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

Question

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

Thank you in advance!

Answer 1

$A$ 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

• Proving any linear transformation can be represented as a matrix