Capstone — Linear Algebra in Practice

A sprite sits at world position p = [10, 5], facing along f = [1, 0]. The player is at q = [13, 9], so the direction from the sprite to the player is d = q − p = [3, 4], magnitude 5. The dot product f · d = 3 is positive, and cos(θ) = 3/(1×5) = 0.6, giving θ ≈ 53.1° — the player is roughly ahead of the sprite, well within a typical detection cone.

Knowing the player is roughly ahead isn't enough for AI logic — turning left or right needs a side. The 2D pseudo-cross-product from Chapter 3, f₁d₂ − f₂d₁ = (1)(4) − (0)(3) = 4, is positive, meaning d is counter-clockwise from f — the player is to the sprite's left. This single sign is the entire "which way should the AI turn" decision.

This sprite needs to render mirrored (it's facing left in its base art) and scaled up 1.5×. Combining a flip matrix Flip = [[-1, 0], [0, 1]] and a scale matrix Sc = [[1.5, 0], [0, 1.5]] uses Chapter 4's real matrix multiplication, not entry-wise multiplication — and here, Flip × Sc = Sc × Flip = [[-1.5, 0], [0, 1.5]].

Packaging the combined flip-and-scale matrix together with the translation to p = [10, 5] needs Chapter 5's homogeneous coordinates: T = [[-1.5, 0, 10], [0, 1.5, 5], [0, 0, 1]]. Applying T to the sprite's own local origin [0, 0, 1] gives [10, 5, 1] — exactly Step 1's world position p, confirming the transform is correct. Applying it to a corner point of the sprite's local bounding box, [2, 1, 1], gives [7, 6.5, 1] — that corner's actual position in the world.

The renderer needs a mapping from world x-coordinates to screen pixels, pixel = m·world + c. Two known calibration points — world 2 → pixel 100, world 5 → pixel 250 — give the system 2m + c = 100, 5m + c = 250. Gaussian elimination (R2 → R2 − 2.5·R1) gives 0m − 1.5c = 0 → c = 0, then back-substitution gives 2m = 100 → m = 50. So pixel = 50 × world — checked against both original points.

A player clicks at world point [7, 6.5] — Step 4's own transformed corner. Before inverting, check the sprite's transform isn't degenerate: det([[-1.5, 0], [0, 1.5]]) = -2.25, nonzero — safe to invert. The inverse is [[-2/3, 0], [0, 2/3]]. Subtracting the translation first ([7,6.5] − [10,5] = [-3, 1.5]) and applying the inverse gives [2, 1] — exactly Step 4's original local corner. The click round-trips perfectly back to sprite-local space.

The control scheme offers three inputs: up = [0, 1], right = [1, 0], and a diagonal up-right = [1, 1] (both keys held at once). Row-reducing the three as rows shows up + right = [1, 1] = up-right exactly — the third row reduces to [0, 0]. Rank 2, not 3: the diagonal input adds no genuinely new reachable direction beyond combining the other two, which is exactly what a player pressing both keys already produces.

A UI element sits at the sprite's own world position, [10, 5] (Step 1). Four recorded player clicks, centered around it, come out (relative to that center) as (1,1), (-1,-1), (2,2), (-2,-2) — every click lies exactly on the line y = x. Computing the covariance matrix from these points: Cov = [[2.5, 2.5], [2.5, 2.5]].

Solving the characteristic equation (trace 5, det 0): λ² − 5λ = 0 → λ = 0 or λ = 5.