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I'm trying to understand the effect of certain operations on a matrix on eigenvectors and eigenvalues.

Let $X$ be a square matrix, I need to understand how eigenvectors change if;

  1. Each column of $X$ is normalized, such that $\hat{x}_{ij} = \frac{x_{ij}}{\Sigma_i x_{ij}}$ which means that each term is divided by the sum of its column, thus making every column total to 1.
  2. Off-diagonal terms are multiplied by $-1$.
  3. A diagonal matrix $D$ is subtracted from $X$

My main purpose is to understand how various dimensionality reduction techniques (such as Laplacian Eigenmaps, Multidimensional scaling, PCA etc.) differ from each other by gaining intuition about the effect of these operations on matrices. I've looked for books, articles on such theoretical results but could not find any.

I appreciate any comments/articles/books on this problem.

Thanks,

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What does it mean to normalize a column of a matrix? –  Gerry Myerson Apr 10 '12 at 12:17
    
Have you tried working out any examples? Take some $2\times2$ matrix, do one of your operations to it, see what happens to the eigenstuff? –  Gerry Myerson Apr 10 '12 at 12:22
    
that's actually what i'm doing from a practical perspective but i'm asking for any theory, work dealing with the mentioned problem. –  Goker Erdogan Apr 10 '12 at 17:19
    
"I need to understand how eigenvectors change if..." Perhaps you want to understand too much? Those operations do not have simple effect on eigenvalues/eigenvectors, in general. –  leonbloy Apr 10 '12 at 17:31
    
So, from a practical perspective, have you noticed any patterns emerging? or have you done enough experiments to be convinced that there are no patterns there? or what? –  Gerry Myerson Apr 11 '12 at 5:26

1 Answer 1

None of the operations 1,2,3 have simple effect eigenvalues/eigenvectors, in general (already noted by leonbloy). But there is a useful special case of 3. If we subtract a scalar matrix $\alpha I$ from $X$, the eigenvectors remain the same, and every eigenvalue $\lambda$ is replaced by $\lambda-\alpha$.

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