Kernel. Prop-Defn. Let T : V W be a linear map.

Transcription

1 Kernel Aim lecture: We examine the kernel which measures the failure of uniqueness of solns to linear eqns. The concept of linear independence naturally arises. This in turn gives the concept of a basis which allows us to construct co-ordinate systems. Recall For a fn f : X Y and subset Y Y, the inverse image of Y is the set f 1 (Y ) = {x X f (x) Y }. Prop-Defn Let T : V W be a linear map. 1 For any subspace W W, the inverse image T 1 (W ) is a subspace of V. 2 In particular, the kernel of T, defined to be ker T = T 1 (0) is a subspace of V. Proof. Just check closure axioms noting T 0 V = 0 W = 0 T 1 (W ) and for v, v T 1 (W ), β F, we know βv + v T 1 (W ) since by closure axioms for W. T (βv + v ) = βt v + T v W Daniel Chan (UNSW) Lecture 14:Kernel & linear independence Semester / 11

2 Examples E.g. 1 We saw last lecture that the set of solns V to the DE d 2 y dx + y = 0 is a 2 subspace of C (R) by showing it is the span of some set of vectors. We can also see immediately it is a subspace by recognising it as the kernel of the linear map + id : C (R) C (R). d 2 dx 2 In fact, many subspaces naturally arise or can be described this way. E.g. 2 Let v R n. Then the hyperplane v is a subspace of R n since it is the kernel of the following map, Daniel Chan (UNSW) Lecture 14:Kernel & linear independence Semester / 11

3 Kernel measures failure of injectivity You should already know the following. Prop Let T : V W be a linear map & w W. 1 Given a particular soln v = v p to the eqn T v = w, we obtain the complete set of solns as T 1 (w) = {v p + v h v h ker T }. 2 In particular, the soln to T v = w is unique (assuming it exists) iff ker T = 0. 3 T is injective iff ker T = 0. Proof. 2)& 3) immediately follow from 1) which we now prove. Given v h ker T note that T (v p + v h ) = T v p + T v h = w + 0 = w so v p + v h T 1 (w). Hence it remains to show that any soln v T 1 (w) has the form v p + v h for some v h ker T. Now v h = v v p ker T since T v h = T (v v p ) = T v T v p = w w = 0. Thus v = v p + v h has the desired form. Daniel Chan (UNSW) Lecture 14:Kernel & linear independence Semester / 11

4 Injectivity of linear maps C : F n V We have seen that any linear map C : F n V determined by the row matrix (v 1... v n ) V n is surjective iff V = Span(v 1,..., v n ). We determine the condition for injectivity below. Prop-Defn With the above notn, the map C is injective iff for any β 1,..., β n F, β 1 v β n v n = 0 = 0 = β 1 =... = β n. In other words, the only linear combn of the v i s which is zero, is the trivial linear combn. In this case, we say that the set {v 1,..., v n } is linearly independent. Otherwise, we say it is linearly dependent. Proof. Clear. Daniel Chan (UNSW) Lecture 14:Kernel & linear independence Semester / 11

5 Ordered bases & co-ordinate systems We now give a way of constructing co-ordinate systems. Defn Let V = F-space. A basis for V is a linearly independent spanning set for V. From the results we have already proved, we immediately see Prop-Defn Let C : F n V be the linear map given by the row matrix (v 1... v n ) V n. Then C defined a co-ordinate system iff {v 1,..., v n } is a basis for V. In other words, co-ordinate systems correspond to ordered bases for V. The co-ordinates of v V wrt C is C 1 v = (x 1,..., x n ) T. In this case, v = x 1 v x n v n is the unique way of writing v as a linear combn of {v 1,..., v n }. Daniel Chan (UNSW) Lecture 14:Kernel & linear independence Semester / 11

6 Example of co-ordinate systems E.g. Show that the matrix C = (1 + x 2 + 3x) : C 2 C[x] 1 defines a co-ordinate system on C[x] 1. Find the co-ordinates of 3 + 4x wrt C. Daniel Chan (UNSW) Lecture 14:Kernel & linear independence Semester / 11

7 Co-ordinate systems of direct sums. Prop 1 For i = 1,..., r, let T i : V i W i be isomorphisms. Then i T i : i V i i W i is also an isomorphism. 2 In particular, given co-ordinate systems C i : F n i W i, we obtain a co-ordinate system i C i : F n i W i where n = i n i. 3 Consider an internal direct sum W = i W i & bases B i W i for W i. Then i B i is a basis for W. Proof. 1) We need only show i T 1 i is the inverse for i T i. For ease of writing we do this when r = 2. ( ) ( ) (T 1 T 2 ) (T 1 1 T 1 T1 0 T 1 2 ) = T 2 0 T 1 = Sim, (T 1 1 T 1 2 ) (T 1 T 2 ) = id so 1) holds. It s clear 1) = 2). Also 3) follows from 2) as can be seen easily from any example. 2 Daniel Chan (UNSW) Lecture 14:Kernel & linear independence Semester / 11

8 A matrix example E.g. We have seen already that V = M 22 (R) is the internal direct sum of the subspaces V +, V of symmetric & anti-symmteric matrices resp. Now any symmetric (resp anti-symmetric) matrix can be written uniquely in the form ( ) ( ) a b 0 d, resp b c d 0 so we obtain the following bases for V +, V, V. Daniel Chan (UNSW) Lecture 14:Kernel & linear independence Semester / 11

9 Alternate characterisation of linear dependence Prop A finite set S = {v 1,..., v n } of vectors is linearly dependent iff we can write one of them as a linear combn of the others. Proof. Suppose that S is lin depedent so there is a non-trivial linear relation say β 1 v β n v n = 0 with say β i 0. Then v i is a linear combn of the others as we may re-write v i = β 1 i β j v j. j i Reversing the above computation gives the converse. Daniel Chan (UNSW) Lecture 14:Kernel & linear independence Semester / 11

10 Alternate characterisation of bases Theorem Let V = F-space & B V be finite. The following are equiv conds on B. 1 B is a basis for V. 2 B is a minimal spanning set for V. 3 B is a maximal linearly independent set in the sense that, B is linearly independent but B is linearly dependent for any set B strictly containing B. Proof. The equivalence of 1) & 2) follows from the alternate characterisation of linear dependence. The proof of 1) 3) follows easily (ex) from Lemma Suppose B = {v 1,..., v n } is lin independent & let v V. Then B {v} is lin indep iff v / Span(B). Daniel Chan (UNSW) Lecture 14:Kernel & linear independence Semester / 11

11 proof continued To see (= ), we prove the contrapositive. If v Span(B) then B {v} must be linearly dependent by our alternate charn of linear dependence. For ( =) we also prove the contrapositive & suppose that B {v} is lin dependent. Hence there is a non-trivial linear relation of form βv + β 1 v β n v n = 0. Now β 0 as B is linearly independent. Hence we may solve to see v is a linear combn of the v i so lies in Span(B). Corollary Any vector space that is spanned by a finite set S has a basis consisting of a subset of S. Daniel Chan (UNSW) Lecture 14:Kernel & linear independence Semester / 11