By Wu X.

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The matrix (k 0 1 — T) is therefore singular. 64) with k = k o , we are forced to conclude that T=k oI. Conversely, it is obvious that every matrix of the form kI (k commutes with all the A(x). E K) Corollary to Schur's Lemma. Let A (x) be an irreducible matrix representation of G over an algebraically closed field. Then the only matrices which commute with all the matrices A(x) (x E G) are the scalar multiples of the unit matrix. 8. The commutant (endomorphism) algebra It would be interesting to know whether the converse of the corollary to Schur's Lemma is true, that is whether the fact that only the scalar multiples of the unit matrix commute with A(x) implies that A(x) is irreducible.

X 2 , x ', x ° (=1), x, x 2 ,... - - be the infinite cyclic group generated by x. ) is a representation of G, because, as is easily verified, A(x h )A(x k )=A(x h+k ). But when h # 0, it is impossible to transform A(x h ) into diagonal form; for both latent roots of A (x h ) are equal to unity so that the diagonal form would have to be diag(1, 1)= I, which in turn would lead to the absurd conclusion that A (x h equals I. However, it is a remarkable fact that the diagonal form is attainable under very general conditions, which will be satisfied in all cases henceforth considered in this book.

Let us formulate this result in the language of module theory. Suppose the G-module V affords the reducible representation A (x). Then V possesses a proper submodule. 53). In our previous discussion (p. 14) the interpretation of D(x) required the use of V/U. 30) are equal to zero so that s ' W hX = (C L dhk(x)wk (h = 1, 2, . . , s). k =1 This means that W=[wl,w2, ... ,ws i is a G-module which affords the representation D(x). In these circumstances it is customary to write V =UO+ W and to say that V is the direct sum of the G-modules U and W.