gusl: (Default)
[personal profile] gusl
Here's something I should have learned in Linear Algebra class, if it is true:

For every linear transformation T that is not projecting into fewer dimensions, there exists an *orthogonal* basis B, such that T takes each basis vector to a multiple of itself, i.e. forall b_i in B, T(b_i) = lambda_i (b_i). In this basis, T is represented by a diagonal matrix, namely the one containing lambda_i. (I can't prove this)

If this is right, there should be something called the "orthogonal eigenbasis algorithm", but I can find no such thing... (doesn't PCA do this?)

(Also, a linear combination of two eigenvectors is an eigenvector iff the two eigenvalues are the same)

I trust that my geeky readership will correct me if this is wrong.

I should read about characteristic polynomials.

(no subject)

Date: 2007-02-12 07:12 am (UTC)
From: [identity profile] smandal.livejournal.com
Well, it's certainly true (http://mathworld.wolfram.com/EigenDecompositionTheorem.html).

If T is square and fully of rank N, where N is the dimension of the space, then the eigenvalues and eigenvectors are easy to find. Examples here (http://en.wikipedia.org/wiki/Eigenvalue). (PCA is a generalization to the Factor Analysis described in the article.)

(no subject)

Date: 2007-02-12 08:29 am (UTC)
From: [identity profile] rdore.livejournal.com
What you're talking about is called an orthoganal diagonalization. Being orthogonally diagonalizable is equivalent to being symmetric (equal to one's transpose).

(no subject)

Date: 2007-02-12 03:05 pm (UTC)
From: [identity profile] jcreed.livejournal.com
Isn't it enough to demand that the matrix is nonsingular if you let the eigenvalues be complex?

(no subject)

Date: 2007-02-12 07:10 pm (UTC)
From: [identity profile] rdore.livejournal.com
No.
  1. All eigenvalues in the field iff there's a Jordan normal form.
  2. Basis of eigenvectors iff diagonalizable.
  3. Symmetric and all eigenvalues in field (missed before) iff orthogonally diagonalizable.
  4. (For complex matrices) Hermitian iff unitarily diagonalizable.

(no subject)

Date: 2007-02-12 10:57 pm (UTC)
From: [identity profile] mauitian.livejournal.com
The eigenvectors are only orthogonal if T is symmetric (or Hermitian), but even if it's not there's still a spectral decomposition (eigen decomposition).

Here are my notes on it:

http://deferentialgeometry.org/#eigen

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