Why is the geometric multiplicity of an eigenvalue less than
or equal to its algebraic multiplicity?
Let A be an n × n matrix, and let λ be an eigenvalue of A. Let k denote the dimension of the
eigenspace for eigenvalue λ, and let { v1 , v2 , . . . , vk } be a basis for this eigenspace. Consider any
basis for Rn that contains v1 , v2 , . . . , vk , say { v1 , v2 , . . . , vk , u1 , u2 , . . . , un−k }. Let P denote the
n × n matrix whose columns are, respectively, v1 , v2 , . . . , vk , u1 , u2 , . . . , un−k . Then P has rank n
and so P is invertible. Since P −1 AP − xIn = P −1 AP − xP −1 In P = P −1 (A − xIn )P , we have
det(P −1 AP − xIn ) = det(P −1 (A − xIn )P ) = det(P −1 ) det(A − xIn ) det(P )
= det(P −1 ) det(P ) det(A − xIn ) = det(P −1 P ) det(A − xIn )
= det(In ) det(A − xIn ) = det(A − xIn ),
and thus the characteristic polynomial of P −1 AP is exactly the same as the characteristic polynomial
of A. We are trying to show that the power of (x − λ) in the characteristic polynomial of A is at
least k, and this requires that (x − λ)k be a factor of det(P −1 AP − xIn ). Consider
AP = [Acol1 (P )|Acol2 (P )| · · · |Acolk (P )|colk+1 (AP )| · · · |coln (AP )]
= [λcol1 (P )|λcol2 (P )| · · · |λcolk (P )|B]
where B = [colk+1 (AP )| · · · |coln (AP )] is an n × (n − k) matrix. Thus
P −1 AP = [λP −1 col1 (P )| · · · |λP −1 colk (P )|P −1 B] = [λe1 |λe2 | · · · |λek |P −1 B]
and so






−1
P AP − xIn = 




λ−x
0
0
···
0
0
..
.
λ−x
..
.
0
···
0
..
.
0
0
···
O

*
λ−x
C − xIn−k










where C is the lower (n − k) × (n − k) block of P −1 AP . Thus expanding along the first column for
the first k steps, we have
det(P −1 AP ) = (λ − x)k det(C − xIn−k ).
Thus (x − λ)k is a factor of det(P −1 AP − xIn ) = det(A − xIn ), and so the algebraic multiplicity of
λ; that is, the power to which x − λ appears in det(A − xIn ), is at least as large as k, the geometric
multiplicity of λ.