Diagonalizability: Theorems Part 2

DIAGONALIZABILITY

THEOREMS PART 2

Theorem 16

Let T be a linear operator, and let λ₁, λ₂ ... λ_k be distinct eigenvalues of T. For each i = 1,2,3... k, let v_i ∈ E_λi, the eigen space corresponding to λ_i. If v₁+v₂+v₃+...+v_n = 0 then v_i= 0 for all i.

Proof:

Suppose otherwise. By renumbering if necessary, suppose that, for 1≤m≤k, we have v_i ≠ 0 for 1 ≤i≤m and v_i = 0 for i>m. Then, for each i≤m, v_i is an eigenvector of T corresponding to λ_i and v₁+v₂ + v₃ + …..+ v_n = 0. But this is a contradiction which states that these v_í's are linearly independent. We conclude therefore, that v_i = 0 for all i.

Theorem 17

Let T be a linear operator on a vector space V, and let λ₁, λ₂, λ₃ ... λ_k be distinct eigenvalues of T. For each i=1, 2, 3,... k, let S_i be a finite linearly independent subset of the eigen space E_λi. Then S=S₁∪S₂∪S₃ ... ∪S_k is a linearly independent subset of V.

Proof:

  Suppose that for each i, S_i = {v_i1, v_i2, v_i3, ... v_ini }. Then S = { v_ij ; 1 ≤ j ≤n_i; and 1≤i≤k}.

Consider any scalars {a_ij} such that 

For each i, let w_i =  a_jj v_ij

Then we E_i ∈ E_λi for each i and w₁+w₂ + w₃ + ... + w_k = 0.

Therefore, w_i = 0 for all i. But each S_i is linearly independent, and a_ij=0 for all j.

. S is linearly independent.

Theorem 18

Let T be a linear operator on a finite dimensional vector space V such that the characteristic polynomial of T splits. Let λ₁, λ₂, λ₃ ... λ_k be the distinct eigenvalues of T. Then

(a) T is diagonalizable if and only if the multiplicity of λ_i is equal to dim (E_λi) for all i.

(b) If T is diagonalizable and β_i is an ordered basis for E_λi for each i, then β = β₁∪β₂ ∪ β₃ ... ∪ β_k is an ordered basis for V consisting of eigenvectors of T.

Proof:

For each i, let m_i denote the multiplicity of λ_i, d_i = dim (E_λi) and n = dim (V).

Suppose that T is diagonalizable.

Let β be a basis for V consisting of eigenvectors of T. For each i, let β_i=β∪ E_λi, the set of vectors in β that are eigenvectors corresponding to λ_i, and let n_i denote the number of vectors in β_i. Then n_i ≤ d_i for each i because β_i is a linearly independent subset of a subspace of dimension d_i, and d_i≤m_¡. The n_i's sum to n because β contains n vectors. The m_i's also sum to n because the degree of the characteristic polynomial T is equal to the sum of the multiplicities of the eigenvalues.

Thus n = n_i ≤ d_i ≤ m_i = n_i

(m_i‒d_i) = 0

Since (m_i ‒ d_i) ≥ 0 for all i, we conclude that m_i = d_i for all i.

Conversely, suppose that m_i = d_i for all i. T is diagonalizable. For each i, let β_i be an ordered basis for E_λi and let β= β₁ ∪ β₂ ∪ β₃ ….. ∪ B_k. Then β is linearly independent. Since d_i = m_i for all i, β contains

 di = mi=n

 β is an ordered basis for V consisting of eigenvectors of V and we conclude that T is diagonalizable.

Test for diagonalization

Let T be a linear operator on an n−dimensional vector space V, then T is diagonalizable if both of the following conditions satisfy.

(i) The characteristic polynomial of T splits.

(ii) For each eigenvalue λ of T, the multiplicity of λ equals

  n− [rank (T−λI)]