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I am trying to follow the proof of Theorem 6.8 in Linear Algebra Done Wrong

Let $A : X → Y$ be a linear transformation. Then $A$ is invertible if and only if for any right side $b ∈ Y$ the equation $$Ax = b$$ has a unique solution $x ∈ X$.

I cannot understand the following part

$B$ is defined as $B:Y \mapsto X$ where $B(y)$ is the unique solution $x \in X$.

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I don't know how to jump from $A(\alpha x_1 + \beta x_2) = \alpha Ax_1 + \beta Ax_2 = \alpha y_1 + \beta y_2$ s to $B(\alpha y_1 + \beta y_2) = \alpha B(y_1) + \beta B(y_2)$.

Thank you in advance!

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    $\begingroup$ How is $B$ defined? $\endgroup$
    – G. Gare
    Oct 24, 2019 at 6:42

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$A$ is the matrix which represents the linear transformation then

$$A(\alpha x_1 + \beta x_2) = \alpha Ax_1 + \beta Ax_2 = \alpha y_1 + \beta y_2 \iff B(\alpha y_1 + \beta y_2) = \alpha B(y_1) + \beta B(y_2)$$

Refer also to the related

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  • $\begingroup$ Yes. I know that a matrix is a linear transformation. And I can also follow the left side of the equation. But the right side, I am not OK. $\endgroup$
    – JOHN
    Oct 24, 2019 at 7:02
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    $\begingroup$ @JOHN We are using that $$Ax=y \iff B(x)=y$$ then $$A(\alpha x_1 + \beta x_2) = \alpha y_1 + \beta y_2 \iff B(\alpha y_1 + \beta y_2) = \alpha B(y_1) + \beta B(y_2)$$ $\endgroup$
    – user
    Oct 24, 2019 at 7:08

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