We have seen that a projection P:V V on a finite vector space can be represented by a matrix of the form

Transcription

1 14A. Rank of an Endomorphism 1 Rank of an endomorphism We have seen that a projection P:V V on a finite vector space can be represented by a matrix of the form 1 L 0 O M 1 M = 0 M O 0 L 0 with respect to some basis {V 1,V 2,,V n } of V. It turns out that any endomorphism T:V V on such a vector space has a similar matrix representation; we only have to choose our bases wisely: Theorem Let T:V V be an endomorphism on the n-dimensional vector space V. Then there is some pair of bases of V with respect to which T is represented by a diagonal matrix of the form

2 14A. Rank of an Endomorphism 2 1 L 0 O M 1 M = 0. M O 0 L 0 Proof Let {U 1,U 2,,U k } be a basis for ker(t). Since the U s are a linearly independent set in V, we can extend this set to a basis for V. Suppose then that the vectors V 1, V 2,, V r are such that {V 1,V 2,,V r,u 1,U 2,,U k } is a basis for V. Since k = dim(ker(t)), then by the Rank-Nullity Theorem, r = dim(im(t)). Also, the vectors {T(V 1 ),T(V 2 ),,T(V r )} are linearly independent in V, for otherwise we can find scalars a 1,a 2,,a r so that 0 = a 1 T(V 1 ) +a 2 T(V 2 )+L+ a r T(V r ) = T(a 1 V 1 +a 2 V 2 +L+a r V r ) which implies that a 1 V 1 +a 2 V 2 +L+a r V r ker(t), or, that there exist scalars b 1,b 2,,b k for which

3 14A. Rank of an Endomorphism 3 a 1V 1 +a 2 V 2 +L+a r V r = b 1 U 1 +b 2 U 2 +L+b k U k ; since {V 1,V 2,,V r,u 1,U 2,,U k } is a basis for V, this is not possible unless all the a s and b s equal 0. It follows that {T(V 1 ),T(V 2 ),,T(V r )} is a basis for Im(T). As we did in the domain of T, so can we also do in the codomain: we can choose vectors W 1,W 2,, W k in V so that {T(V 1 ),T(V 2 ),,T(V r ), W 1,W 2,, W k } is a basis for V. Using the basis {V 1,V 2,,V r,u 1,U 2,,U k } for the domain and {T(V 1 ),T(V 2 ),,T(V r ), W 1,W 2,, W k } for the codomain of T, we represent T by the matrix M as stated above. // We can write the matrix in the statement of the theorem in simpler block form as M = I r 0 ; the 0 0 number r = dim(im(t)) is called the rank of the transformation T. This theorem shows that to any endomorphism T:V V there correspond four subspaces of V. Two of these we are already familiar with, ker(t) (represented by L(U 1,U 2,,U k ) in the proof above),

4 14A. Rank of an Endomorphism 4 a subspace of the domain of dimension k, and Im(T) (represented by L(T(V 1 ),T(V 2 ),,T(V r )) above), a subspace of the codomain of dimension r; but there is also the subspace L(V 1,V 2,,V r ) of the domain of dimension r, which we call it the coimage of T, and the subspace L(W 1, W 2,,W k ) of the codomain of dimension k, which we call the cokernel of T. We observe that the domain space satisfies V = ker(t ) coim(t ), that the codomain space satisfies and that V = coker(t) Im(T ), dim(ker(t)) = dim(coker(t)), dim(im(t)) = dim(coim(t)). Further, the following properties characterize the four subspaces:

5 14A. Rank of an Endomorphism 5 cokernel: T avoids the cokernel, in the sense that there is no vector X in the domain for which T(X) coker(t ); kernel: when T is restricted to the kernel, it corresponds to the zero map; image: T sends all vectors of the domain space into Im(T); coimage: every vector Y in the image is the image of a unique vector X in the coimage (unique because: if X, X coim(t) satisfy T(X) = Y = T( X ), then X X coim(t) satisfies T(X X ) = 0, so X X ker(t), forcing X X = 0.) The theorem shows that bases can be chosen for the four subspaces which combine to form bases for the domain and codomain of T in order that the matrix of T relative to these bases is M = I r 0 ; 0 0 restricting T to the coimage subspace produces a transformation (which we also call T) that is one-toone onto the image subspace. That is, T: coim(t) Im(T)

6 14A. Rank of an Endomorphism 6 is an isomorphism of vector spaces, and is representable relative to certain bases of these two subspaces by the identity matrix I r. Note, however, that this restriction of T is not necessarily the identity function, since the coimage and image spaces are not necessarily the same subspace of V; rather, the most we can say is that this restriction map T: coim(t) Im(T) is an invertible transformation (i.e., an isomorphism). This very powerful theorem also allows us to give a straightforward proof of a result you learned in your first linear algebra course: Corollary The row rank of any matrix in Mat n,n (i.e., the number of row vectors of the matrix that are linearly independent in R n ) also equals its column rank (i.e., the number of column vectors of the matrix that are linearly independent in R n ). Proof Suppose A Mat n,n is the given matrix. Let T:R n R n be the linear transformation represented in terms of the standard basis for R n by the matrix A. That is, T(X) = AX. Then the column rank of A, being the number of column vectors of the matrix that are linearly independent,

7 14A. Rank of an Endomorphism 7 is precisely the number of linearly independent vectors amongst T(E 1 ),T(E 2 ),,T(E n ), as these are precisely the column vectors of A. Since Im(T) = L(T(E 1 ), T(E 2 ),, T(E n )), there must be exactly r = dim(im(t )) such vectors, where by the theorem, r is the rank of T. So the column rank of A equals the rank of T. Next, let S:R n R n be the linear transformation represented in terms of the standard basis for R n by the matrix A tr. Then the column rank of A tr, being the number of column vectors of the matrix that are linearly independent, is the same as the row rank of A. By the same argument as above, there are dim(im(s)) such vectors, equal to the rank of S. However, the theorem concludes that T is represented with respect to some pair of bases for R n by the matrix M = I r 0 ; 0 0 this means that there are invertible change-of-basis matrices P and Q in Mat n,n for which A = PMQ 1. Since A tr = (PMQ 1 ) tr = (Q 1 ) tr M tr P tr, and both matrices (Q 1 ) tr and P tr are invertible (with

8 14A. Rank of an Endomorphism 8 inverses Q tr and (P 1 ) tr, respectively), S is represented by the matrix M tr with respect to some pair of bases for R n. But M tr = M, so working with respect to these bases, we find that dim(im(s)) = r. So the row rank of A equals r as well, completing the proof. //