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Thread: Prove S is a linear transformation

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    Prove S is a linear transformation

    Suppose that $\displaystyle T:R^n --> R^n$ is a bijection. Then T has an inverse function S (ie $\displaystyle S:R^n-->R^n $ such that $\displaystyle ToS = I_n = SoT$). If T is a linear transformation, show that S is a linear transformation also.
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    Quote Originally Posted by wopashui View Post
    Suppose that $\displaystyle T:R^n --> R^n$ is a bijection. Then T has an inverse function S (ie $\displaystyle S:R^n-->R^n $ such that $\displaystyle ToS = I_n = SoT$). If T is a linear transformation, show that S is a linear transformation also.
    Let $\displaystyle \bold{x},\bold{y}\in\mathbb{R}^n$. Since $\displaystyle T$ is surjective we know that $\displaystyle T(\bold{x}')=\bold{x},T(\bold{y}')=\bold{y}$ for some $\displaystyle \bold{x}',\bold{y}'\in\mathbb{R}^n$. Therefore, $\displaystyle T^{-1}(\bold{x}+\bold{y})=T^{-1}\left(T(\bold{x}')+T(\bold{y}')\right)$$\displaystyle =T^{-1}\left(T\left(\bold{x}'+\bold{y}'\right)\right)=\ bold{x}'+\bold{y}'$. But, $\displaystyle T^{-1}(\bold{x})+T^{-1}(\bold{y})=\bold{x}'+\bold{y}'$. Therefore $\displaystyle T^{-1}$ is additive.

    Also, using the same idea $\displaystyle T^{-1}(a\bold{x})=T^{-1}\left(aT(\bold{x}')\right)=T^{-1}\left(T(a\bold{x}')\right)=a\bold{x}'=aT(\bold{x })$
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