L E C T U R E 2 1 : P R O P E RT I E S O F M AT R I X T R A N S F O R M AT I O N S I I. Wednesday, November 30

Transcription

1 L E C T U R E 2 1 : P R O P E RT I E S O F M AT R I X T R A N S F O R M AT I O N S I I Wednesday, November 30 1 the range of a linear transformation Let s begin by defining the range of a linear transformation. Definition 1.1. Let T : V Ñ W be a linear transformation of vector spaces. Then the range of T is the image of V under T, that is, the set rangeptq tw P W : w Tpvq for some v P Vu. Let s determine the range for the examples of linear transformations we considered in the previous lecture. Example 1. The range of the zero transformation T 0 : V Ñ W defined by T 0 pvq 0 W for all v P V, is the zero vector space t0 W u. Notationally, rangept 0 q t0 W u. Example 2. The identity transformation I V : V Ñ V is defined by I V pvq v for all v P V. Hence, rangepi V q V. Example 3. Differentiation T : PpRq Ñ PpRq is defined by Tpppxqq p 1 pxq. Since any polynomial in PpRq is the derivative of another polynomial in PpRq, we have rangeptq PpRq. Example 4. For any two real numbers a ď b we have the linear transformation T a,b : PpRq Ñ R defined by T a,b pppxqq ż b a ppxq dx. If a b, the integral of r{pb aq for any real number r, is exactly r, and we see that rangept a,b q R. If a b we have rangept a,b q t0u. 1

2 Example 5. The backward shift linear transformation T : R 8 Ñ R 8 is defined by We have rangeptq R 8. Tppa 1, a 2, a 3, a 4,... qq pa 2, a 3, a 4,... q. The forward shift linear transformation T : R 8 Ñ R 8 by Tppa 1, a 2, a 3,... qq p0, a 1, a 2, a 3,... q. We have rangeptq tp0, a 1, a 2, a 3,... q P R 8 u. Example 6. The transposition linear transformation T : M n prq Ñ M n prq is defined by TpMq M T. Since pm T q T M, we see that rangeptq M n prq. Example 7. The reflection about x-axis linear transformation T : R 2 Ñ R 2 is defined by Tpx, yq px, yq. We see that rangeptq R 2. Example 8. Let n ě 2 be an integer. The projection linear transformation T : R n Ñ R n 1 is defined by Tpa 1,..., a n q pa 1,..., a n 1 q. We have rangeptq R n 1. As was mentioned earlier, the range of a linear transformation T : V Ñ W is a subspace of W. Let s prove this. Theorem 1.2. Let T : V Ñ W be a linear transformation of vector spaces. Then rangeptq is a subspace of W. Proof. Since Tp0 V q 0 W, we know that 0 W P rangeptq. Hence, rangeptq is nonempty. Next, let w, z P rangeptq and a P F. Then there exists v, u P V such that Tpvq w and Tpuq z. Consequently, aw ` z atpvq ` Tpuq Tpav ` uq P rangeptq. Hence rangeptq is closed under addition and scalar multiplication. Recall that the kernel gave us about whether the associated linear transformation was one-to-one. The range also tells us something about the linear transformation T, it tells us whether the linear transformation is onto. 2

3 Definition 1.3. Let T : V Ñ W be a linear transformation of vector spaces. We say that T onto if for every vector w P W, there exists a vector v P V such that Tpvq w. Note that a linear transformation of vector spaces T : V Ñ W being onto is equivalent to the range of T being all of W, that is, rangeptq W. Example 9. The range of the zero transformation T 0 : V Ñ W defined by T 0 pvq 0 W for all v P V, is the zero vector space t0 W u. Notationally, rangept 0 q t0 W u. So T 0 is not onto. Example 10. The identity transformation I V : V Ñ V is defined by I V pvq v for all v P V. Hence, rangepi V q V. So I V is onto. Example 11. Differentiation T : PpRq Ñ PpRq is defined by Tpppxqq p 1 pxq. Since any polynomial in PpRq is the derivative of some other polynomial in PpRq, we have rangeptq PpRq. So T is onto. Example 12. For any two real numbers a ď b we have the linear transformation T a,b : PpRq Ñ R defined by T a,b pppxqq ż b a ppxq dx. If a b, the integral of r{pb aq for any real number r, is exactly r, and we see that rangept a,b q R. If a b we have rangept a,b q t0u. So T a,b is onto if a b, and is not if a b. Example 13. The backward shift linear transformation T : R 8 Ñ R 8 is defined by Tppa 1, a 2, a 3, a 4,... qq pa 2, a 3, a 4,... q. We have rangeptq R 8. So T is onto. The forward shift linear transformation T : R 8 Ñ R 8 by Tppa 1, a 2, a 3,... qq p0, a 1, a 2, a 3,... q. We have rangeptq tp0, a 1, a 2, a 3,... q P R 8 u. So T is not onto. 3

4 Example 14. The transposition linear transformation T : M n prq Ñ M n prq is defined by TpMq M T. Since pm T q T M, we see that rangeptq M n prq. So T is onto. Example 15. The reflection about x-axis linear transformation T : R 2 Ñ R 2 is defined by Tpx, yq px, yq. We see that rangeptq R 2. So T is onto. Example 16. Let n ě 2 be an integer. The projection linear transformation T : R n Ñ R n 1 is defined by Tpa 1,..., a n q pa 1,..., a n 1 q. We have rangeptq R n 1. So T is onto. In the last example, notice that we could have considered the similar linear transformation T : R n Ñ R n is defined by Tpa 1,..., a n q pa 1,..., a n 1, 0q. Note that in this case, T is not onto. Putting all of this together, if the null space of T is the zero vector space t0 V u, then T is injective. If the range of T is the whole vector space W, then T is surjective. This hints at a relationship between the kernel and range of a linear transformation, which will explore in the next section. 2 relating the kernel and range In order to understand the relationship between the null space and the range we have to consider their dimensions. Definition 2.1. Let T : V Ñ W be a linear transformation of vector spaces. If nullpt q and rangept q are finite dimensional, we refer to dimpkerptqq as the nullity of T and dimprangeptqq as the rank of T. Now, if you look at the examples of linear transformations that have been given, you see that the larger the nullity the smaller that range, and vice versa. This isn t a coincidence. 4

5 Theorem 2.2 (Dimension Theorem). Let T : V Ñ W be a linear transformation of vector spaces, with V a finite dimensional vector space. Then dimpkerpt qq ` dimprangept qq dimpvq. Proof. Since V is a finite dimensional vector space, we know that the subspace kerptq is also finite dimensional. Let tv 1,..., v m u be a basis for kerptq. Then as was shown in a previous lecture, we can extend this set to a basis for V. Suppose tv 1,..., v m, u 1,..., u n u is an extension of the basis of kerptq to a basis for V. Then we see that dimpvq m ` n dimpkerptqq ` n. Therefore, in order to prove the theorem we just need to show that dimprangeptqq n. To do so, we will show that ttpu 1 q,..., Tpu n qu is a basis for rangeptq. Recall that in order to show ttpu 1 q,..., Tpu n qu is a basis for rangeptq, we need to show two things. First, we need to show that this set generates rangeptq, and second, we need to show it is linearly independent. We begin by proving that ttpu 1 q,..., Tpu n qu generates rangeptq. Let w P rangeptq. Then w Tpvq for some vector v P V. Since tv 1,..., v m, u 1,..., u n u is a basis for V, we know that v a 1 v 1 ` ` a m v m ` b 1 u 1 ` ` b n u n. Hence, w Tpvq Tpa 1 v 1 ` ` a m v m ` b 1 u 1 ` ` b n u n q b 1 Tpu 1 q ` ` b n Tpu n q. Hence, the set ttpu 1 q,..., Tpu n qu generates rangeptq. Next, we want to show the set ttpu 1 q,..., Tpu n qu is linearly independent. Let a 1,..., a n P R such that a 1 Tpu 1 q ` ` a n Tpu n q 0. We want to show that a 1 a n 0. We have 0 a 1 Tpu 1 q ` ` a n Tpu n q Tpa 1 u 1 ` ` a n u n q. This implies that a 1 u 1 ` ` a n u n P kerptq. Since tv 1,..., v m u is a basis for kerptq, this would imply that a 1 u 1 ` ` a n u n b 1 v 1 ` ` b m v m 5

6 for some b i P F. In turn, this implies that a 1 u 1 ` ` a n u n b 1 v 1 b m v m 0. Since the set tv 1,..., v m, u 1,..., u n u is linearly independent, we must have a 1 a n b 1 b m 0. Let s look at a few examples. Example 17. Suppose V is a finite dimensional vector space with dimension n. Recall the zero transformation T 0 : V Ñ W defined by T 0 pvq 0 W for all v P V. Then kerpt 0 q V and rangept 0 q t0 W u. Recalling that the zero vector space t0 W u has dimension 0 (the empty set is a basis), we have dimpvq n n ` 0 dimpkerpt qq ` dimprangept qq. Example 18. Suppose V is a finite dimensional vector space with dimension n. The identity transformation I V : V Ñ V is defined by I V pvq v for all v P V. Then kerpi V q t0 V u and rangepi V q V. Hence, we have dimpvq n 0 ` n dimpkerpt qq ` dimprangept qq. Example 19. Let P n prq denote the set of polynomials with real coefficients whose degree is at most n. Note that P n prq is a finite dimensional vector space having dimension n ` 1 (a basis is t1, x,..., x n u). Differentiation T : P n prq Ñ P n 1 prq is defined by Tpppxqq p 1 pxq. Then kerptq tconstant polynomialsu and rangeptq P n 1 prq. Since kerptq tconstant polynomialsu has basis t1u and dimpp n 1 prqq n, we have dimpp n prqq n ` 1 dimpkerptqq ` dimprangeptqq 6

7 Example 20. Let n ě 1. For any two real numbers a ă b we have the linear transformation T a,b : P n prq Ñ R defined by T a,b pppxqq ż b a ppxq dx. In this case we have rangept a,b q R and kerpt a,b q tppxq P P n prq : ş b a ppxq dx 0u. By the dimension formula dimpkerptqq dimpp n prqq dimprangeptqq n ` 1 1 n. Example 21. Let n ě 1. The backward shift linear transformation T : R n Ñ R n is defined by Tppa 1,..., a n qq pa 2,..., a n, 0q. We have rangeptq tpa 1,..., a n 1, 0q P R n u and kerptq tpa 1, 0,..., 0q P R n u. Therefore, dimpr n q n 1 ` n 1 dimpkerptqq ` dimprangeptqq. 3 isomorphisms In this section we want to consider what it means if a linear transformation is both one-to-one and onto. Definition 3.1. If a linear transformation T : V Ñ W is both one-toone and onto, we say that T is an isomorphism and that V and W are isomorphic. If two vector spaces V and W are isomorphic, it means that, as vector spaces, they are the same. They may look very different, but this difference is superficial and the two vector spaces are really the same vector space in different guises. Let s consider and example. Example 22. Consider the linear transformation T : R 3 Ñ P 2 prq defined by Tpa, b, cq ax 2 ` bx ` c. Let s show that this map is an isomorphism. 7

8 First we ll know that T is a linear transformation. Let pa, b, cq, pd, e, fq P R 3 and α P R. Then Tpαpa, b, cq ` pd, e, fqq Tpαa ` d, αb ` e, αc ` fq pαa ` dqx 2 ` pαb ` eqx ` pαc ` fq αpax 2 ` bx ` cq ` pdx 2 ` ex ` fq αtpa, b, cq ` Tpd, e, fq. From the above, we see that T is in fact a linear transformation. Next we ll show that T is an isomorphism. To do so, suppose pa, b, cq P kerptq. This means 0 Tpa, b, cq ax 2 ` bx ` c. In order for the right hand side to be zero, it must be the case that a b c 0. Hence, we ve shown that kerptq is contained in the set tp0, 0, 0qu. Since the set tp0, 0, 0qu is contained in kerptq, we know that kerptq tp0, 0, 0qu. By Proposition??, we know that kerptq tp0, 0, 0qu implies that T is one-to-one. All that remains to be shown is that T is also surjective. There are several ways that one could show this, but perhaps the simplest is to use the Dimension Theorem (Theorem 2.2). We know that dimpr 3 q 3 and that dimpkerpt qq dimptp0, 0, 0quq 0. Therefore, the Dimension Theorem tells us that dimprangeptqq dimpr 3 q dimpkerptqq Since rangeptq is a subspace of P 2 prq having dimension 3, and the whole vector space P 2 prq also has dimension 3, we know that rangeptq P 2 prq. Hence, T is onto. Since T is both one-to-one and onto, we know that T is an isomorphism. 8