For all z in R^m, since A is onto, there exists y in R^n s.t. Ay=z, and since B is onto, there exists x in R^r s.t. Bx=y. Hence, for all z in R^m, z=Ay=A(Bx)=Cx, 所以C=AB會是onto
對了, 你上面那個(a)小題也有些問題, 因為那個ker(A)不會是{0}, 且事實上那個命題應該是對的, C 會是one-to-one, 所以應該找不到反例, 你可以證看看
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For all z in R^m, since A is onto, there exists y in R^n s.t. Ay=z, and since B is onto, there exists x in R^r s.t. Bx=y. Hence, for all z in R^m, z=Ay=A(Bx)=Cx, 所以C=AB會是onto
對了, 你上面那個(a)小題也有些問題, 因為那個ker(A)不會是{0}, 且事實上那個命題應該是對的, C 會是one-to-one, 所以應該找不到反例, 你可以證看看
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