## The Posts

The purpose of this post is to provide guidance to a reader who has just discovered that I have a large pile of posts about principal components / factor analysis. This pile of posts might seem a very jungle, without any map.

Here, have a map.

As I finalize this post, it will be number 52 in PCA / FA. Here’s a list of the 52 posts, including the dates spanned by any group, and the number of posts in that group. (When the picture was taken, I didn’t know when this would be published. In fact, post 51 was scheduled but not yet published. Even more, post 51 did not even exist when the first picture was created.)

transition/attitude matrices is a post that is sometimes relevant when we discuss “new data” in PCA, but it is not in the PCA / FA category.

“tricky prepro” is short for “tricky preprocessing”, and discusses the combination of constant row sums and covariance or correlation matrix.
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## PCA/FA Answers to some Basilevsky questions

Let us look at three of the questions I asked early in February, and answer two of them.

## First, what do we know? What have we done?

We assume that we have data X, with the variables in columns, as usual. In fact, we assume that the data is at least centered, and possibly standardized.

We compute the covariance matrix

$c = \frac{X^T X}{N-1}\$,

then its eigendecomposition

$c = v\ \Lambda^2\ v^T\$,

where $\Lambda^2$ is the diaginal matrix of eigenvalues. We define the $\sqrt{\text{eigenvalue}}$-weighted matrix

$A = v\ \Lambda\$.

Finally, we use A as a transition matrix to define new data Z:

$X^T = A\ Z^T\$.

We discovered two things. One, the matrix A is the cross covariance between Z and X:

$A = \frac{X^T Z}{N-1}\$.

I find this interesting, and I suspect that it would jump off the page at me out of either Harman or Jolliffe; that is, I suspect it is written there but it didn’t register.

Two, we discovered that we could find a matrix Ar which is the cross covariance between Zc and Xs. Read the rest of this entry »

## Introduction

I am looking into Basilevsky because he did something I didn’t understand: he normalized the rows of the Ac matrix (which I denoted Ar). We discussed that, and we illustrated the computations, in the previous two posts. But we did those computations without having any data. I want to take a closer look, with data.

In contrast to As and Ac, which are eigenvector matrices, his Ar matrix is not. Nevertheless, as I said, his Ar is not without some redeeming value. In fact, all three of the A matrices have the same redeeming value.

I will show, first by direct computation and then by proof, that each of these A matrices is the cross covariance between X data and Z data.
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## PCA / FA example 9: centered and raw, 3 models

What follows is simple computation, solely to show us exactly what happens. It continues the work of the previous post, which did my default calculations for standardized data. Here I do the same calculations for centered and raw data.

## Centered

The raw data is still

$\text{raw} = \left(\begin{array}{lll} 2.09653 & -0.793484 & -7.33899 \\ -1.75252 & 13.0576 & 0.103549 \\ 3.63702 & 29.0064 & 8.52945 \\ 0.0338101 & 46.912 & 19.8517 \\ 5.91502 & 70.9696 & 36.0372\end{array}\right)$

I will center the data and call it Xc. Get the column means…

$\{1.98597,\ 31.8304,\ 11.4366\}$

and subtract them from each column to get Xc:
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## Introduction

edited 16 Jan 2009: I found a place where I called F^T the loadings instead of the scores. That’s all.

I want to run thru what is admittedly a toy case, but this seems to be where I stand on the computation of PCA / FA.

Recall the raw data of example 9:

$\text{raw} = \left(\begin{array}{lll} 2.09653 & -0.793484 & -7.33899 \\ -1.75252 & 13.0576 & 0.103549 \\ 3.63702 & 29.0064 & 8.52945 \\ 0.0338101 & 46.912 & 19.8517 \\ 5.91502 & 70.9696 & 36.0372\end{array}\right)$

Get the mean and variance of each column. The means are

$\{1.98597,\ 31.8304,\ 11.4366\}$

and the variances are

$\{8.99072,\ 796.011,\ 291.354\}$

We see that the raw data will differ from the centered data, and that will differ from the standardized data. Let’s do the standardized data first, because that’s what we’ve been doing most recently.

Here’s what I’m going to do. For a data matrix X

1. get the SVD, $X = u\ w\ v^T$
2. get the eigenvalues $\lambda\ \text{of } X^T\ X/\left(N-1\right)$ (in 2 cases, that’s the correlation matrix or the covariance matrix)
3. form the diagonal matrix $\Lambda\ \text{of } \sqrt{\lambda}$
4. form the weighted eigenvector matrix $A = v\ \Lambda$
5. form the loadings scores $F^T= \sqrt{N-1}\ u$
6. form the new data Y wrt v, Y = u w
7. form Davis’ loadings $A^R = v\ w^T$
8. form Davis’ scores $S^R = X\ A^R\$.

I want to look at reconstituting the data. Equivalently, I want to look at setting successive singular values to zero.

This example was actually built on the previous one. Before I set the row sums to 1, I had started with

$t1 = \left(\begin{array}{lll} 1 & 1 & -3 \\ -1 & 2 & -2 \\ 1 & 3 & -1 \\ -1 & 4 & 1 \\ 1 & 5 & 4\end{array}\right)$

I’m going to continue with Harmon’s & Bartholomew’s model: Z = A F, Z = X^T, X is standardized, A is an eigenvector matrix weighted by the square roots of the eigenvalues of the correlation matrix of X.

I want data with one eigenvalue so large that we could sensibly retain only that one. Let me show you how I got that.
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## introduction

Recall Harman’s or Bartholomew’s model

Z = A F

with $Z = X^T\$, X standardized, and A a $\sqrt{\text{eigenvalue}}\$-weighted eigenvector matrix, with eigenvalues from the correlation matrix.

We saw how to compute the scores $F^T$ in the case that A was invertible (here). If, however, any eigenvalues are zero then A will have that many columns of zeroes and will not be invertible.

What to do?

One possibility – shown in at least one of the references, and, quite honestly, one of the first things I considered – is to use a particular example of a pseudo-inverse. I must tell you up front that this is not what I would recommend, but since you will see it out there, you should see why I don’t recommend it.

(Answer: it works, it gets the same answer, but computing the pseudo-inverse explicitly is unnecessary. In fact, it’s unnecessary even if we don’t have the Singular Value Decomposition (SVD) available to us.)
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