math111_logo Eigenspace and Multiplicity

1. Direct sum of eigenspaces

The diagonalization process is actually the process of computing the eigenspaces Eλ = nul(A - λI). The following result says that the different eigenspaces are independent. See here for the definition of direct sum and the relation to linear independence.

Let λ1, λ2, ..., λk be distinct eigenvalues of A. Then the sum Eλ1 + Eλ2 + ... + Eλk of the corresponding eigenspaces is a direct sum.

Proof We prove by induction. Let us assume the statement is true for k - 1 eigenspaces. We prove the statement is still true for k eigenspaces. Specifically, for v1Eλ1, v2Eλ2, ..., vkEλk satisfying

v1 + v2 + ... + vk = 0,

we need to prove v1 = v2 = ... = vk = 0.

Multiplying λ1 and applying A to the equality, respectively, we get

λ1v1 + λ1v2 + ... + λ1vk = 0, λ1v1 + λ2v2 + ... + λkvk = 0.

Substracting the two equalities, we get

(λ1 - λ2)v2 + ... + (λ1 - λk)vk = 0.

Since the eigenspaces are subspaces, we have v'2 = (λ1 - λ2)v2Eλ2, ..., v'k = (λ1 - λk)vkEλk. By the inductive assumption, Eλ2 + ... + Eλk is already a direct sum. Thus from

v'2 + ... + v'k = (λ1 - λ2)v2 + (λ1 - λk)vk = 0,

we have v'2 = ... = v'k = 0. Since λ1, λ2, ..., λk are distinct, this further implies v2 = ... = vk = 0. Finally, from v1 + v2 + ... + vk = 0, we have v1 = 0.

By this property of direct sum, the result above has the following consequence (Bi consisting of one vector).

Eigenvectors with distinct eigenvalues must be linearly independent.

Recall that for an n by n matrix A, det(A - λI) is a polynomial of degree n. If the polynomial has no multiple root, then the n distinct roots are n distinct eigenvalues λ1, λ2, ..., λn. Then each eigenvalue λi must yield at least one eigenvector vi. Thus we have n eigenvectors v1, v2, ..., vnRn with distinct eigenvalues, which must be linearly independent and form a basis of Rn. The discussion leads to the following conclusion.

The characteristic polynomial has no multiple roots ⇒ The matrix is diagonalizable.

In this example, once we know the eigenvalues of the 2 by 2 matrix are 5 and 15, we may conclude, without further computation, that the matrix has a basis of eigenvectors.

Example In an earlier example, we know the eigenvalues of A must be from -5, -3, -1, 1, 3. Moreover, we also know that the eigenvalues must include -1, at least one from 1, -3, and at least one from 3, -5. Thus if A is 3 by 3, then A must have three distinct eigenvalues and must be diagonalizable.

Example The following lower triangular matrix

[ 2.3 0 0 0 ]
2 π 0 0
1.7 100 π2 0
π2 5 0.01 11

has 4 distinct eigenvalues 2.3, π, π2, 11 and has a basis of eigenvectors.


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