Linear Algebra

Let V be an n-dimensional vector space over a field F, and let T be a linear transformation of V to itself. Let us say a vector v in V is cyclic for T if {v,Tv,T²v,...,T^n-1v} is a basis of V. I put it to you that if T has a cyclic vector, then the only other linear transformations with which T commutes are polynomials in T.

Let S be a linear transformation with ST=TS. Since {v,Tv,T²v,...,T^n-1v} is a basis for V, whatever Sv might be, we can expand it as Sv=a₀v+a₁Tv+...+a_n-1T^n-1v=(a₀+a₁T+...+a_n-1T^n-1)v=p(T)v, where p(x) is the polynomial a₀+a₁x+...+a_n-1x^n-1. Now let us compute the matrix for S in our cyclic basis: the i^th column of this matrix S' is the image of the i^th basis vector under S.

S'=[Sv|S(Tv)|S(T²v)|...|S(T^n-1v)]
=[Sv|T(Sv)|T²(Sv)|...|T^n-1(Sv)]
=[p(T)v|Tp(T)v|T²p(T)v|...|T^n-1p(T)v]
=[p(T)v|p(T)(Tv)|p(T)(T²v)|...|p(T)(T^n-1v)]

S', the matrix for S in our basis, is thus, we see, precisely the matrix for the linear transformation p(T) in our basis. Therefore, S=p(T): S is a polynomial in T.

This cyclic basis puts T into rational canonical form, with a single block: the matrix for T in this basis is [Tv|T(Tv)|...|T(T^n-1v)] or

0	0	...	0	?
1	0	...	0	!
0	1	...	0	*
.	.	...	.	.
0	0	...	1	#

Conversely, suppose M is any linear transformation whose matrix is in rational canonical form with a single block. Then there is a basis {e₁,e₂,...,e_n} of V such that the matrix of M in this basis, which is [Me₁|Me₂|...|Me_n] is precisely of the above form: thus Me₁=e₂, Me₂=e₃, and so forth down the line to Me_n-1=e_n. Thus we have a basis of V of the form {e₁,Me₁,...,M^n-1e₁}, so e₁ is cyclic for M. Thus a linear transformation has a cyclic vector if and only if its matrix in rational canonical form consists of a single block.

Now suppose does not have a cyclic vector: then, in rational canonical form, T consists of at least two blocks. Let S be a transformation with a block diagonal matrix in this basis with blocks the same sizes as those of T. Then the product of the matrices is the block diagonal matrix whose blocks are the products of the corresponding blocks of S and T, so S and T commute if and only if each block in S is a polynomial in the corresponding block of T (since each rational canonical block is cyclic). If T contains any zero blocks, let the corresponding blocks of S be diagonal with all entries nonzero and distinct, and let the other blocks be zero. ST=TS=0 so S and T commute, but S is not a polynomial in T, since any such would have all its diagonal entries in T's zero blocks equal. If not...

This may be wrong; I just came up with it. Let A and B be the final two blocks of T; then the minimal polynomial of A divides the minimal polynomial of B, which is the minimal polynomial of T. Let S be block diagonal, with all its blocks zero but the last two, which are A and B+tI for some scalar t. ST=TS, but I put it to you that S is not a polynomial in T. Suppose that it were: q(T)=S. Then q(A)=A, and q(B)=B+tI. Then A satisfies the polynomial q(x)-x, and B satisfies the polynomial q(x)-x-t. Therefore, the minimal polynomial of A must divide q(x)-x, and must also divide the minimal polynomial for B, which divides in turn q(x)-x-t. Choose t so that q(x)-x and q(x)-x-t are relatively prime; we reach a contradiction. Therefore T has a cyclic vector iff the only operators it commutes with are polynomials in T.

Posted by aloysius at July 17, 2003 01:31 PM | TrackBack |