and can the composition of a one-to-one linear transformation and a linear transformation that is not one-to-one bo one-to-one? Justify your answer
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1. Yes. Let S and T be one-to-one linear transformations. Suppose S(T(x))=0. Since S is one-to-one, this implies T(x)=0. Since T is one-to-one, this in turn implies x=0. Therefore S o T has trivial kernel, so it is one-to-one.
2. It depends on the order you compose them. Suppose that S and T are linear transformations, and that T is not one-to-one. Then S o T cannot be one-to-one; any nonzero x in the kernel of T must also lie in the kernel of S o T. However, it is possible for T o S to be one-to-one. For example, define S:R-->R^2 and T:R^2-->R by S(x)=(x,0) and T(x,y)=x. Clearly T is not one-to-one, but T(S(x))=T(x,y)=x, so T o S is one-to-one.
2. It depends on the order you compose them. Suppose that S and T are linear transformations, and that T is not one-to-one. Then S o T cannot be one-to-one; any nonzero x in the kernel of T must also lie in the kernel of S o T. However, it is possible for T o S to be one-to-one. For example, define S:R-->R^2 and T:R^2-->R by S(x)=(x,0) and T(x,y)=x. Clearly T is not one-to-one, but T(S(x))=T(x,y)=x, so T o S is one-to-one.