# 1. Affine transformations

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

Prove or disprove: if f:ℝ

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

^{n}→ℝ^{n}is an affine transformation and f^{-1}exists, then f^{-1}is also an affine transformation.

## 1.1. Solution

- 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).
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^{-1}y - A^{-1}b. 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^{-1}y - A^{-1}b 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 ℝ^{n} 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.)

Prove or disprove: If f:ℝ

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

^{n}→ℝ^{m}is a linear transformation, and f(S) is convex, then S is convex.

## 3.1. Solution

- 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).
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 ℝ

^{1}), giving a counterexample to the claim.