From: Izmirlian, Grant (NIH/NCI) [E] <izmirlig_at_mail.nih.gov>

Date: Mon 14 Aug 2006 - 12:39:22 EST

R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Mon Aug 14 12:46:05 2006

Date: Mon 14 Aug 2006 - 12:39:22 EST

Ok, I had a look at it. It seems like awefully far to dig for the main point which is easily summarized in a few sentences.

If we super-impose the pre-image and image spaces (plot the input and output in the same picture), then in 1 dimension, a linear function, say 'a x', takes its input, x, and stretches it by a factor |a|. If 'a' is negative, then the direction that 'x' points is reversed.

Understanding several dimensions, as is usually the case, requires us to refine our understanding of the 1-dimensional case. In several dimensions, a linear function, say 'A x' (where 'A' is an m by m matrix and 'x' is an 'm' vector) will result in the stretching of the input, 'x', along the direction its pointing, by a factor 'a'. However, this is the case _only_ if 'x' lies in one of the 'characteristic directions' corresponding to 'A'. Since 'A' is an m by m matrix, there will be at most m such 'characteristic directions'. Each of the characteristic directions has its associated stretching factor. The characteristic directions are called eigenvectors and the corresponding stretching factors are called eigenvalues.

Think about what this means in 1-dimension (hint: there's only one dimension so only one possible direction).

The number of linearly independent characteristic directions (eigenvectors) is called the rank of the matrix, A. If you understand the concept of 'basis' then you know that any m vector can be expressed in terms of the basis of eigenvectors of 'A' (that is unless A is not of 'full rank' and has less than m linearly independent eigenvectors, in which case we decomponse 'x' into two orthogonal components, one as a linear combination of the eigenvectors of A and the other gets mapped to 0 by A.)

Thus to each input 'x' is assigned an output 'y' which is the sum of coefficients in the eigenvector basis representation of 'x' times corresponding eigenvalues. This can be understood as the diagonalization of 'A'. By the way, the referenced page was in error because the singular value decomposition (I think the page actually called it the single value decomposition...free translation(s).com anyone) is not the same thing as the diagonalization.

There, it took a little more than a few sentences, but at least by the close of the second paragraph one gets the basic idea.

Now, in closing, Arun, please spend some time thinking about the answer to your question before you cut and paste it into your homework assignment.

-----Original Message-----

From: Dirk Enzmann [mailto:dirk.enzmann@uni-hamburg.de]
Sent: Sat 8/12/2006 7:01 AM

To: r-help@stat.math.ethz.ch

Cc: arun.kumar.saha@gmail.com

Subject: Re: [R] Geometrical Interpretation of Eigen value and Eigen vector

Arun,

have a look at:

http://149.170.199.144/multivar/eigen.htm

**HTH,
**

Dirk

"Arun Kumar Saha" <arun.kumar.saha@gmail.com> wrote:

> It is not a R related problem rather than statistical/mathematical. However

*> I am posting this query hoping that anyone can help me on this matter. My
**> problem is to get the Geometrical Interpretation of Eigen value and Eigen
**> vector of any square matrix. Can anyone give me a light on it?
*

R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Mon Aug 14 12:46:05 2006

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