linearly mapping an interval onto another

basically the idea is that, given a real interval $\text{[math]}$ and the number $\text{[math]}$ such that $\text{[math]}$ and another interval $\text{[math]}$ , we wanna find the number $\text{[math]}$ such that $\text{[math]}$ . we need to isolate $\text{[math]}$ : $\text{[math]}$

extending to higher dimensions

assume a vector space $\text{[math]}$ with a dimension of $\text{[math]}$ , given the 2 sets of real intervals:
$\text{[math]}$ and a vector $\text{[math]}$ :
$\text{[math]}$ where $\text{[math]}$ for all $\text{[math]}$ is bound in the interval $\text{[math]}$ , i.e.
$\text{[math]}$ we wanna find the vector $\text{[math]}$ :
$\text{[math]}$ such that
$\text{[math]}$ so we write the linear transformation (might not be necessarily linear but affine):
$\text{[math]}$ obviously this function doesnt preserve the origin as $\text{[math]}$ , so its an affine transformation, we separate the intercept so we can drop it and add it later:
$\text{[math]}$ we drop the intercept and find the transformation matrix of the function without it, which is a square matrix of size $\text{[math]}$ :
$\text{[math]}$

im taking it for granted that the $\text{[math]}$ without the intercept is linear, in reality we need to check if the properties of a linear function are preserved in $\text{[math]}$

then we add the intercept with projective coordinates to get the final matrix, which is a square matrix of size $\text{[math]}$ :
$\text{[math]}$ so to conclude, the formula
$\text{[math]}$ transforms the vector $\text{[math]}$ from a given set of intervals onto another, resulting in $\text{[math]}$ .