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2. Square Matrices
30
, which is nilpotent. Let us also write
Let us define N λ = M λ − λI n λ
,... ),
=
D
diag(... ,λI n λ
diag(... ,N λ ,... ),
N
=
−1
−1
D P, N = P
N P. The matrices D ,N are respec-
and then D = P
tively diagonal and nilpotent. Moreover, they commute with each other:
D N = N D . One deduces the following result.
Proposition 2.7.2 If K is algebraically closed, every matrix M ∈ M n (K)
decomposes as a sum M = D+N,where D is diagonalizable, N is nilpotent,
DN = ND,and Sp(D) = Sp(M).
Let us continue this analysis.
Lemma 2.7.1 Every nilpotent matrix is similar to a strictly upper
triangular matrix (and also to a strictly lower triangular one).
Proof
Let us consider the nondecreasing sequence of linear subspaces E k =
k
n
ker N .Since E 0 = {0} and E r = K for a suitable r, one can find a basis
1 n n 1 j
{x ,... ,x } of K such that {x ,... ,x } is a basis of E k if j =dim E k
(use the theorem that any linearly independent set can be enlarged to
j
a basis). Since N(E k+1 )= E k , Nx ∈ E k .If P is the change-of-basis
matrix from this basis to the canonical one, then PNP −1 is strictly upper
triangular.
Let us return to the decomposition PMP −1 = D + N above. Each N λ
can be written, from the lemma, in the form R −1 T λ R λ ,where T λ is strictly
λ
upper triangular. Then R λ (D λ + N λ )R −1 = D λ + T λ is triangular. Let us
λ
set
R = diag(... ,R λ ,... ).
Then (RP)M(RP) −1 is block-diagonal, with the diagonal blocks upper
triangular, and hence this matrix is itself upper triangular.
Theorem 2.7.1 If K is algebraically closed, then every square matrix is
similar to a triangular matrix (one says that it is trigonalizable).
More generally, if the characteristic polynomial of M ∈ M n(K) splits as
the product of linear factors, then M is trigonalizable.
A direct proof of this theorem that does not use the three previous
statements is possible. Its strategy is used in the proof of Theorem 3.1.3
2.8 Irreducibility
A square matrix A is said reducible if there exists a nontrivial partition
{1,... ,n} = I ∪ J such that (i, j) ∈ I × J implies a ij =0. It is irreducible