The algebraic multiplicity of an eigenvalue 𝜆 of a matrix M is the largest power of x-𝜆 that will divide the characteristic polynomial |xI-M|.
Let M be any n×n complex square matrix. It turns out that such a matrix will always have n eigenvalues. However, some of these eigenvalues may be repeated. Let us take the simple example of a diagonal matrix:D=a
5
5
5
7
7
D has two distinct eigenvalues: 5 and 7. The eigenvalue 5 is repeated thrice and the eigenvalue 7 is repeated twice. We say that the algebraic multiplicity of the eigenvalue 5 is three and that of 7 is two. The reason this is called the algebraic multiplicity has to do with the characteristic polynomial, an algebraic structure:a
|xI-D|
=(x-5)3(x-7)2
The exponents of x-5 is the algebraic multiplicity of 5. In general, any complex matrix M will have n eigenvalues, not necessarily distinct. Let the number of distinct eigenvalues be k. Let the ith eigenvalue and its algebraic multiplicity be 𝜆i and mi. Then, the characteristic polynomial of M can be written as:|xI-M|=k∏i=1(x-𝜆i)mi2. Geometric Multiplicity
The geometric multiplicity of an eigenvalue 𝜆 of a matrix M is the dimension of N(M-𝜆I), where N(.) is the nullspace (kernel) of the matrix.
N(M-𝜆I) is the eigenspace corresponding to 𝜆. The eigenspace is a subspace of Rn that contains all the eigenvectors of M corresponding to 𝜆. For example, consider the diagonal matrix D again:D=a
5
5
5
7
7
If ei=a
0
⋯
1⋯
0
T is the ith unit vector in the standard basis for R5, then the eigenspaces corresponding to the eigenvalues 5 and 7 are:span({e1,e2,e3}),span({e4,e5})The geometric multiplicity of the eigenvalues 5 and 7 are the dimensions of these two spaces, which are 3 and 2 respectively. The geometric multiplicity need not be equal to the algebraic multiplicity. Take the case of this matrix:M=a
3
2
0
3
Its eigenvalues are:a
|xI-M|
=(x-3)2
=0
x
=3,3
The eigenvector:a
0
2
0
0
a
x1
x2
=a
0
0
𝛼⋅a
1
0
is the only type of eigenvector. The eigenspace for this eigenvalue is span({a
1
0
}) and the geometric multiplicity is 1. The algebraic multiplicity on the other hand is equal to 2. For this case, we see that the geometric multiplicity is less than the algebraic multiplicity. We will see a stronger version of this statement in the next section.3. Inequality
For any eigenvalue 𝜆 of a matrix M, its geometric multiplicity is always less than or equal to its algebraic multiplicity.
Consider an eigenvalue 𝜆 with geometric multiplicity p. Then, we can find p linearly independent eigenvectors v1,⋯,vp with eigenvalue 𝜆 for M. Let us use this as the first p elements in a basis for Rn and call the resulting basis B2, with B1 being the standard basis. If T denotes the linear transformation corresponding to M, then we can represent the following matrices in terms of the transformation:M=[T]B1B1,P=[T]B1B2,L=[T]B2B2here, P is the change of basis matrix (from B2→B1) and L is the same transformation represented in the new basis. We have the following result:L=P-1MPL is similar to M and hence both have the same characteristic polynomial and the same eigenvalues. Studying L is the same as studying M. The eigenvector vi in basis B2 will have coordinates that correspond to those of the unit vector ei in the standard basis B1:L[vi]B2=𝜆[vi]B2=a
0
⋮
𝜆
⋮
0
(𝜆 is in position i)Therefore, the first p columns will have zeros in all but the diagonal positions. We can represent L as a block matrix:L=a
𝜆Ip×p
Xp×n-p
0n-p×p
Yn-p×n-p
Expanding the determinant along the first column, the characteristic polynomial will be of the form:|xI-L|=(x-𝜆)p|xI-Y|We see that the highest power of x-𝜆 is at least p. Hence, the algebraic multiplicity is at least as high as the geometric multiplicity.4. Diagonalizable Matrices
A complex matrix M is diagonalizable if and only if for every eigenvalue 𝜆 of M, its geometric and algebraic multiplicities are equal.
• M is diagonalizable⟹Multiplicities are equalLet 𝜆i have geometric and algebraic multiplicities of gi and ai respectively. Each eigenvalue can contribute at most gi linearly independent eigenvectors. If there are k distinct eigenvalues, then the total number of linearly independent eigenvectors available isk∑i=1giThe algebraic multiplicity is an upper bound on the geometric multiplicity:gi⩽ai, 1⩽i⩽kThe sum of the algebraic multiplicities is:k∑i=1ai=nUsing these three results:n∑i=1gi⩽nwhere equality occurs if and only if gi=ai. If M is diagonalizable, it has n linearly independent eigenvectors. Therefore, we have strict equality and hence gi=ai for 1⩽i⩽k.• Multiplicities are equal⟹M is diagonalizableAs before, let us assume that there are k distinct eigenvalues: Let the multiplicity of 𝜆i be mi:k∑i=1mi=nEach eigenvalue can contribute mi linearly independent eigenvectors. Let vji denote the jth contribution from the ith eigenvalue:v1i,⋯,vji,⋯,vmiiConsider the following set of n eigenvectors :B={a
v11,
⋯,
vm11,
v12,
⋯,
vm22,
⋮
v1k,
⋯,
vmkk
}It is clear that each row is a linearly independent collection by construction. It remains to see if the entire set is linearly independent. Let us consider any linear combination of these n vectors and set it to zero:k∑i=1mi∑j=1𝛼jivji=0Let us define the inner sum to be:wi=mi∑j=1𝛼jivjiIf wi≠0, then it is an eigenvector for eigenvalue 𝜆i. The entire sum can now be written as:k∑i=1wi=0Each element in the set {w1,⋯,wk} is either a zero-vector or an eigenvector. If every vector is a zero-vector, then we are done. If not, let us throw away the zero vectors and only consider the subset of eigenvectors. This subset is a collection of eigenvectors corresponding to distinct eigenvalues and hence linearly independent. The sum of these vectors cannot be zero. Therefore, every vector wi=0, 1⩽i⩽k. This means:mi∑j=1𝛼jivji=0, 1⩽i⩽kWe already know that the set of vectors {v1i,⋯,vmii} is linearly independent. Hence 𝛼ji=0, 1⩽i⩽k, 1⩽j⩽mi, showing that B is a linearly independent collection of n eigenvectors of M. It follows that M is diagonalizable.5. References• Reference-1, for the inequality• Reference-2, for the result on diagonalizability• Reference-3, an example for a defective matrix