# what does eigenvalues expres in the covariance matrix?

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Mohamed Moawed
on 23 Apr 2013

Commented: TUSHAR MURATKAR
on 18 Feb 2020

is there a relationship between a covariance matrix and eigenvalues? like an example

Let us consider a 321 × 261 image dimention 321 × 261 = 83781. We have only 32 observations and 83781 unknowns then we have a matrix of (32 row X 83781 column)

then we will calculate the covariance matrix (32 X 32) so we get 32 eigenvalues the question is: does these eigenvalues express the 32 images? or there is no any relationship between eigenvalues and images

thanks for you,

##### 8 Comments

Vincent Spruyt
on 10 Mar 2015

The eigenvalues in this case represent the magnitude of the spread in the direction of the principal components. If you data has a diagonal covariance matrix (covariances are zero), then the eigenvalues are equal to the variances:

If the covariance matrix is not diagonal, then the eigenvalues still define the variance of the data along the the principal components, whereas the covariance matrix operates along the axes:

Here is an article (and the source of the above images) that discusses this in more detail: http://www.visiondummy.com/2014/04/geometric-interpretation-covariance-matrix/

### Accepted Answer

Kye Taylor
on 23 Apr 2013

Edited: Kye Taylor
on 23 Apr 2013

Long story short: The eigenvalues of the covariance matrix encode the variability of the data in an orthogonal basis that captures as much of the data's variability as possible in the first few basis functions (aka the principle component basis).

For example, this code creates an ellipse, whos major axis is the x-axis, and whos minor axis is the y-axis.

t = linspace(0,2*pi,256);

data = [cos(t);0.2*sin(t)];

plot(data(1,:),data(2,:),'.')

axis([-1,1,-1,1])

Now, compute the variance of the data's coordinates

% tranpose to get variance down each column

% computes variance of each coordinate of the data

v = var(data')

Finally, observe the eigenvalues of the covariance matrix are equal to the variance of the data's coordinates

% need to transpose since input to cov must have

% rows = observations

% columns = variables

l = eig(cov(data'))

In other words, v and l contain the same values. The order may be different because the eig function returns eigenvalues in no particular order.

Note that if the data is rotated so that the major and minor axes are no longer the x and y axes, then var(data') no longer computes the variance about the coordinates from their principle axes, but eig(cov(data')) does automatically since the eigenvectors are principle components.

For example

% rotate everything

r = [cos(pi/4),-sin(pi/4);sin(pi/4),cos(pi/4)]; % rotation matrix

rData = r*data;

hold on

plot(rData(1,:),rData(2,:),'r.') % plot rotated data

vr = var(rData') % different

lr = eig(cov(rData')) % should be same as l

##### 4 Comments

### More Answers (1)

Shashank Prasanna
on 23 Apr 2013

Edited: Shashank Prasanna
on 23 Apr 2013

Essentially what you are describing are the principal components of your data.

https://en.wikipedia.org/wiki/Principal_component_analysis#Computing_PCA_using_the_covariance_method

Its a popularly used dimensionality reduction technique, for example to make your image smaller such that it still retains most of its variance.

The PCA command in MATLAB does all this for you directly.

##### 2 Comments

TUSHAR MURATKAR
on 18 Feb 2020

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