Given a dataset that consists of the following points below: A=(2, 3), B=(5, 5), C=(6, 6), D=(8,9) 1. Calculate the covariance matrix for the dataset. 2. Calculate the eigenvalues and eigenvectors of the covariance matrix请给我这两个问题的答案

1. To calculate the covariance matrix, we first need to calculate the mean values for x and y:

mean_x = (2+5+6+8)/4 = 5.25
mean_y = (3+5+6+9)/4 = 5.75

Then, we calculate the deviations from the mean for each point:

A: (2-5.25, 3-5.75) = (-3.25, -2.75)
B: (5-5.25, 5-5.75) = (-0.25, -0.75)
C: (6-5.25, 6-5.75) = (0.75, 0.25)
D: (8-5.25, 9-5.75) = (2.75, 3.25)

Using these deviations, we can calculate the covariance matrix:

| var(x) cov(x,y) |
Cov = | cov(x,y) var(y) |

where

var(x) = sum((xi - mean_x)^2)/n
var(y) = sum((yi - mean_y)^2)/n
cov(x,y) = sum((xi - mean_x)*(yi - mean_y))/n

Plugging in the values, we get:

var(x) = (3.25^2 + 0.25^2 + 0.75^2 + 2.75^2)/4 = 4.1875
var(y) = (2.75^2 + 0.75^2 + 0.25^2 + 3.25^2)/4 = 4.1875
cov(x,y) = (-3.25*-2.75 + -0.25*-0.75 + 0.75*0.25 + 2.75*3.25)/4 = 3.9375

So the covariance matrix is:

| 4.1875 3.9375 |
Cov = | 3.9375 4.1875 |

2. To calculate the eigenvalues and eigenvectors of the covariance matrix, we first need to solve for the characteristic equation:

| 4.1875-lambda 3.9375 | | x | | 0 |
| 3.9375 4.1875-lambda | * | y | = | 0 |

Expanding the determinant, we get:

(4.1875-lambda)^2 - 3.9375^2 = 0

Solving for lambda, we get lambda1 = 8.375 and lambda2 = 0

For lambda1 = 8.375, we need to solve for the eigenvector:

(4.1875-8.375)x + 3.9375y = 0
3.9375x + (4.1875-8.375)y = 0

Simplifying, we get:

-4.188x + 3.938y = 0
3.938x - 4.188y = 0

This system has infinitely many solutions, but we can choose any non-zero vector that satisfies it. One such solution is (1, 1), so the corresponding eigenvector is:

v1 = (1, 1)

For lambda2 = 0, we similarly get:

(4.1875-0)x + 3.9375y = 0
3.9375x + (4.1875-0)y = 0

Simplifying, we get:

4.188x + 3.938y = 0
3.938x + 4.188y = 0

This system has a unique solution (x=0, y=0), which means that there is only one linearly independent eigenvector for lambda2. We can choose any non-zero vector that is orthogonal to v1, such as (-1, 1), so the corresponding eigenvector is:

v2 = (-1, 1)

Therefore, the eigenvalues and eigenvectors of the covariance matrix are:

lambda1 = 8.375, v1 = (1, 1)
lambda2 = 0, v2 = (-1, 1)