Diagonalizability: Theorems Part 1

DIAGONALIZABILITY

THEOREMS PART 1

Theorem 12

Let T be a linear operator on a vector space V and let λ₁, λ₂, ... λ_k be distinct eigenvalues of T. If v₁, v₂, v₃, ... v_k are eigenvectors of T such that λ_i corresponds to v_i; i=1 to k then {v₁, v₂, … v_k} is linear independent.

Proof:

The proof is by mathematical induction on k, suppose that on k = 1, then v ≠ 0 since v₁ is an eigenvector corresponding to λ₁.

Hence {v₁} is linearly independent.

Now assume that the theorem holds for k−1 distinct eigenvalues.

(ie). { v₁, v₂, ... v_k−1} is linearly independent.

Now we have to prove that

{v₁, v₂, ... v_k } is linearly independent.

Let a₁v₁ + a₂v₂ + …. + a₂v₂ = 0

                   ……………(1)

where a_i's are scalars.

Applying T−λ_kI on both sides

 (T−λ_kI) (α₁v₁ + a₂v₂ + ... + a_k−1v_k−1 + a_kv_k) = (T−λ_kI) (0)

 (T − λ_kI) (a₁v₁) + (T − λ_kI)(a₂v₂) + ... + (T − λ_kI)(a_k−1v_k−1) + (T−λ_kI)(a_kv_k) = 0

 T(a₁v₁) − λ_kα₁v₁ + T(a₂v₂) − λ_ka₂ v₂ +...

 a₁T (v₁) − λ_ka₁v₁ + a₂T(v₂) − λ_ka₂v₂ + ...

 a_k−1T(v_k−1) − λ_ka_k−1 ‒ V_k−1+ a_kT(v_k) − λ_ka_kv_k = 0

 a₁λ₁v₁ − λ_ka₁v₁ + a₂λ₂v₂ − λ_ka₂v₂ + a_k−1λ_k−1v_k−1 − λ_ka_k−1v_{k – 1} + a_kλ_kv_k − λ_ka_kv_k = 0

 a₁v₁(λ₁− λ_k) + a₂v₂(λ₂− λ_k) + ... + a_k−1v_k−1(λ_k−1 − λ_k) = 0

by the induction hypothesis { v₁, v₂, ... v_k−1) is linearly independent and hence

 a₁ (λ₁− λ_k) = a₂ (λ₂ − λ_k) = ... = a_k−1(λ_k−1 − λ_k) = 0

Since λ₁, λ₂ ... λ_k are distinct

 a₁=a₂= ….. = a_k−1 = 0

From (1) ⇒  a_kv_k=0

 a_k=0

 {v₁, v₂, ... v_k−1, v_k} is linearly independent.

Theorem 13

Let T be a linear operator on a n−dimensional vector space V if T has n−distinct eigenvalues then T is diagonalizable.

Proof:

Suppose that T has n distinct eigenvalues λ₁, λ₂, λ₃ ... λ_n. For each i, choose an eigenvector v_i corresponding to λ_i. Then we know that {v₁, v₂, v₃ ... v_n} is linearly independent, and since dim (V)=n, this set is a basis for V. Thus T is diagonalizable.

Note

If the matrix is diagonalizable, the eigenvalues need not be distinct.

(ie) The converse of theorem is not true.

Definition:

A polynomial f(t) in P(F) splits over F if there are scalars c, a₁, a₂, ... a_n (not necessarily distinct) in F such that

   f(t) = c(t−a₁) (t−a₂) ... (t−a_n).

Theorem 14

The characteristic polynomial of any diagonalizable matrix linear operator on a vector space V over a field F splits over F.

Proof:

Let T be a diagonalizable linear operator on the n−dimensional vector space V and let β be an ordered basis for V such that

 [T]_β = D where D is the diagonal matrix such that

 The characteristic polynomial is |D−tI|=0.

  f(t) = | D−tI | 

 | D−tI | = 0

 (λ₁ − t) (λ₂ − t) ... (λ_n − t) = 0

 T is a diagonalizable linear operator on an n−dimensional vector space that fails to have distinct eigenvalues, then the characteristic polynomial of T must have repeated zeros.

The converse of the theorem is false. The characteristic polynomial of T may split, but T need not be diagonalizable.

Definition:

Algebraic multiplicity: Let λ be an eigenvalue of a linear operator or matrix with characteristic polynomial f(t). The algebraic multiplicity of λ is the largest positive integer k for which (t−λ)^k is a factor of f(t).

Definition:

Eigenspace: Let T be a linear operator on a vector space V and let λ be an eigenvalue of T.

 Define E_λ = { x ∈ V; T(x) = λx) = N(T−λI_V).

The set E_λ is called the eigenspace of T, corresponding to the eigenvalue λ. The eigenspace of a square matrix A corresponding to the eigenvalue λ to be the eigenspace of L_A corresponding to λ.

Theorem 15

Let T be a linear operator on a finite dimensional vector space V, and let λ be an eigenvalue of T having multiplicity m, then 1 ≤ dim (E_λ) ≤ m.

Proof:

Choose an ordered basis { v₁, v₂, v₃ ... v_n} for E_λ, extend it to an ordered basis β = { v₁, v₂, v₃ ... v_p, v_p+1 ... v_n } for V. Let A = [T]_β. Then v_i(1≤ i ≤p) is an eigenvector of T corresponding to λ and therefore,

The characteristic polynomial of T is

 f(t) = |A − tI_n| = 

= |(λ − t) I_p| |C − tI_n−p|

= (λ−t)^Pg(t).

where g(t) is a polynomial. Thus (λ−t)^P is a factor of f(t), hence the multiplicity of λ is atleast p.

But dim (E_λ) = p and so dim (E_λ) ≤ m.