Let M be an n×n matrix of complex elements. A complex number λ and vector X, other than the zero vector 0, such that

MX = λX

are called, repectively, an eigenvalue and eigenvector of M. The set of all eigenvectors associated with λ is called the eigenspace associated with that eigenvalue. Let the eigenspace of λ be denoted as Λ_λ. Let V_λ be the null space of the matrix (M-λI), where I is the n×n identity matrix. V_λ is a vector subspace of the n dimensional vector space. V_λ is just the eigenspace of λ with the zero vector 0 adjoined; i.e.,

V_λ = Λ_λ∪{0}

The dimension of V_λ is equal to the multiplicity of λ as a root of the n-th degree polynomial equation

det(M-λI) = 0

Let {λ_j; j=1, …,m} be the distinct eigenvalues of M with {Λ_λj; j=1, …,m} as their associated eigenspaces. Then {Λ_λj; j=1, …,m} are linearly independent.

Proof:

Let X_j be any element from Λ_λj for j=1, …,m. Suppose {X_j; j=1, …,m} are linear dependent. This would mean that there exists coefficients {c_j; j=1, …,m}, not all zero, such that

Σc_jX_j = 0

where the summation is from j=1 to m. Some of the coefficients, but not all, might be zero. Let q be the smallest number of linearly dependent elements of the set {X_j; j=1, …,m} and let the vectors be numbered such that the linearly dependent vectors are numbered 1 through q. Then

Σc_jX_j =0
for j=1 to q

Now multiply this equation through by M. The result is

Σc_jλ_jX_j =0
for j=1 to q

Now multiply the prior equation by any of the λ's, say λ₁ to obtain

Σc_jλ₁X_j =0
for j=1 to q

This equation may be subtracted from the previous equation to obtain

Σc_j(λ_j−λ₁)X_j =0

where now the summation runs from j=2 to q, since the term for j=1 cancels out.

None of the terms c_j(λ_j−λ₁) are zero so this means there is a contradiction of the assumption that the set {X_j; j=1, …,q} had the least number of dependent variables. Therefore there is no set of linearly dependent vectors among the vectors chosen from the distinct eigenspaces.