is a linear subspace of V and rng L is a linear subspace of W .

1 Proof. Suppose 0 is the zero vector of V ; then L(0) = L(0 + 0) = L(0) + L(0) so L(0) = 0 so 0 ∈ ker L and 0 ∈ rng L. (Make sure you understand here and in what follows which 0 is which!). Suppose c ∈ R and v ∈ ker L; then L(cv) = cL(v) = c0 = 0 so cv ∈ ker L. Suppose v1, v2 ∈ ker L; then L(v1 + v2) = L(v1) + L(v2) = 0 + 0 = 0 so v1 + v2 ∈ ker L. Suppose c ∈ R and w ∈ rng L; then there is v ∈ V such that w = L(v). Then cw = cL(v) = L(cv) so cw ∈ rng L. Suppose w1, w2 ∈ V ; then there are v1, v2 ∈ V such that L(v1) = w1 and L(v2) = w2 so w1 + w2 = L(v1) + L(v2) = L(v1 + v2) so w1 + w2 ∈ rng L.

Theorem. Suppose V and W are vector spaces, L ∈ L(V,W ) and v0 ∈ V and w0 = L(v0). Then

{v ∈ V : L(v) = w0} = {u + v0 : u ∈ ker L}.

Proof. Suppose v ∈ V and L(v) = w0. Let u = v − v0. Then u + v0 = v and L(v − v0) = L(v) − L(v0) = 0 − 0 = 0 so u ∈ ker L. Suppose u ∈ ker L. Then L(u + v0) = L(k) + L(v0) = 0 + w0 = w0.

Theorem. Suppose V and W are vector spaces and L ∈ L(V,W ). Then ker L = {0} if and only if, for each independent subset S of V , {L(v): v ∈ S} is an independent subset of W with the same number of elements as S. Proof. Suppose ker L = {0}; S is an independent subset of V ; n is a positive integer; c1, . . . , cn are scalars; v1, . . . , vn are distinct members of S and

c1L(v1) + ··· + cnL(vn) = 0.

Then, as L is linear, L(c1v1 + ··· + cnvn) = c1L(v1) + ··· + cnL(vn) = 0

so c1v1 +···+cnvn = 0 and therefore, as S is independent and the v1, . . . , vn are distinct, c1 = 0, . . . , cn = 0. It follows that {L(v): v ∈ S} is an independent subset of W having the same number of members as S. On the other hand, suppose u ∈ ker L ∼ {0} and let S = {u}. Then S is an independent subset of V but {L(v): v ∈ S} = {L(u)} = {0} so {L(v): v ∈ S} is not an independent subset of W .

Theorem. Suppose V and W are vector spaces. Then L(V,W ) is a linear subspace of W V . Proof. Simple exercise. You do it.

Definition. Suppose V and W are vector spaces. We say L is a (linear) isomorphism from V onto W if L ∈ L(V,W ), ker L = {0} and rng L = W. We let Iso(V,W ) be the set of L such L is linear isomorphism from V onto W . In case not both V and W are trivial we have 0 6∈ Iso(V,W ) so Iso(V,W ) is not a linear subspace of L(V,W ).

Theorem. Suppose V and W are vector spaces. and L ∈ L(V,W ). Then

(a) L(v1) 6= L(v2) whenever v1, v2 ∈ V and v1 6= v2 if and only if

(b) ker L = {0}.

Proof. Suppose (a) holds and v ∈ V and v 6= 0. Then L(v) 6= L(0) = 0 so (b) holds.

2 Suppose (b) holds, v1, v2 ∈ V and v1 6= v2. Then v1 − v2 6∈ ker L so

0 6= L(v1 − v2) = L(v1) − L(v2)

so L(v1) 6= L(v2) so (a) holds.

Corollary. Suppose V and W are vector spaces. and L ∈ Iso(V,W ). Then Then L−1 ∈ Iso(W, V ). Proof. Simple exercise for the reader.

Definition. Suppose V and W are vector spaces. For c ∈ R and (v, w) ∈ V × W we let

c(v, w) = (cv, cw) ∈ V × W.

For (v1, w1), (v2, w2) ∈ V × W we let

(v1, w1) + (v2, w2) = (v1 + v2, w1 + w2) ∈ V × W.

We leave as a simple exercise for the reader to verify that these operations make V × W into a vector space.

Definition. Suppose V and W1,...,WK are vector spaces and Lκ ∈ L(V,Wκ), κ = 1,...,K. We define

(L1,...,LK ): V → W1 × · · · × WK by letting (L1,...,LK )(v) = (L1(v),...,LK (v)) whenever v ∈ V . We leave it a simple exercise for the reader to verify that

(L1,...,LK ) ∈ L(V,W1 × · · · × WK ).

Question. Is it ever the case that

{(L1,...,LK ): Lκ ∈ L(V,Wκ), κ = 1,...,K} = L(V,W1 × WK )?

Theorem. Suppose V and W1,...,WK are vector spaces and Lκ ∈ L(V,Wκ), κ = 1,...,K. Then

K ker (L1,...,LK ) = ∩κ=1ker Lκ.

Proof. Simple exercise for the reader.

The existence and uniqueness theorem for LDO’s.

Suppose n is a positive integer and −∞ ≤ a < b ≤ ∞.

Definition. We say L is a linear differential operator (LDO) on Cn(a, b) of order n if

L : Cn(a, b) → C(a, b)

and there are qn, . . . , q1, q0 ∈ C(a, b) such that Xn (m) (n) 0 Ly(x) = qm(x)y (x) = q(x)ny (x) + ... + q1(x)y (x) + q0(x)y(x) whenever a < x < b. m=0

3 We say L is nonsingular if qn(x) 6= 0 whenever a < x < b.

Theorem. Suppose L is an LDO. The L is linear. Proof. Suppose c ∈ R and y ∈ C(a, b). Then

Xn Xn Xn (m) (m) (m) L(cy) = qm(cy) = qm(cy ) = c qmy = cLy. m=0 m=0 m=0

n Suppose y1, y2 ∈ C (a, b). Then

Xn Xn Xn Xn (m) (m) (m) (m) (m) L(y1 + y2) = qm(y1 + y2) = qm(y1 + y2 ) = qmy1 + qmy2 = Ly1 + Ly2. m=0 m=0 m=0 m=0

Definition. For each x0 ∈ (a, b) we define

n n Ix0 : C (a, b) → R

by letting   y(x0)    0   y (x0)  I (y) =   whenever y ∈ C(a, b). x0  .   .  (n−1) y (x0)

Proposition. Ix0 is linear. Proof. I hope this is as obvious to you as it is to me.

Remark. Suppose y1, . . . , yn and x ∈ (a, b). Then

(1) W (y1, . . . , yn)(x) = det [ Ix(y1) ··· Ix(yn)] .

n Proposition. Suppose y1, . . . , yn ∈ C (a, b) and x ∈ (a, b). Then

W (y1, . . . , yn)(x) 6= 0 if and only if n {Ix(y1),...,Ix(yn)} is a basis for R . Proof. This is a consequence of the fact that the determinant of an n × n matrix is nonzero if and only if its column space equals Rn.

Henceforth we assume that

L is an nonsingular LDO on (a, b) of order n.

The fundamental existence and uniqueness theorem for nonsingular LDOs. Suppose

x0 ∈ (a, b).

4 Then n n (L, Ix0 ) ∈ Iso(C (a, b),C(a, b) × R ). Proof. This will come later.

n n Corollary. Existence. Suppose g ∈ C(a, b) and Y = (Y0,...,Yn−1) ∈ R . Then there is y ∈ C (a, b) such that

Ly = g and Ix0 (y) = Y.

Proof. This is just a restatement of rng (L, Ix0 ) = C(a, b). n Corollary. Uniqueness. Suppose y1, y2 ∈ C (a, b). Then

y1 = y2 ⇔ Ly1 = Ly2 and Ix0 (y1) = Ix0 (y2).

Proof. We have

(L, Ix0 )(y1 − y2) = (L(y1 − y2),Ix0 (y1 − y2)) = (0, 0) so, as ker (L, Ix0 ) = {0}, we find that y1 = y2.

n Corollary. Suppose g ∈ C(a, b), yp ∈ C (a, b) and Ly = g. Then

n n {y ∈ C (a, b): Ly = g} = {z + yp : z ∈ C (a, b) and Lz = 0}.

Proof. This follows directly from a previous Theorem.

n Corollary. Ix0 |ker L ∈ Iso(ker L, R ). Moreover, if

y1, . . . , yn ∈ ker L and B = {y1, . . . , yn} then the following statements are equivalent:

(i) B is a basis for ker L.

(ii)

{Ix0 (y1),...,Ix0 (yn−1)} n is a basis for R for some x0 ∈ (a, b).

(iii)

{Ix0 (y1),...,Ix0 (yn−1)} n is a basis for R for all x0 ∈ (a, b).

(iv) W (y1, . . . , yn)(x0) 6= 0 for some x0 ∈ (a, b).

(v) W (y1, . . . , yn)(x0) 6= 0 for all x0 ∈ (a, b).

Proof. This follows directly from a previous Theorem.

Definition. Suppose L is an LDO. A basis for ker L is called a fundamental set of solutions.

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