I have the following points:
$$
\left\{ (1.1,1.4),(1.5,2.1),(1,1.6),(2,2.1),(2.3,3.2),(3.1,3.5),(1.9,2.7),(2.2,3.4),(0.5,1.2),(2.5,2.9)\right\}
$$
I'm trying to figure what the get_covariance() of from sklearn.decomposition import PCA as sklearnPCA does. From what I understand, it first normalize:
$$
\begin{cases}
\overline{x}=\frac{1.1+1.5+1+2+2.3+3.1+1.9+2.2+0.5+2.5}{10}=1.81\\
\overline{y}=\frac{1.4+2.1+1.6+2.1+3.2+3.5+2.7+3.4+1.2+2.9}{10}=2.41
\end{cases}
$$
Then you do $\left(x_{i}^{norm},y_{i}^{norm}\right)=\left(x_{i}-\overline{x},y_{i}-\overline{y}\right)$, so you get:
$$
\left\{ (-0.71,-1.01),(-0.31,-0.31),(-0.81,-0.81),(0.19-0.31),(0.49,0.79),(1.29,1.09),(0.09,0.29),(0.39,0.99),(-1.31,-1.21),(0.69,0.49)\right\}
$$
But now, how do you calculate the covariance matrix? It returns:
But how do I calculate it manually?
EDIT: My question is that I don't get how they got $Var(X)=0.616$. I get:
$$
Var(X)=\frac{1.1^{2}+1.5^{2}+1^{2}+2^{2}+2.3^{2}+3.1^{2}+1.9^{2}+2.2^{2}+0.5^{2}+2.5^{2}}{10}-(\frac{1.1+1.5+1+2+2.3+3.1+1.9+2.2+0.5+2.5}{10})^2=0.5549
$$
How did they got that number? Also I get:
$$
Var(Y)=(\frac{1.4^2+2.1^2+1.6^2+2.1^2+3.2^2+3.5^2+2.7^2+3.4^2+1.2^2+2.9^2}{10})-(\frac{1.4+2.1+1.6+2.1+3.2+3.5+2.7+3.4+1.2+2.9}{10})^2=0.6449
$$
and not $Var(Y)=0.7165$.
The code in python if someone wants to try:
POINTS = [[1.1,1.4], [1.5,2.1], [1,1.6], [2,2.1], [2.3,3.2], [3.1,3.5], [1.9,2.7], [2.2,3.4], [0.5,1.2], [2.5,2.9]]
clf = sklearnPCA(n_components=1)
pca_transformed = clf.fit_transform(POINTS)
covariance_matrix = clf.get_covariance()
print(covariance_matrix)
Each component of the matrix $a_{ij}$ will be: $$a_{ij} = E\left[(x_i - \mu_i)(x_j - \mu_j)\right]$$
In your case, $x_1$ and $x_2$ are $x$ and $y$, and I defined $\mu_i = E(x_i)$.