Page 45 - Matrices theory and applications

2. Square Matrices
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2.6 Diagonalization
If λ ∈ K is an eigenvalue of M, one calls the linear subspace E K (λ)=
n
ker(M − λI n )in K the eigenspace associated to λ.It isformed of eigen-
vectors associated to λ on the one hand, and of the zero vector on the other
hand. Its dimension is nonzero. If L is a field containing K (an “extension”
of K), then dim K E K (λ)=dim L E L (λ). This equality is not obvious. It
follows from the third canonical form with Jordan blocks, which we shall
see in Section 6.3.3.
If λ 1 ,... ,λ r are distinct eigenvalues, then the eigenspaces are in direct
sum. That is,
(x 1 ∈ E K (λ 1 ),... ,x r ∈ E K (λ r ),x 1 +···+x r =0) =⇒ (x 1 = ··· = x r =0).
As a matter of fact, if there existed a relation x 1 + ··· + x s =0 where
x 1 ,... ,x s did not vanish simultaneously (we say that it has length s), one
could choose such a relation of minimal length r. One then would have
r ≥ 2. Multiplying this relation by M − λ r I n , one would obtain
(λ 1 − λ r )x 1 + ··· +(λ r−1 − λ r )x r−1 =0,
which is a nontrivial relation of length r − 1 for the vectors (λ j − λ r )x j ∈
E K (λ j ). This contradicts the minimality of r.
If all the eigenvalues of M are in K and if the algebraic and geometric
multiplicities coincide for each eigenvalue of M, the sum of the dimensions
of the eigenspaces equals n. Since these linear subspaces are in direct sum,
one deduces that
n
K = E(λ 1 ) ⊕ ··· ⊕ E(λ r ).
Thus one may choose a basis of K n formed of eigenvectors. If P is the
change-of-basis matrix from the canonical basis to the new one, then
M = P −1 MP is diagonal, and its diagonal terms are the eigenvalues,

repeted with their multiplicities. One says that M is diagonalizable in K.
A particular case is that in which the eigenvalues of M are in K and are
simple.
Conversely, if M is similar, in M n(K), to a diagonal matrix M =

P −1 MP,then P is a change-of-basis matrix from the canonical basis to
an eigenbasis (that is, a basis composed of eigenvectors) of M. Hence, M
is diagonalizable if and only if the algebraic and geometric multiplicities of
each eigenvalue coincide.
Two obstacles could prevent M from being diagonalizable in K.The
first one is that an eigenvalue of M does not belong to K.One can always
overcome this difficulty by moving towards M n (K). Thesecond oneis more
serious: In K, the geometric multiplicity of an eigenvalue can be strictly
less than its algebraic multiplicity. For instance, a triangular matrix whose
diagonal vanishes has only one eigenvalue, zero, of algebraic multiplicity
n. Such a matrix is nilpotent. However it is diagonalizable only if it is 0 n ,