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In this picture you can see the formula (red rectangle added by me for emphasis): $$ \textbf{V}^\intercal\textbf{HV} = \textbf{D} $$

Should not this rather be (eigenvalue decomposition): $$ \textbf{V}^\intercal\textbf{DV} = \textbf{H} $$

The first two lines of equation 21 would make little sense otherwise.

The full paper can be found here: http://jov.arvojournals.org/article.aspx?articleid=2192836

Attila Kun
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  • I tend to agree with you. From the context of that sentence, it should be as you wrote. Btw, did you get any interesting results with Slow Feature Analysis? I applied it on some time series data but did not get anything conclusive... – Vladislavs Dovgalecs Jul 30 '15 at 23:11
  • @xeon: I do not have much experience with it in practice yet, so can not comment :) – Attila Kun Jul 31 '15 at 21:42

1 Answers1

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Just to be sure we are talking about the same thing, the eigendecomposition of a square matrix with $n$ linearly independent eigenvectors is

$$H=VDV^{-1}\qquad\qquad\qquad (1)$$

which for real symmetric matrices

$$H=VDV^T\qquad\qquad\qquad (2)$$

where $D$ is diagonal, and $V$ is the matrix of eigenvectors, such that $V^T=V^{-1}$, so you have $VV^T=I$ and $V^TV=I$. If that's the decomposition we're talking about, then

$$V^THV=V^TVDV^TV=D\,.$$

Glen_b
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