How to check $B$ is linear in the proof of Theorem 6.8 in Linear Algebra Done Wrong

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 $$x \in X$$.

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 $$B(\alpha y_1 + \beta y_2) = \alpha B(y_1) + \beta B(y_2)$$.

• How is $B$ defined? Oct 24, 2019 at 6:42

$$A$$ is the matrix which represents the linear transformation then
$$A(\alpha x_1 + \beta x_2) = \alpha Ax_1 + \beta Ax_2 = \alpha y_1 + \beta y_2 \iff B(\alpha y_1 + \beta y_2) = \alpha B(y_1) + \beta B(y_2)$$
• @JOHN We are using that $$Ax=y \iff B(x)=y$$ then $$A(\alpha x_1 + \beta x_2) = \alpha y_1 + \beta y_2 \iff B(\alpha y_1 + \beta y_2) = \alpha B(y_1) + \beta B(y_2)$$