Let T be a linear operator on an n-dimensional vector space V such that I has n distinct
eigenvalues. Prove that V is a T-cyclic subspace of itsself.
Solution
Proof. If V is T-cyclic, then PT (X) = (1)
dim(V )mT (X). Since T is diagonal-izable,
mT (X) = (X 1)(X 2)....(X n);
where the i
\'s are the distinct eigenvalues of T. In other words, the algebraic
multiplicity mi of each eigenvalue i
is 1, and since the geometric multiplicity
of i respects the inequality
1 dim(ker (T- iI)) mi;
we see that dim(ker (T iI)) = 1 for all i.
Conversely, suppose that all eigenspaces have dimension 1. Since T is diag-onalizable, the
minimal polynomial splits as
mT (X) = (X 1)(X 2)....(X n)
and V decomposes asV = ker (T- 1I) ker (T- 2I) ... ker (T- nI)where the i\'s are the distinct
eigenvalues of T. Let vi be an eigenvector as-sociated to i for each i, and let v = v1 + ... + vn.
Since each eigenspace hasdimension 1, fvig is a basis for ker (T iI), and fv1; v2;....vng is a basis
for V .Now let c0; c1; ...; cn1 be scalars such thatc0v + c1T(v) + ... + cn1Tn1(v) = 0:From this
equation, we have0 = c0(v1 + v2 + .. + vn) + .... + cn1(n11v1 + n12v2... + n1nvn)= (c0 + c11 +
... + cn1n11)v1 + ... + (c0 + c1n + ... + cn1n1n)vn:By the linear independence of the vi\'s, each
iis a root of the polynomialg(X) = c0 + c1X + ... + cn1Xn1.
Since there are n distinct eigenvalues,the degree n 1 polynomial g(X) has n roots and is therefore
zero. Hence,c0 = c1 = .... = cn1 = 0, and fv; T(v); ....; Tn1(v)g is a basis for V .
Thus, theT-cyclic subspace generated by v is none other than V ..
Let T be a linear operator on an n-dimensional vector space V such t.pdf
1. Let T be a linear operator on an n-dimensional vector space V such that I has n distinct
eigenvalues. Prove that V is a T-cyclic subspace of itsself.
Solution
Proof. If V is T-cyclic, then PT (X) = (1)
dim(V )mT (X). Since T is diagonal-izable,
mT (X) = (X 1)(X 2)....(X n);
where the i
's are the distinct eigenvalues of T. In other words, the algebraic
multiplicity mi of each eigenvalue i
is 1, and since the geometric multiplicity
of i respects the inequality
1 dim(ker (T- iI)) mi;
we see that dim(ker (T iI)) = 1 for all i.
Conversely, suppose that all eigenspaces have dimension 1. Since T is diag-onalizable, the
minimal polynomial splits as
mT (X) = (X 1)(X 2)....(X n)
and V decomposes asV = ker (T- 1I) ker (T- 2I) ... ker (T- nI)where the i's are the distinct
eigenvalues of T. Let vi be an eigenvector as-sociated to i for each i, and let v = v1 + ... + vn.
Since each eigenspace hasdimension 1, fvig is a basis for ker (T iI), and fv1; v2;....vng is a basis
for V .Now let c0; c1; ...; cn1 be scalars such thatc0v + c1T(v) + ... + cn1Tn1(v) = 0:From this
equation, we have0 = c0(v1 + v2 + .. + vn) + .... + cn1(n11v1 + n12v2... + n1nvn)= (c0 + c11 +
... + cn1n11)v1 + ... + (c0 + c1n + ... + cn1n1n)vn:By the linear independence of the vi's, each
iis a root of the polynomialg(X) = c0 + c1X + ... + cn1Xn1.
Since there are n distinct eigenvalues,the degree n 1 polynomial g(X) has n roots and is therefore
zero. Hence,c0 = c1 = .... = cn1 = 0, and fv; T(v); ....; Tn1(v)g is a basis for V .
Thus, theT-cyclic subspace generated by v is none other than V .
