Question

10 9. Let U be a finite-dimensional vector space and TE LU). Prove the following statements. (a) (5 pts) Let λ be an eigenvalue T whose geometric multiplicity is m, and algebraic multiplicity is ma. Then (b) (5 pts) Let u be a cyclic vector of T of period k 2 2 (such that T*(u) 0 but T-(u) 0). Then are linearly independent.

Answer 1

Part-A:

Let $\lambda$ be an eigen value of with algebraic multiplicity $m_a$ and geometric multiplicity
mg
.

Let be a basis with respect to which $\lambda$ is an eigen value of .

Let $E_\lambda$ be the eigen space associated with $\lambda$.

Since dimension of $E_\lambda$ is
mg
there are
mg
linearly independent eigen vectors of associated with $\lambda$.

Let the linearly independent eigen vectors be denoted by

UTiq

Since they are linearly independent so they can be extended to form another basis of say $B^{'}$.

Surely B is similar to $B^{'}$ i.e there exist non-singular matrices such that

SB S

Since characteristic polynomial of a matrix does not change when basis is similar we find  $\lambda$ forms an eigen value of with respect to $B^{'}$ with algebraic multiplicity at least $m_a$.

Since $B^{'}$ has
UTiq
as basis vectors so $m_g\le m_a$

Part-B:

Assume that

-------------(2)

k-1

Since

$T^{k}(u)=0\\ \implies T^{k+p}(u)=0\forall p\ge 1$-------------(1)

Applying $T^{k-1}$ on (2) we get

--------(Using (1))

k-1 k-1

Since

Tk-1 (11)メ0

$c_0T^{k-1}(u)=0\implies c_0=0$-------(3)

Again applying $T^{k-2}$ on (2) we get

$T^{k-2}(c_0u+c_1T(u)+c_2T^2(u)+\cdots +c_{k-1}T^{k-1}(u))=0\\ \implies c_1T^{k-1}(u)=0$------(Using (3))

Since

Tk-1 (11)メ0

$c_1T^{k-1}(u)=0\implies c_1=0$-------(4)

Similarly we can obtain that $c_i=0\forall i\ge 0$

Thus the set is linearly independent