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Let $X,Y,Z$ be vector spaces and let $S:X\to Y,T:X\to Z$ be linear transformations. I have got in a book that if ker$S\subset $ ker$T$, then there exists a linear transformation $R:Y\to Z$ uniquely determined over $S(X)$ such that $R\circ S=T$. I couold not prove this. Can anyone please suggest me anything?

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1 Answer 1

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We want $R\circ S=T$, so let's define

$R:Y\rightarrow Z$ by $R(y)=T(x)$, if $\exists x\in S$ such that $S(x)=y\in Y$, else define $R(y)=0$.

This map obviously satisfies the condition that $R\circ S=T$, but firstly we need to check that it is $well-defined$. So if $y=S(x_1)=S(x_2)\implies S(x_1-x_2)=0\implies (x_1-x_2)\in ker(S)\subset ker(T)$. Thus $R$ is well-defined.

You can check that $R$ is linear too.

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