PCA: Eigenvectors of opposite sign and not being able to compute eigenvectors with `solve` in R

Question

I'm learning PCA in R language. I met two problems right now that I don't understand.

1. I am performing a PCA analysis in R on a 318×17 dataset using some custom code. I take eigen function in R to find eigenvalues and eigenvectors. But my 1st and 3rd eigenvectors are of the opposite sign to my handbook. My second eigenvectors is almost the same.

2. I know that given a square matrix A, the condition that characterizes an eigenvalue, $\lambda$, is the existence of a nonzero vector $x$ such that $Ax=\lambda x$; this equation can be rewritten as follows: $(A - \lambda)x=0$.

   Now I calculate covariance of my data and have eigenvalues. I want to solve this linear combination equation to find $x$ and compare with initial eigenvectors. When I take solve function in R, my $x$ vector is always zero.

Here are my questions: Why the sign is different? How to use solve function in R to find a non-zero vector $x$?

Answer 1

1) The definition of eigenvector $Ax = \lambda x$ is ambidextrous. If $x$ is an eigenvector, so is $-x$, for then

$$A(-x) = -Ax = -\lambda x = \lambda (-x)$$

So the definition of an eigenbasis is ambiguous of sign.

2) It's hard to know for sure, but I have a strong suspicion of what is happening here. Your equation

$$ (A - \lambda)x = 0 $$

is technically incorrect. The correct equation is

$$ (A - \lambda I)x$$

The first equation is often used as a shorthand for the second. In general, this is unambiguous, because there is no real mathematical way to subtract a vector from a square matrix, but it is abuse of notation. In R though, you have broadcasting. So if you do

> M <- matrix(c(1, 1, 1, 1), nrow=2)
> M - .5
     [,1] [,2]
[1,]  0.5  0.5
[2,]  0.5  0.5

its not really what you want. The proper way would be

> M - diag(.5, 2)
     [,1] [,2]
[1,]  0.5  1.0
[2,]  1.0  0.5

The reason you are getting zero solutions is that the matrix you are starting with $A$ is invertible. More than likely (almost surely), the matrix you get by subtracting the same number from every entry will also be invertible. For invertible matrices, the only solution to $Ax = 0$ is the zero vector.