Introduction:
suppose you have a linear mapping from an euclidean vector space V to an euclidean vector space w.
This mapping can be represented using a matrix A.
for all x out of v that are not zero, you can evaluate |Ax|/|x|.
because of linearity, if two nonzero vectors x and x' out of v are collinear, |Ax|/|x|=|Ax'|/|x'|.
x'=kx (collinear vectors)
therefore:
|Ax'|/|x'|=|A(kx)|/|kx|=|k(Ax)|/|kx|=k|(Ax)|/(k|x|)=|Ax|/|x|
but this implies that the image set of f : v -> R, with f(x)=|Ax|/|x| is the same as the image set of g : {x out of v| |x|=1} -> R, with g(x)=f(x)=|Ax|/|x| especially, this image set is a closed intervall on R, reaching from some lower bound a to some higher bound b. (Proof?)
So here is the question:
How to calculate those bounds (a and b) given the matrix representation A of some linear mapping?
or more precisely:
given two euclidean Vector spaces v and w,
and a matrix A, representing a linear mapping from v to w,
there exists a function l : v/{0} -> R,
given by l(x)=|Ax|/|x|.
the image set of l is an intervall [a,b]
how to calculate a and b given the matrix A?
why I am asking this:
Doing 3D graphics, I want to implement some specular lighting.
In my approach, I have a surface wich is projected onto another.
when approximating it, one may model this projection as consisting of two compounds, a "squeezing", attributed to the angle between the surfaces, and an "uniform scale", wich is attributed to the distance between the surfaces.
I can barely get the hands on the projection matrix A (a 2x2 matrix, invertibe most of the times), and now i am struggling to decompose it into these two compounds.
calculating these bounds, i could evaluate the "uniform scale" part, and removing that part, i could get my hands on the other...
however, since the matrix is different for every fragment, (fragment is something like a "pixel", a smallest unit) I need a very efficient formula taking the entries of the 2x2 matrix describing the projection and calculating the upper bound of the "image set interval" associated with this matrix (as described above).