math111_logo Eigenspace and Multiplicity

3. Multiplicity and diagonalization

For an n by n matrix A, we first find eigenvalues λ1, λ2, ..., λk and then a basis Bi = {vi1, vi2, ..., vimiG} for each eigenspace Eλ = nul(A - λI). Since Eλ1 + Eλ2 + ... + Eλk is a direct sum, by this result, the collection B = B1B2∪...∪Bk = {vij: all i, j} of all the eigenvectors we computed is linearly independent. The number of vectors in B is the sum m1G + m2G + ... + mkG of geometric multiplicities.

The diagonalizability of A is equivalent to whether B is a basis of Rn. Since B is already linearly independent, by this result, we have the following condition for diagonalizability.

An n by n matrix is diagonalizable ⇔ The sum of geometric multiplicities is n.

We already know the sum of algebraic multiplicities must be n. In view of the statement above, it is natural to ask how the algebraic and the geometric multiplicities are related.

The algebraic multiplicity is no less than the geometric multiplicity: miAmiG.

Proof Let us consider the first eigenvalue λ1 of A and abbreviate its multiplicities: m = m1G, m' = m1A. Then we have m linearly independent eigenvectors B1 = {v1, v2, ..., vm} with eigenvalue λ1. Extend the vectors to a basis {v1, v2, ..., vm, wm+1, ..., wn} of Rn. Then

P = [v1 v2 ... vm wm+1 ... wn] = [P1 P2]

(P1 is the first m columns, P2 is the last n - m columns) is an invertible matrix. We partition P-1AP accordingly (m and n - m columns, also m and n - m rows):

P-1AP = [ B11 B12 ].
B21 B22

Comparing

AP = [λ1v1 λ1v2 ... λ1vm Awm+1 ... Awn] = [λ1P1 AP2]

with

P[ B11 B12 ] = [P1B11 + P2B21 P1B12 + P2B22],
B21 B22

we get B11 = λ1I, B21 = O. Thus

P-1AP = [ λ1I B12 ]
O B22

and (note that λ1I is an m by m matrix)

det(A - λI) = det(P-1AP - λI) = det[ (λ1 - λ)I B12 ] = (λ1 - λ)m det(B22 - λI).
O B22 - λI

In particular, λ1 appears as a root of the characteristic polynomial for at least m times. Since the algebraic multiplicity m' is the total number of times λ1 appears as a root, we conclude that m'm.

Based on m1A + m2A + ... + mkA = n and miAmiG, we conclude that m1G + m2G + ... + mkG = nmiA = miG. Thus the earlier result on the diagonalizability has the following reformulation.

A square matrix is diagonalizable ⇔ Algebraic and geometric multiplicities are equal.


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