applet-magic.com Thayer Watkins
Silicon Valley & Tornado Alley
U.S.A.

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

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_{j}X_{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_{j}X_{j} =0 for j=1 to q

Now multiply this equation through by M. The result is

Σc_{j}λ_{j}X_{j} =0 for j=1 to q

Now multiply the prior equation by any of the λ's, say λ_{1}
to obtain

Σc_{j}λ_{1}X_{j} =0 for j=1 to q

This equation may be subtracted from the previous equation to obtain

Σc_{j}(λ_{j}−λ_{1})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}−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.