12. Special Transformations 1

Transcription

1 12. Special Transformations 1 Projections Take V = R 3 and consider the subspace W = {(x, y,z) x y z = 0}. Then the map P:V V that projects every vector in R 3 orthogonally onto the plane W is a linear transformation (why is it linear?). See Fig for a picture of what such a map looks like. We want to produce a formula for P. If X is any vector in W, then certainly, P(X) = X. So this holds for the two vectors A = (1,1,0) and B = (1,0,1), both of which lie in W. Indeed, these vectors are a linearly independent set of vectors, and W is 2-dimensional, so they form a basis for W. Further, C = (1, 1, 1) is orthogonal to both A and B (its dot product with both is 0), so P(C) = 0. It is easy to see that {A, B, C} remains linearly independent, so these three vectors form a basis for V. Relative to this basis (used for both domain and codomain), the matrix for P is simply M =

2 12. Special Transformations 2 Must we compute the image of a typical vector, like X = (1,2,3), by first expressing it in terms of the basis {A, B, C}, applying the matrix representation M, then translating back into the standard basis representation? Essentially, yes. But the work is best done by expressing the standard basis vectors {E 1,E 2,E 3 } in terms of {A = (1,1,0), B = (1,0,1), C = (1, 1, 1)}: using the relations we see that 3E 1 = (3,0,0) = A + B +C 3E 2 = (0,3,0) = 2A B C 3E 3 = (0,0,3) = A + 2B C 3P(E 1 ) = P(A) + P(B) + P(C) = A + B = (2,1,1) 3P(E 2 ) = 2P(A) P(B) P(C) = 2A B = (1,2, 1) 3P(E 3 ) = P(A) + 2P(B) P(C) = A + 2B = (1, 1,2) so that the matrix for P relative to the standard basis is N = ; 1 1 2

3 12. Special Transformations 3 that is, P(X) = = Is it possible to extend the notion of a projection map to a general abstract vector space? The important feature of projection maps is that there is some subspace of the domain on which the map is the identity, and every other vector is mapped into this subspace; that is, any vector that is the projection of some vector is itself fixed by the projection. In symbol form: P(P(X)) = P(X) for every X. This leads to the following definition: an endomorphism P:V V is called a projection map if it satisfies the condition P 2 = P. While this definition does allow us to use the notion of a projection map on any type of vector space, the essential features of its behavior carry over from the geometric setting in R 3.

4 12. Special Transformations 4 Theorem Let P:V V be a projection map on the n-dimensional vector space V. Then there is some (ordered) basis {V 1,V 2,,V n } of V for which P is the linear extension of the assignments P(V i ) = V i i = 1,2,,r 0 i = r +1,,n where r = dim(im(p)). Consequently, the matrix of P relative to the basis {V 1,V 2,,V n } (in both domain and codomain) is the diagonal matrix 1 L 0 O M 1 M = 0. M O 0 L 0 Proof Let {X 1,X 2,, X r } be a basis for Im(P) and let {Y 1,Y 2,,Y k } be a basis for ker(p). We claim that {X 1,X 2,, X r,y 1,Y 2,, Y k } is a basis for V. To do this, we first show that these vectors are linearly independent: if the scalars a 1,a 2,,a r,b 1,b 2,,b k satisfy a 1 X 1 +L+ a r X r +b 1 Y 1 +L+b k Y k = 0, then applying P yields

5 12. Special Transformations 5 0 = P(a 1 X 1 +L+a r X r +b 1 Y 1 +L+b k Y k ) = a 1 P(X 1 ) +L+a r P(X r ) +b 1 P(Y 1 ) +L+b k P(Y k ) = a 1 P(X 1 ) +L+a r P(X r ) since the Y s are in the kernel of P. But each of the X s is in the image, so there are vectors W 1,W 2,, W r in V for which X i = P(W i ), i = 1,,r. Therefore, P(X i ) = P 2 (W i ) = P(W i ) = X i since P is a projection map. Substituting this above, we find that 0 = a 1 X 1 +L+a r X r. But {X 1,X 2,, X r } is a basis for Im(P), so the X s are linearly independent, whence all the a s must equal 0. This means that b 1 Y 1 +L+b k Y k = 0; but {Y 1,Y 2,,Y k } is a basis for ker(p), so the Y s are linearly independent, and the b s must all equal 0. This completes the proof that {X 1,X 2,, X r,y 1,Y 2,, Y k } is linearly independent in V. Then, since there are r +k = dim(im(p)) +dim(ker(p )) = dimv vectors in the set, they must form a basis for V. So if we put V 1 = X 1,, V r = X r,v r +1 = Y 1,,V n = Y k, we will have

6 12. Special Transformations 6 P(V i ) = P(X i) = X i = V i P(Y i r ) = 0 i = 1,2,,r i = r +1,,n exactly as we require. That the matrix of P in the basis {V 1,V 2,,V n } is M is clear. // (It is worth mentioning that not every projection map on R n is an orthogonal projection of vectors onto a smaller dimensional hyperplane. Smith describes (p. 188) a skew projection map on R 3 in which all vectors in space are projected onto a planar subspace in a direction different from the perpendicular to this plane.) Proposition Let P:V V be a linear transformation on the n-dimensional vector space V. Then P is a projection map if and only if it is represented (relative to the same basis for V) by an idempotent matrix M (one for which M 2 = M). //

7 12. Special Transformations 7 Nilpotent maps An endomorphism N:V V is called nilpotent of index k if the map N k is the zero map, but N k 1 is not the zero map, on V. The prototype of nilpotent maps is the differential operator D:P n (R) P n (R) (whose index n + 1 = dim(p n (R)) ). But a simpler example is offered by the shift operator S:R n R n, defined as S(x 1, x 2,, x n ) = (0, x 1, x 2,, x n 1 ). This shift operator has index n = dim(r n ). Not all nilpotent maps have index equal to the dimension of the underlying vector space (consider, for instance, the second derivative map on P n (R)). But when this property does hold, there is a particularly simple canonical form for representing the map by a matrix. Theorem Let N:V V be an endomorphism on the n-dimensional vector space V which is nilpotent of index n. Then there is a vector V V so that {V, N (V ), N 2 (V),, N n 1 (V)} is a basis for V. Relative to this basis, N has matrix

8 12. Special Transformations L L 0 M = L 0. O 0 L Proof Since N has index of nilpotency n, there must be some vector V V so that N n 1 (V) 0 (otherwise N n 1 is the zero map and N has index of nilpotency smaller than n). And as all the powers of N are also endomorphisms of V which would have to carry 0 to 0, none of the vectors V, N(V), N 2 (V),, N n 1 (V) is the zero vector. Indeed, these vectors are linearly independent, for if there were a linear relation of the form a 1 V + a 2 N (V)+ a 3 N 2 (V) +L+a n N n 1 (V) = 0, applying N repeatedly to both sides of the equation produces the relations

9 12. Special Transformations 9 a 1 N(V) +a 2 N 2 (V)+ a 3 N 3 (V) +L+a n 1 N n 1 (V) = 0 a 1 N 2 (V) +a 2 N 3 (V)+L+ a n 2 N n 1 (V) = 0 M a 1 N n 1 (V) = 0 The last of these equations forces a 1 = 0 since N n 1 (V) 0. Applying N to each of the remaining equations ultimately produces the equation a 2 N n 1 (V) = 0, from which we deduce a 2 = 0. Continuing in this fashion, we show that each of the a s is 0, whence {V, N (V ), N 2 (V),, N n 1 (V)} is a linearly independent set of vectors in V. Since V has dimension n and this set contains n vectors, it must be a basis for V. Finally, it is clear that with respect to this basis, N is represented by the matrix M above. //

10 12. Special Transformations 10 Cyclic transformations An endomorphism C:V V on a vector space is called cyclic if there exists a vector V V such that V and its successive image vectors C(V),C 2 (V), under C span all of V. That is, V = L(V,C(V),C 2 (V), ) The vector V is called a cyclic vector for C. Notice that any nilpotent endomorphism of index equal to the dimension of the underlying vector space is necessarily cyclic. It is also easy to identify the cyclic transformations on R 2 : Proposition The cyclic endomorphisms of the plane C:R 2 R 2 are precisely those which are not a multiple of the identity map. Proof If C is cyclic with cyclic vector V, then R 2 = L(V,C(V), ). So C(V) cannot be a scalar multiple of V, else all the successive image vectors C(V),C 2 (V), are multiples of V as well, whence R 2 = L(V,C(V), ) is one-dimensional. This implies that C is not a multiple of the identity map on R 2.

11 12. Special Transformations 11 Conversely, if C:R 2 R 2 is not cyclic, then it has no cyclic vector; that is, every vector V in R 2 fails to satisfy R 2 = L(V,C(V), ). This means that for every vector V, C(V) must be a scalar multiple of V. In particular, this holds for the three vectors (1,1), (1,0) and (0,1); that is, C(1,1) = a(1,1), C(1,0) = a 1 (0,1), C(0,1) = a 2 (0,1). But then (a,a) = C(1,1) = C(1,0) +C(0,1) = (a 1,a 2 ), so a = a 1 = a 2, whence in general, C(x, y) = xc(1,0)+ yc(0,1) = x(a,0) + y(0,a) = (ax,ay) = a(x, y). That is, C is a multiple of the identity map. // There is a general form for matrices that represent cyclic transformations.

12 12. Special Transformations 12 Theorem Let C:V V be a cyclic endomorphism on the n-dimensional vector space V. If V is a cyclic vector for C, then {V,C(V),C 2 (V),,C n 1 (V)} is a basis for V and the matrix of C relative to this basis has the form a a 1 M = O O M, 1 0 a n a n 1 where C n (V) = a 0 V +a 1 C(V) +L+a n 1 C n 1 (V). Proof Since V is n-dimensional and V is a cyclic vector for C, then {V,C(V),C 2 (V),,C n 1 (V)} is a spanning set of n vectors in V. Thus, it must also be linearly independent, hence a basis for V. That M is the matrix for C relative to this basis is clear. //