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The Linear Independence of the Eigenspaces of a Matrix 

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
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 (MλI), 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.,
The dimension of V_{λ} is equal to the multiplicity of λ as a root of the nth degree polynomial equation
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
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
Now multiply this equation through by M. The result is
Now multiply the prior equation by any of the λ's, say λ_{1} to obtain
This equation may be subtracted from the previous equation to obtain
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}−λ_{1}) 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.
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