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# 7.1 Linear Transformations

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1. 7.1 Linear Transformations

2. Notation f: X  Y, “function from X to Y” takes input values from X and assigns them a value in Y. If V and W are two vector spaces, a function T: V  W is called a linear transformation if it satisfies the axioms: T1. T(v + v1) = T(v) + T (v1) for all v and v1 in V (T preserves vector addition) T2. T(rv) = rT(v) for all v in V and r in . (T preserves scalar multiplication) The linear transformation T: V  V is called a linear operator on V

3. Example T: 2  3 by for all in 2 Show that T is a linear transformation. Sol’n: let

4. Definition Notice that we could have written the transformation in the example using matrices: Def: If A is (mxn), the matrix transformation TA: n m is defined by TA(X) =AX for all columns X in n

5. Theorem 1 TA: n m is a linear transformation for each A (mxn) Proof: TA(X+X1) = A(X+X1) = AX + AX1 = TA(X) + TA(X1) TA(rX) = A(rX) = r(AX) = rTA(X) 

6. Elementary Matrices TE exchanges the x and y points (like a 2 dimensional inverse of a function)-- reflection over the line y = x. Action on a graph of a rectangle: (see p. 342) TF doubles all of the x values and keeps the same y, so this is X expansion (X compression if value <1) (see p. 342) TG changes x to x+y while keeping y the same (X shear)--every x value is moved by y units (could be right or left) (see p. 342)

7. Example R: 2 2 is a counterclockwise rotation of the plane about the origin through an angle . Show that R is a linear operator. To show first: R(X+Y) = R(X) + R(Y) R(X+Y)R(X) Y X+Y R(X+Y) rotates X+Y by . As we do, we rotate the entire parallelogram. Then R(Y) X we have R(X) and R(Y) also. Diagonal then is both R(X+Y) and R(X) + R(Y) by parallelogram rule. So R(X+Y)=R(X) + R(Y) . (Need to show R(rX)=r R(X)--same idea)

8. Other important examples Given vector spaces V and W, the following are LT’s: Identity operator on V: 1V: V V where 1V(v) = v  v in V Zero transformation V  W: 0: V  W where 0(v) = 0 Scalar operatorV  V: a: V V where a(v) = av  v in V (all fairly easy to prove)

9. Example Transposing is a linear transformation: T: Mmn  Mnm where T(A) = AT for all A in Mmn T1: (A+B)T = AT + BT T2: (rA)T = r(AT) (Trace is also a linear transformation)

10. Example-Projections U is a subspace of n. Then projection on U is an LT. T: n  U is defined by T(X) = proju(X) for all X in n. {E1,E2,…,Em} is an orthonormal basis of U. L T(X) = (X•E1)E1 + (X•E2)E2+…+ (X•Em)Em T1: T(X+Y)= ((X+Y)•E1)E1 +…+ ((X+Y)•Em)Em = ((X•E1)+(Y•E1))E1 +…+ ((X•Em)+(Y•Em))Em = (X•E1)E1 +(Y•E1)E1 +…+(X•Em)Em +(Y•Em)Em = T(X)+T(Y) (T2 similar idea)

11. Example Differentiation and Integration are linear transformations: D: PnPn-1 where D[p(x)]=p’(x) for all p(x) in Pn I: Pn Pn+1 where I[p(x)] = for all p(x) in Pn For D: T1: (p(x)+q(x))’=p’(x)+q’(x) T2: [rp(x)]’ = rp’(x) For I: T1: T2:

12. Theorem 2 T: V  W is a linear transformation. Then 1. T(0) = 0 2. T(-v) = -T(v) for all v in V 3. T(r1v1 + …+rkvk) = r1T(v1)+…+rkT(vk) for all vi in V and all ri in  Proof: 1) T(0) = T(0v)=0T(v) = 0 2) T(-v) = T[-1(v)]=(-1)T(v) = -T(v) 3) by induction (next page)

13. Theorem 2 (proof cont) 3) by induction: If k=1, T(r1v1)=r1T(v1) by axiom 2. Assume true for k, then for k+1: T(r1v1 + …+rkvk+rk+1vk+1)= T(r1v1 + …+rkvk)+T(rk+1vk+1) = r1T(v1)+…+rkT(vk)+rk+1T(vk+1) 

14. Example T: VW is a LT. If T(v-3v1) = w and T(2v-v1) = w1, find T(v) and T(v1) in terms of w and w1. T(v) - 3T(v1) = w 2T(v) - T(v1) = w1 2R1-R2: -5T(v1) = 2w-w1

15. Example T: 3  is a LT. T(3,-1,2) = 5, T(1,0,1) = 2. Find T(-1,1,0) Sol’n: We can use thm 2 if (-1,1,0) is some linear combination of the vectors given: a(3,-1,2) + b(1,0,1) = (-1,1,0) 3a + b =-1 -a = 1 2a + b = 0  b = 2 which also works in eq’n 1, so -(3,-1,2) + 2(1,0,1) = -5+4=-1.

16. A nice extension Notice that, using the idea of the last example with use of thm 2: 1) If we know T(v1),…,T(vn) 2) If V = span{v1,…,vn} Then we can write any v in V as a linear combo of the vi and we can then find T(v) !!

17. Theorem 3 T: V W and S: V  W. Suppose that V = span{v1,…,vn} If T(vi) = S(vi), then S = T. Proof: for v in V: v = t1v1+…+tnvn (any vector in V) Then T(v) = t1T(v1) + …+ tnT(vn) = t1S(v1) + … + tnT(vn) = S(v) (We’ve shown true for any v in V). 

18. Theorem 4 Let V and W be vector spaces and let {e1,…,en} be a basis of V. Given any vectors w1,…,wn in W (don’t have to be distinct), there exists a unique LT T: V W satisfying T(ei)=wi for each i =1,…,n. The action of T is: Given v = v1e1 + …+ vnen in V, then T(v) = T(v1e1 + … + vnen) = v1w1 +…+ vnwn

19. Theorem 4-Proof 1) If T exists, it is unique: Say there also exists S also satisfies S(ei)=wi=T(ei) for all i, then thm 3 says S=T. 2) Does such a T exist?: Given v in V, we choose T(v) in W. Since {e1,…,en} is a basis of V, we know v = v1e1+…+vnen where v1,…,vn are uniquely determined (5.3 thm2) So T: V  W by T(v) = T(v1e1+…+vnen) = v1T(e1)+…+vn T(en) = v1w1+…+vnwn for all v in V. So T(ei) = wi for each i. Need to show that linear trans: T1,T2

20. So what? We can now easily define linear transformations: 1) Just take a set of basis vectors of V and assign them values in W. Every ordered set in W will have some unique linear transformation. (See next example)

21. Example Find an LT T: 3 2 such that: Since we have a basis of 3, we can use thm 4.

22. Continued...

23. Theorem 5 T: n m is a LT. Give vectors in n as column vectors. 1. There exists an (mxn) matrix A such that T(X) = AX for all columns in n ; that is T = TA. 2. The columns of A are T(E1),…,T(En) where {E1,…,En} is the standard basis of n . So A written in terms of its columns is: A = [T(E1) … T(En)] (called standard matrix of T)

24. Theorem 5--Proof Proof: Show that the matrix given in (2) satisfies (1): Given X in n, let X = x1E1 + … + xnEn (an LC of basis vectors) Use thm 2 to find T(X): So T=TA. 

25. Example Find the standard matrix of T: 32 when Easy to see matrix A: Could also use Thm 5 pt 2 to find columns of A (if it were tougher to see):

26. Example Find the standard matrix of the following linear transformations: 1. Rotation R about the origin through the angle . 2. Projection Pm on the line y = mx 3. Reflection Sm in the line y = mx Sol’n: 1. {E1,E2} is the standard basis: E2 R(E2) R(E1) E1

27. Example Sol’n: 2. {E1,E2} is the standard basis: y=mx Sm(X) Pm(X) X D is direction vector of line y=mx.

28. Example Sol’n: 3. y=mx Sm(X) Pm(X) X Sm(X)=X+2[Pm(X)-X]=2Pm(X)-X