can only hit 3 points in the codomain. Hence, f is not surjective. For another example, if n = 4

Transcription

1 .. Conditions for Injectivity and Surjectivity In this section, we discuss what we can say about linear maps T : R n R m given only m and n. We motivate this problem by looking at maps f : {,..., n} {,..., m} and then we realize analogous results hold for linear maps T : R n R m. We then discuss equivalent versions in terms of span and linear independence given the number of vectors and the number of components, which we prove. What we can say about maps f : {,..., n} {,..., m} given only n and m? For example, if n = and m =, then any map f : {,, } {,,, }, such as can only hit points in the codomain. Hence, f is not surjective. For another example, if n = and m =, then any map f : {,,, } {,, }, such as has outputs among points. At least two outputs must be the same, so f is not injective. For a final example, if n = m =, any map f : {,,, } {,,, }, such as must be surjective if it is injective. If f is injective, each of outputs must hit different points in a codomain of size, so each point in the codomain is an output, and f is surjective. On the other hand, any map f : {,,, } {,,, }, such as

2 cannot be surjective if it is not injective. If f is not injective, each of outputs cannot all points in the codomain. Else, all outputs would be different, and f would be injective, a contradiction. Hence, all points in the codomain cannot be hit, so f is not surjective. There is nothing special about and here, and these arguments only depend on the relative sizes of the domain and codomain. Here is what we can say about maps f : {,..., n} {,..., m} given only m and n. Proposition... Suppose f : {,..., n} {,..., m} is any map. (a) If n > m, then f is not injective. (b) If n < m, then f is not surjective. (c) If n = m, then f is injective if and only if f is surjective. Proof. Problem. We have a name for maps that are both injective and surjective, namely bijective. Injectivity and surjectivity are both properties of maps that we desire, so bijectivity is even more desirable. Definition... A map f : X Y is bijective if it is both injective and surjective. Why did we discuss when maps f : {,..., n} {,..., m} are injective and surjective anyway? What does it have to do with linear maps? It turns out that exactly the same properties in Proposition.. hold for linear maps T : R n R m! Theorem... Suppose T : R n R m is a linear map. (a) If n < m, then T is not surjective. (b) If n > m, then T is not injective. (c) If n = m, then T is injective if and only if T is surjective. In order to prove Theorem.., we translate these properties into properties of list of vectors. Let T : R n R m be a linear map, and write T = a... a n. By Proposition..7 and Proposition..6, Theorem.. is equivalent to Theorem.. below. Thus, proving Theorem.. will prove Theorem... We also summarize the m = n case in the Invertible Matrix Theorem. The reason for this name will become clear in Section.6. Theorem... Suppose a,..., a n R m. (a) If n < m, then ( a,..., a n ) does not span R m. (b) If n > m, then ( a,..., a n ) is linearly dependent. (c) If n = m, then ( a,..., a n ) spans R m if and only if ( a,..., a n ) is linearly independent. Proof. Let A = a... a n and B = RREF(A), which are both m n matrices. (a) Suppose n < m. Since all of the leading nonzero entries in each row in B lie in different columns, B has at most n < m rows with leading nonzero entries. Thus, B, such as , , , must have a row of all zeros. Therefore, by Proposition..8, ( a,..., a n ) does span R m.

3 (b) Suppose n > m. Since each row gives rise to at most pivot column, B has at most m < n pivot columns. Thus, B, such as 0 5, , , must have a free column. Therefore, by Proposition..7, ( a,..., a n ) is linearly dependent. (c) Finally, suppose m = n. If B = I n =......, then B has neither a row of all zeros nor a free column. So by Proposition..8 and Proposition..7, ( a,..., a n ) is linearly independent and spans R n. Else, B I n. Observe that if none of the n rows in B is all zeros, then the leading s must appear in each row and proceed on the main diagonal, which forces B = I n by properties of RREF. Thus, B, such as 0 0 6, , , must have a row of all zeros. Then, B also has a free column since each of the pivot columns of B corresponds with a nonzero row of B, of which there at most n. So by Proposition..8 and Proposition..7, ( a,..., a n ) is linearly dependent and does not span R n. Theorem..5 (Invertible Matrix Theorem). Given a,..., a n R n, let T : R n R n be linear with T = A = a... a n. The following are equivalent: (a) ( a,..., a n ) spans R n. (b) T is surjective (c) ( a,..., a n ) is linearly independent. (d) T is injective. (e) RREF(A) = I n. Proof. If (e) is true, then (a) and (c) are true by the proof of Theorem..(c). Then, (b) and (d) are true by Proposition..7 and Proposition..6. If (e) is false, then (a) and (c) are false by the proof of Theorem..(c). Then, (b) and (d) are false by Proposition..7 and Proposition..6. Keep in mind that Theorem..5 requires n vectors in R n. If the number of vectors and number of components do not agree, (a) - (e) need not be equivalent. Let us realize via examples that any combination of ( a,..., a n ) spans R m and ( a,..., a n ) is linearly independent can be true, where m is the number of components each of the vectors have. We also note the necessary, but not sufficient, conditions that have to hold about m, n to make the given properties hold.

4 ( a,..., a n ) spans R m ( a,..., a n ) is LI Example Necessary conditions 0, n = m 0 ) 0,, n > m 0 ) n < m 0), n, m 0 0) We leave it to the reader to verify that these necessary conditions are necessary for the given properties to hold, Problems 6-9. Note that these necessary conditions are not sufficient. For example, for the (, ) case,, has the same number of vectors as components, but 0 0), neither spans R 0 0) n nor is linearly independent. Exercises:. Determine if the following lists of vectors span R m, where m is the number of components they each have. Also, determine if the list is linearly independent. (a) ) ) (b), 5 7 (c),, 7) (d), (e),, (f),, (g),,, (h),,, (i),, 5,

5 . Find linear maps T : R R so that the following hold, if possible. If it is not possible, explain why. (a) T is both injective and surjective. (b) T is injective but not surjective. (c) T is not injective, but is surjective. (d) T is neither injective nor surjective.. Do the same as Exercise but for T : R R.. Do the same as Exercise but for T : R R. Problems:. () Prove Proposition.... (+) Show that if f : X Y is bijective, then for every y Y, there exists a unique x X so that f(x) = Y.. () Show that if X, Y are finite sets, then there exists a bijection f : X Y if and only if X, Y have the same size.. () Show that every map f : {,..., n} {,..., n} is either bijective or neither injective nor surjective. 5. () Show that every map T : R n R n is either bijective or neither injective nor surjective. 6. () Show that if ( a,..., a n ) spans R m and is linearly independent, n = m. 7. () Show that if ( a,..., a n ) spans R m and is linearly dependent, n > m. 8. () Show that if ( a,..., a n ) does not span R m and is linearly independent, n < m. 9. () Given any n, m, find a linearly dependent list of vectors ( a,..., a n ) in R m which does not span R m. 0. () Show that if A is an m n matrix with m > n, then RREF(A) has at least m n rows of all zeros. While the following questions fall into the area of combinatorics rather than linear algebra, they are a good tool for understanding maps f : {,..., n} {,..., m}.. () How many maps f : {,..., n} {,..., m} are there?. () How many maps f : {,..., n} {,..., m} are injective? Assume n m. (Look at some examples to get ideas!). () How many maps f : {,..., n} {,..., m} are not injective? Assume n m. Note: Counting how many maps f : {,..., n} {,..., m} are surjective requires a more advanced technique called the Principle of Inclusion-Exclusion.. () How many maps f : {,..., n} {,..., n} are bijective? 5. () How many maps f : {,..., n} {,..., n} are neither injective nor surjective? 5