is injective because f(a) = f(b) only when a = b. On the other hand, 1

Transcription

1 2.. Injective maps, Kernel, and Linear Independence In this section, we introduce the concept of injectivity for maps. Being injective is a very important property for maps and it appears everywhere in more advanced mathematics and other areas. In order to understand being injectivity a linear map, we define the kernel of a linear map. This leads directly to definition of linear independence for list of vectors. Definition 2... A map f : X Y is injective if f(a) = f(b) only when a = b for all a, b X. Rephrasing this, f : X Y is NOT injective if there exists some a, b X with a b so that f(a) = f(b). So, f is not injective if there are any instances of a f(a) = f(b) b For example, is injective because f(a) = f(b) only when a = b. On the other hand, is not injective because f(2) = f(), but 2. In order to characterize when linear maps are injective, we define the kernel of a linear map. Kernels show up in many areas of mathematics. Given a linear map, its kernel is one of the most important objects associated to it. Definition Given a linear map T : R n R m, the kernel of T, denoted ker(t ), is the set of vectors in R n that T sends to : ker(t ) = { x R n T ( x) = }. For any linear map T, we have T ( ) =, so ker(t ). We warn you that T ( x) = does NOT mean x =. Instead, T ( x) = means x ker(t ), and ker(t ) can include more vectors than just, which we will soon. As we shall see, it is only when T is injective that T ( x) = implies x =. Example 2... Find the kernel of the following maps. Express your answer as the span of a finite set of vectors. 2 (a) T : R 2 R 2 with [T ] =. 4 4

2 2 (b) T : R 2 R with [T ] = [ 2 ] 2 (c) T : R R 2 with [T ] = (d) T : R 4 R with [T ] = Solutions: Finding ker(t ) means solving T ( x) =. Then, we express the solution as a vector and write it as a linear combination of constant vectors where the coefficients are free variables. The kernel will be the span of these vectors. (a) We solve so x = x 2 =. Therefore, (b) We solve T ( x) = so x = 2x 2, and x 2 is free. Therefore, x 2x2 2 = = x x 2 x 2 2 (c) We solve T ( x) = 2 4 ker(t ) = { } = span( ). T ( x) = [ 2 ],, = ker(t ) = span () 2, 2 so x = x, x 2 = 2x, and x is free. Therefore, x x 2 = x 2x = x 2 = ker(t ) = span 2. (d) We solve x x 4 x x T ( x) = , so x = 2x 5x 4, x 2 = 4x + 7x 4, and x, x 4 are free. Therefore, x 2x 5x x 2 = 4x + 7x 4 x = x 4 + x 7 4 = ker(t ) = span 4, 7. x 4 It is essential that we write x as a linear combination with only free variables as coefficients. If our linear combination includes a variable which is not free, then we cannot choose our coefficients freely, so the kernel is not all linear combinations of this list of vectors. For example, consider Example 2..(b) where T : R 2 R is linear with [T ] = [ 2 ] x. For ker(t ), although we x 2 can write x = x x + x 2 2,

3 ([ [ ([ [ ker(t ) span, because not all linear combinations of, are allowed. The ] ]) ] ]) coefficients x, x 2 must satisfy x + 2x 2 =. Once we account for this by setting x = 2x 2 and letting x 2 be free, we get the correct answer: () x 2x2 2 2 = = x x 2 x 2 = ker(t ) = span, 2 because x 2 is free. Next, we proceed to the main result of this section and one of the main reasons we care about the kernel. Theorem A linear map T : R n R m is injective if and only if ker(t ) = { }. Proof. First, suppose that T is injective. Suppose x ker(t ). Then, T ( x) = = T ( x) = T ( ) = x =, the last implication following from T being injective. So, the only vector that can be in ker(t ) is. Since ker(t ), ker(t ) = { }. Secondly, suppose that ker(t ) = { }. How do we show that T is injective? Since T being injective means if T ( x) = T ( y), then x = y, we should start with T ( x) = T ( y) and end with x = y. For all x, y R n, T ( x) = T ( y) = T ( x) T ( y) =, = T ( x y) =, = x y ker(t ) = x y = = x = y. This shows T is injective. Linearity is used in the second implication, T ( u) = means u ker(t ) is used in the third implication, and ker(t ) = { } is used in the fourth implication. What is happening in terms of the columns of [T ] when T is injective? Suppose T : R n R m with [T ] = a... a n, so that x T. = x a + + x n a n. x n Then, () x ker(t ) = { x R n T ( x) = } =. x a + + x n a n =, x n so ker(t ) is the set of solutions to x a + + x n a n =. Hence, by Theorem 2..4, T is injective when the only solution to x a + + x n a n = is x =,..., x n =. We call this property of the list of column vectors linear independence. We summarize in Proposition 2..6.

4 4 Definition Given a,..., a n R m, ( a,..., a n ) is linearly independent (LI) if (2) x a + + x n a n = ONLY when x =,..., x n =. Otherwise, ( a,..., a n ) is linear dependent (LD). We define the empty list () to be linearly independent. The solution x =,..., x n = is called the trivial solution to (2). Proposition Suppose T : R n R m is linear with [T ] = [ a... a n ]. Then, T is injective ( a,..., a n ) is linear independent. We say ( a,..., a n ) IS linearly independent because a list like ( a,..., a n ) is a singular object. The trivial solution is always a solution to (2), but it is the only solution if and only if ( a,..., a n ) is linearly independent. Thus, ( a,..., a n ) is linearly independent means If x a + + x n a n =, then x =,..., x n =. In order to show ( a,..., a n ) linearly independent, one must start with x a + + x n a n =, and then show x =,..., x n =. In order to show ( a,..., a n ) linearly dependent, one needs to show x a + + x n a n = has a nontrivial solution. From Example 2.., we conclude ([ [ 2 (a), is linearly independent, ] 4]) (b) ([ ], [ 2 ]) is linearly dependent, ([ [ [ 2 (c),, is linearly dependent, 4] 5] 6]) (d),, 2 4, 5 7 is linearly dependent. We now characterize linear independence of the columns of a matrix A in terms of RREF(A). Free variables come from free columns, and free variables are the only way for a linear system to have more than one solution. This realization leads to Proposition Proposition Suppose a,..., a n R m, and let A = [ a... a n ]. Then, ( a,..., a n ) is linearly dependent if and only if RREF(A) has a free column, and ( a,..., a n ) is linearly independent if and only if RREF(A) has no free columns. Proof. By definition, ( a,..., a n ) is linearly dependent if and only if () x a + + x n a n = has a nontrivial solution. Converting () to augmented matrix and row reducing, [ A ] [ RREF(A) ]. Since the augmented column of () is all zeros, () has a nontrivial solution if and only if a solution description for it has a free variable, which is if and only if RREF(A) has a free column. Thus, ( a,..., a n ) is linearly dependent if and only if RREF(A) has a free column. Equivalently, ( a,..., a n ) is linearly independent if and only if RREF(A) has no free columns. We next state an alternative characterization of linear dependence involving span. It is useful to have different characterizations of the same thing not only to improve our understanding of that concept, but because different characterizations can be useful in different circumstances.

5 Proposition A list ( a,..., a n ) of vectors in R m is linearly dependent if and only if one of the vectors is in the span of the others. Proof. Assume n 2. First, suppose that one of the vectors, say a n, is in the span of the others. If it is one besides a n instead, the proof will work similarly. Then, a n span( a,..., a n ) = a n = c a + + c n a n = c a + + c n a n a n =, for some scalars c,..., c n. Since the coefficient of a n is, a nontrivial linear combination of ( a,..., a n ) gives. Hence, ( a,..., a n ) is linearly dependent. Secondly, suppose ( a,..., a n ) is linearly dependent. By definition, we have c a + + c n a n + c n a n = for some scalars c,..., c n, not all zero. Thus, at least one of the coefficients, say c n, is nonzero. If it is one besides c n that is not zero, the proof will work similarly. Because c n, we can divide by c n : c a + + c n a n + c n a n = = c a c n a n = c n a n = a n = c a c n c n c n = a n span( a,..., a n ). When n =, ( a ) is linearly dependent if and only if a =, Problem, which agrees with a being in the span of the others, since span( ) = { }. To be clear, Proposition 2..8 does not say a n is always in the span of the others, because our proof depended on c n. By the technique in Proposition 2..8, we can only say a j is in the span of the others if c j. In fact, it is possible for any number of vectors in a list to be in the span of the others, Problem 22. We illustrate this for vectors in R. We list vectors in order ( a, a 2, a ): a, a 2, a a n In Span of the Others NOT in Span of the Others Example a, a 2 a a a 2, a a, a 2, a,,, 2,,,,, The third example where only a is in the span of the others is especially interesting since we can only have a =, Problem 2. Linear independence of ( a,..., a n ) tells us that x a + +x n a n = only has the trivial solution. But what about inhomogeneous linear systems? Does linear independence of ( a,..., a n ) tell us anything about the solutions to x a + + x n a n = v. In fact, it does. 5

6 6 Proposition Suppose ( a,..., a n ) is linearly independent list of vectors in R m. Then, for all v R m, (4) has at most solution. Proof. Suppose that (5) (6) x a + + a n a n = v v = c a + + c k a n, v = d a + + d k a n are two solutions to (4). In order to use linear independence of ( a,..., a n ), we want a linear combination of ( a,..., a n ) resulting in. We can get such an equation by subtracting (6) from (5), yielding (c d ) a + + (c n d n ) a n =. Because ( a,..., a n ) is linearly independent, c d =,..., c n d n = = c = d,..., c n = d n. Thus, the two solutions, (5) and (6), to (4) must be the same. Thus, (4) has at most solution. One can also prove Proposition 2..9 using Proposition 2..7, Problem 8, or the definition of injectivity, Problem 9. It should be noted that if ( a,..., a n ) islinear independent, then x a + +a n a n = v could have or solutions. For example,, is linearly independent, and x + x 2 = has solutions, whereas x + x 2 = 2 has solution, namely x = 2 and x 2 =. Exercises:. For each of the following linear maps T, find ker(t ). Express your answer as the span of a finite set of vectors. (a) T : R 2 R 2 with [T ] = [ 5 ], (b) T : R R with [T ] = [ ], 2 4 (c) T : R R 2 with [T ] =, 5 (d) T : R R 2 given by T ( x) =, (e) T : R R given by T ( x) = x, 2 (f) T : R 2 R 4 with [T ] = 4 5 6, 7 8

7 (g) T : R 4 R with [T ] = Determine if the following lists of vectors are linearly independent. If not, find a linear combination of them that gives. ([ [ 2 (a), ] 5]) (b) (d) ([ [ 2 8, ] 2]) (c) ( ) 2,, (e) 2,, Find a linear map T : R R with ker(t ) = w w 2 w 4w 2 + 6w =. 4. Find a linear map T : R R 2 with ker(t ) = w w 2 w 4w 2 =, w 2 9w =. 5. Find a linear map T : R R 2 with ker(t ) = w w 2 8w + 6w 2 + w =. 6. Find a linear map T : R R 2 with ker(t ) = span,. w w w 7. Find a linear map T : R R with ker(t ) = span, Find a linear map T : R R 2 with ker(t ) = span 4. 8 Problems:. () Show that ( a ) is linearly dependent if and only if a =. 2. () Show that if ( a,..., a n ) is a list of vectors in R m with a i = for some i, then ( a,..., a n ) is linearly dependent.. () Show that if ( a,..., a n ) is a list of vectors in R m with a i = a j for some i j, then ( a,..., a n ) is linearly dependent.

8 8 4. () Show that if ( a,..., a n ) is a list of vectors in R m with a i = c a j for some i j and scalar c, then ( a,..., a n ) is linearly dependent. 5. () Given a list of vectors L = ( u,..., u k ) in R n, let M be any rearrangement of L. Show that L is linearly independent if and only if M is linearly independent. 6. (2) Suppose a, a 2 R m. Prove or give a counterexample: If ( a, a 2 ) is linearly independent, then a = c a 2 for some scalar c. 7. Suppose u,..., u m R n. Let X = ( u,..., u k ) and Y = ( u,..., u m ) with k m. (a) (2) Show that if X is linearly dependent, so is Y. (b) (2) Show that if Y is linearly independent, so is X. 8. (2) Suppose that ( u, u 2,..., u k ) is linearly independent, and c, c 2,..., c k are nonzero scalars. Show that (c u, c 2 u 2,..., c k u k ) is linearly independent as well. 9. (2) Prove or give a counterexample: If T, U : R n R m are linear maps with [T ] [U], then ker(t ) = ker(u).. () Prove or give a counterexample: If T, U : R n R m are linear maps with ker(t ) = ker(u), then [T ] [U].. () Prove or give a counterexample: If T : R n R m is a linear map and ( v,..., v k ) is linearly independent in R n, then (T ( v ),..., T ( v k )) is linearly independent. 2. () Prove or give a counterexample: If T : R n R m is a linear map and v,..., v k R n are such that (T ( v ),..., T ( v k )) is linearly independent, then ( v,..., v k ) is linearly independent.. () Suppose T : R n R m is linear, and ( v,..., v n ) spans R n and is linearly independent. Show that T is injective if and only if (T ( v ),..., T ( v n )) is linearly independent. 4. Suppose T : R n R m is a linear map. Show that (a) () ker(t ). (b) (2) If x, y ker(t ), then x + y ker(t ). (c) (2) If x ker(t ) and c is a scalar, then c x ker(t ). 5. (2+) Suppose T : R n R m is a linear map, and v,..., v k ker(t ). Show that span( v,..., v k ) ker(t ). 6. () Suppose ( u,..., u k ) is a list of vectors in R n. Show that ( u,..., u k ) is linearly dependent if and only if there exists some j =,..., k so that u j span( u,..., u j ). 7. (2) Suppose ( u, u 2,..., u k ) is a linearly independent list of vectors in R n. Show that the following lists are also linearly independent. (a) (2) (c u, c 2 u 2,..., c k u k ) for any nonzero scalars c,..., c k. (b) () ( u, u 2 u, u u,..., u k u ). Here, u is subtracted from every vector except itself. (c) () ( u, u 2 u, u u 2,..., u k u k ). Here, the previous vector is subtracted from each of u 2,..., u k.

9 (d) (+) ( u, u 2 u, u u 2 u,..., u k u k u ). Here, each of the previous vectors is subtracted from each of u 2,..., u k. 8. () Prove Proposition 2..9 using Proposition () Prove Proposition 2..9 using the definition of injectivity, Definition Suppose ( u,..., u k ) is linearly independent in R n, and v R n with v / span( u,..., u k ). (a) () Show that ( u,..., u k, v) is linearly independent. (b) (2) Show that if ( u,..., u k ) is linearly dependent or v span( u,..., u k ), then ( u,..., u k, v) is linearly dependent. (c) (2) Show that if ( u,..., u k, v) is linearly dependent, then ( u,..., u k ) is linearly dependent or v span( u,..., u k ). 2. () Suppose T, U : R n R m are linear with ker(t ) = ker(u), and suppose v,..., v k R n. Show that c T ( v ) + + c k T ( v k ) = if and only if c U( v ) + + c k U( v k ) =. 22. () For each n and k, find a list of vectors ( a,..., a n ) in R n so that a,..., a k lie in the span of the other vectors in ( a,..., a n ), but a k+,..., a n do not. 2. () Suppose a,..., a n R m, and suppose that in the list ( a,..., a n ), only a lies in the span of the others. Show that a =. 24. () Suppose a,..., a n R m, and suppose that in the list ( a,..., a n ), exactly a,..., a k lie in the span of the other vectors in ( a,..., a n ). Show that each vector in a,..., a k lies in the span of the other vectors in ( a,..., a k ). 25. () A list of vectors ( a, a 2,..., a k ) in R n is called affinely independent if the only solution to x a + x 2 a x k a k = satisfying x + + x k = ([ [ [ is x =,..., x k =. For example,,, is affinely independent but not linearly ] ] ]) ([ [ [ 2 independent, and,, is affinely dependent. ] ] ]) Show that ( a, a 2,..., a k ) is affinely independent if and only if ( a 2 a,..., a k a ) is linearly independent. 26. (4) Suppose ( u,..., u k ) is linearly independent in R n. Let c i,j be scalars and consider ( v,..., v m ) given by v i = c,i u + + c k,i u k. Show that ( v,..., v m ) is linearly independent if and only if is linearly independent. c, c 2,. c k,,..., c,m c 2,m. c k,m 9