6.6. EXERCISES 147

Theorem 6.6.1 Let A ∈ L (V,V ) and suppose (ui,λ i) , i = 1,2, · · · ,m are eigen-pairs such that if i ̸= j, then λ i ̸= λ j. Then {u1, · · · ,um} is linearly independent.In words, eigenvectors from distinct eigenvalues are linearly independent.

Hint: Suppose ∑ki=1 ciui = 0 where k is as small as possible such that not all of the

ci = 0. Then ck ̸= 0. Explain why k > 1 and

k

∑i=1

ciλ kui =k

∑i=1

ciλ iui

Nowk

∑i=1

ci (λ k−λ i)ui = 0

Obtain a contradiction of some sort at this point. Thus if the n× n matrix has ndistinct eigenvalues, then the corresponding eigenvectors will be a linearly indepen-dent set and so the matrix will be diagonal and all the Jordan blocks will be singlenumbers.

13. This and the next few problems will give another presentation of the Jordan canon-ical form. Let A ∈L (V,V ) be a nonzero linear transformation where V has finitedimensions. Consider {

x,Ax,A2x, · · · ,Am−1x}

where for k ≤ m−1,Akx is not in,

Akx /∈ span(

x,Ax,A2x, · · · ,Ak−1x)

show that then{

x,Ax,A2x, · · · ,Am−1x}

must be linearly independent. Hint: Letη (λ ) be the minimum polynomial for A. Then let φ (λ ) be the monic polynomial ofsmallest degree such that φ (A)x = 0. Explain why φ (λ ) divides η (λ ). Then showthat if the degree of φ (λ ) is d, then

{x,Ax,A2x, · · · ,Ad−1x

}is linearly independent

and if k is as described above, then k ≤ d. Note: linear dependence implies theexistence of a polynomial ψ (λ ) such that ψ (A)x = 0. An ordered set of vectors ofthe form x,Ax,A2x, · · · ,Am−1x where Amx ∈ span

(x,Ax,A2x, · · · ,Am−1x

)with m as

small as possible is called a cyclic set.

14. ↑Suppose now that N ∈L (V,V ) for V a finite dimensional vector space and the min-imum polynomial for N is λ

p. In other words, N is nilpotent, N p = 0 and p as smallas possible. For x ̸= 0, let β x =

{x,Nx,N2x, · · · ,Nm−1x

}where we keep the order of

these vectors in β x and here m is such that Nmx ∈ span(x,Nx,N2x, · · · ,Nm−1x

)with

m as small as possible.

(a) Show that Nmx = 0. Hint: You know from the assumption that

Nmx ∈ span(x,Nx,N2x, · · · ,Nm−1x

)that there is a monic polynomial η (λ ) of degree m such that η (N)x = 0. Ex-plain why η (λ ) divides the minimum polynomial λ

p. Then η (λ ) = λm. Thus

Nmx = 0.