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Suppose $f$ is a bijection and linear map such that $f \colon E \to F$. Could someone help me prove that the inverse $f^{-1} \colon F \to E$ is also linear?

Thanks a lot!

2 Answers 2

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Here's a hint. Let's call $g$ the inverse of $f$, so that $g(f(x))=x$ and $f(g(y))=y$.

To prove that $g$ is linear you need to prove that $g(ax)=ag(x)$ and $g(x+y)=g(x)+g(y)$.

For the first one of these, we know that $f(ay)=af(y)$ since $f$ is linear. Applying $g$ to both sides we get

$g(f(ay)) = g(af(y))$

which reduces to

$ay = g(af(y))$

and finally defining $y=g(x)$, so that $f(y)=x$, we get

$ag(x) = g(ax)$

Now can you see how to prove the second part?

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Let $y_1, y_2 \in F$. We wish to show $f^{-1}(y_1 + y_2) = f^{-1}(y_1) + f^{-1}(y_2)$.

Since $f$ is a bijection, we know there exist $x_1, x_2 \in E$ with $f(x_1) = y_1$ and $f(x_2) = y_2$. From this and linearity of $f$, you can justify the following: $\begin{align*} f^{-1}(y_1 + y_2) &= f^{-1}\left(f(x_1) + f(x_2)\right)\\ &= f^{-1}\left(f(x_1 + x_2)\right)\\ &= x_1 + x_2\\ &= f^{-1}(y_1) + f^{-1}(y_2) \end{align*}$

Try similar reasoning for the other part (or follow Chris Taylor's advice).