how linear algebra is connected with projective geometry ??

6 years agoProjective geometry is an extension of 'ordinary' geometry in the sense that points 'at infinity' are allowed to be represented. Normally we exclude, implicitly or otherwise, infinity from calculations that don't involve limiting procedures. It turns out that there are many shades of infinity. So if I say to you that "I am infinitely far away", you might reply "well OK, but in which direction?" With that in mind then for projective geometry as applied to 3D space you have four components representing vectors: \[V = (x,y,z,w)\]with rules available to determine the meaning of that representation depending on what 'w' is. The 'x', 'y' and 'z' have the standard sense. If w = 0 then V is understood as the point at infinity in the direction indicated by (x, y, z). If w is non-zero then V is understood to indicate the point at \[(\frac{ x}{w }, \frac{ y }{w }, \frac{ z }{w })= (X, Y, Z)\]and if you think in terms of limits then as w-> 0 each of x/w, y/w and z/w will increase without bound BUT retaining their original proportions amongst themselves. That is X/Y, X/Z and Y/Z - if any of those ratios have meaning - are constant regardless of w. You could visualise this limiting process ( w-> 0) as a vector in the direction (x, y, z) progressively stretching - as (X,Y,Z) - to an arbitrarily large length. With that construct in mind then it is clear that there is an infinite number of points at infinity in 3D space - one for each possible direction from the origin. Now linear algebra as we are studying has concepts of angle and distance, and these don't ( usefully ) translate to projective geometry - which allows, say, parallel lines to meet. So if I have three distinct points all at ( different ) infinities, then what does the triangle inequality mean ? Or an inner product, and thus orthogonality?

6 years ago