1. Affine transformations

An affine transformation is a function f:ℝ^m→ℝⁿ of the form f(x) = Mx + b where M is an n×m matrix and b is a column vector.

1. Prove or disprove: if f:ℝ^m→ℝⁿ and g:ℝⁿ→ℝ^k are both affine transformations, then (g∘f) is also an affine transformation.

2. Prove or disprove: if f:ℝⁿ→ℝⁿ is an affine transformation and f^-1 exists, then f^-1 is also an affine transformation.

1.1. Solution

1. Proof: Let f(x) = Ax+b and g(y) = Cy+d. Then g(f(x)) = g(Ax+b) = C(Ax+b)+d = (CA)x + (Cb+d).

2. Proof: Let f(x) = Ax+b. We'll find an inverse for f by solving the equation y = Ax+b for x in terms of y: this gives x = A^-1(y-b) = A^-1y - A^-1b. This has the right form to be an affine transformation, provided A^-1 exists. If A^-1 does not exist, then either A is not surjective or not injective. If A is not surjective, then there is some y such that no x satisfies Ax = y. But then no x satisfies Ax + b = y+b, which means f isn't surjective either. If A is not injective, then there exist distinct x and y such that Ax = Ay. But then f(x) = Ax + b = Ay + b = f(y), so f is also not injective. So under the assumption that f^-1 exists, A^-1 exists as well, and we have f^-1 = A^-1y - A^-1b is affine.

2. Pythagoras goes mod

Let x and y be vectors in (ℤ_p)ⁿ, where p is a prime.

Show that if (x+y)⋅(x+y) = x⋅x + y⋅y, then either x⋅y = 0 (mod p) or p = 2.

2.1. Solution

Calculate (x+y)⋅(x+y) = x⋅x + x⋅y + y⋅x + y⋅y = (x⋅x + y⋅y) + 2(x⋅y). For this to equal (x⋅x + y⋅y), we need 2(x⋅y) = 0 (mod p), or p|(2(x⋅y)). Since p is prime, it divides a product if and only if it divides one of the factors. If p|2, then p = 2; otherwise, p|(x⋅y), which means (x⋅y) = 0 (mod p).

3. Convexity

A set of points S in ℝⁿ is convex if, for any x,y∈S, and any 0 ≤ λ ≤ 1, the point λx + (1-λ)y is in S. (Intuitively, this means that the line segment between any two points in S is also in S; visually, S has no dimples or holes.)

1. Prove or disprove: If f:ℝⁿ→ℝ^m is a linear transformation, and S is convex, then f(S) = { f(x) | x∈S } is convex.

2. Prove or disprove: If f:ℝⁿ→ℝ^m is a linear transformation, and f(S) is convex, then S is convex.

3.1. Solution

1. This one is just calculation: Let f(x) and f(y) be in f(S), and let 0 ≤ λ ≤ 1. Then λx + (1-λ)y is in S, so f(λx + (1-λ)y) = λf(x) + (1-λ)f(y) is in f(S).

2. Disproof: Let f(x) = 0. Then f is linear (f(ax) = af(x) = 0, f(x+y) = 0 = f(x) + f(y)), and f(S) is convex for all S, since for any x,y∈S we have λf(x) + (1-λ)f(y) = 0 ∈ f(S). But we can easily find a non-convex S (e.g. { (0), (1) } in ℝ¹), giving a counterexample to the claim.