EigenFaces

(squared) Variance A measure of how "spread out" a sequence of numbers are. λ= 1 𝑛 𝑖=1 𝑛 (𝑥 𝑖 −μ ) 2 μ= 1 𝑛 𝑖=1 𝑛 𝑥 𝑖

Covariance matrix A measure of correlation between data elements. Example: Data set of size n Each data element has 3 fields: Height Weight Birth date

Covariance [Collect data from class]

Covariance λ 11 λ 12 … λ 1𝑛 λ 21 λ 22 … λ 2𝑛 ⋮ λ 𝑛1 ⋮ λ 𝑛2 ⋱ ⋮ … λ 𝑛𝑛 λ 𝑖𝑗 = 1 𝑛 𝑘=1 𝑛 ( 𝑥 𝑖 − μ 𝑖 )( 𝑥 𝑗 − μ 𝑗 )

Covariance The diagonals are the variance of that feature Non-diagonals are a measure of correlation High-positive == positive correlation one goes up, other goes up Low-negative == negative correlation one goes up, other goes down Near-zero == no correlation unrelated [How high depends on the range of values]

Covariance You can calculate it with a matrix: Raw Matrix is a p x q matrix p features q samples Convert to mean-deviation form Calculate the average sample Subtract this from all samples. Multiply MeanDev (a p x q matrix) by its transpose (a q x p matrix) Multiply by 1/n to get the covariance matrix.

Covariance [Calculate our covariance matrix]

EigenSystems An EigenSystem is: Such that: 𝐴 𝑣 =λ 𝑣 A vector 𝒗 (the eigenvector) A scalar λ (the eigenvalue) Such that: 𝐴 𝑣 =λ 𝑣 (the zero vector isn't an eigenvector) In general, not all matrices have eigenvectors.

EigenSystems and PCA When you calculate the eigen-system of an n x n Covariance matrix you get: n eigenvectors (each of dimension n) n matching eigenvalues The biggest eigen-value "explains" the largest amount of variance in the data set.

Example Say we have a 2d data set First eigen-pair (v1 = [0.8, 0.6], λ=800.0) Second eigen-pair (v2 = [-0.6, 0.8], λ=100.0) 8x as much variance is along v1 as v2. v1 and v2 are perpendicular to each other v1 and v2 define a new set of basis vectors for this data set. v2 v1

Conversions between basis vectors Let's take one data point… Let's say it is [-1.5, 0.4] in "world units" Project it onto v1 and v2 to get the coordinates relative to (v1, v2 unit-length basis vectors) 𝑛𝑒𝑤𝐶𝑜𝑜𝑟𝑑= 𝑃 ● 𝑣1 𝑃 ● 𝑣2 𝑛𝑒𝑤𝐶𝑜𝑜𝑟𝑑≈ −0.9 0.9 v2 v1 To convert back to "world units": 𝑤𝑜𝑟𝑙𝑑𝐶𝑜𝑜𝑟𝑑= 𝑛𝑒𝑤𝐶𝑜𝑜𝑟𝑑 0 ∗ 𝑣1 𝑛𝑒𝑤𝐶𝑜𝑜𝑟𝑑 1 ∗ 𝑣2

PCA and compression Example: n (the number of features) is high (~100) Most of the variance is captured by 3 eigen-vectors. You can throw out the other 97 eigen-vectors. You can represent most of the data for each sample using just 3 numbers per sample (instead of 100) For a large data set, this can be huge.

EigenFaces Collect database images Subject looking straight ahead, no emotion, neutral lighting. Crop: on the top include all of the eyebrows on the bottom include just to the chin on the sides, include all of the face. Size to 32x32, grayscale (a limit of the eigen-solver) In code, include a way to convert to (and from) a VectorN.

EigenFaces, cont. Calculate the average image Just pixel (Vector element) by element.

EigenFaces, cont. Calculate the Covariance matrix Calculate the EigenSystem Keep the eigen-pairs that preserve n% of the data variance (98% or so) Your Eigen-database is the 32x32 average image and the (here) 8 32x32 eigen-face images.

Eigenfaces, cont. Represent each of your faces as a q-value vector (q = # of eigenfaces). Subtract the average and project onto the q eigenfaces The images I'm showing here are the original image and the 8-value "eigen-coordinates 9.08 187.4 -551.7 -114.4 -328.8 29.2 -371.9 -108.0 1277.0 150.9 -133.6 249.3 338.9 13.14 16.8 3.35

EigenFaces, cont. (for demonstration of compression) You can reconstruct a compressed image by: Start with a copy of the average image, X Repeat for each eigenface: Add the eigen-coord * eigenface to X Here are the reconstructions of the 2 images on the last slide: Original Reconstruction

EigenFaces, cont. Facial Recognition Take a novel image (same size as database images) Using the eigenfaces computed earlier (this novel image is usually NOT part of this computation), compute eigen-coordinates. Calculate the q-dimensional distance (pythagorean theorem in q-dimensions) between this image and each database image. The database image with the smallest distance is your best match.