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2. Square Matrices
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                                                        , 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
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