When one computes the vector of scores (t1) using the principal component (p1) the data is being projected over the direction of biggest variation. One could measure the distance between the point where the data was projected and the origin.
If we do the squared sum of squares of those distances (because they could be negative), we won't obtain the eigenvalue of the eigenvector pointing in the direction of the principal component. Why?
An example, as requested:
The original data:
sample = [[1.343730519 , -.160152268 , .186470243], [-.160152268 , .619205620 , -.126684273], [.186470243 , -.126684273 , 1.485549631]] )
Eigenstuff (from the covariance matrix):
evalues = [2.22044605e-16, 1.67438287, 2.82561713] evectors.T = [ 0.54061848, 0.65888106, 0.52307496], [ 0.68485977, 0.0164023 , -0.72849026], [ 0.48856807, -0.75206829, 0.44237374]]
Score using just the first component (3rd vector from above)
t1 = [1.0619562 , -1.93803314, 0.87607695]
The following is the part that I don't get
The elements in vector
t1 are the 'distances' from the origin to the point where the original data was projected in the direction of
Since the eigenvalue of
p1 is the magnitude of the variance in that direction, I would expect that the sum of squares of the elements in
t1 would yield the same result as the plain eigenvalue. Which is not the case, computing the squared sum of squares (SS) over
SS = 2.3772324776675657
The eigenvalue was:
evalue_p1 = 2.82561713
It is very similar yet not the same, why?