When using eig function in Matlab, it seems that this function has already normalize the values of the eigenvalues. Do we need to write some lines of code to normalize the eigenvalues after using the eig function.
The function eig in MATLAB normalizes the eigenvectors (not the eigenvalues).
See the following from the documentation:
[V,D] = eig(A) returns matrix V, whose columns are the right
eigenvectors of A such that AV = VD. The eigenvectors in V are
normalized so that the 2-norm of each is 1.
Eigenvectors can vary by a scalar, so a computation algorithm has to choose a particular scaled value of an eigenvector to show you. eig chooses 2-norm = 1. Just look at the eigenvector definition to see why: AV=VD. V shows up on both sides, so you can multiple V by anything without affecting the equation.
Eigenvalues do not vary. Look again at AV=VD. D is only on one side, so it can't be scaled.
Suppose there is a matrix B, where its size is a 500*1000 double(Here, 500 represents the number of observations and 1000 represents the number of features).
sigma is the covariance matrix of B, and D is a diagonal matrix whose diagonal elements are the eigenvalues of sigma. Assume A is the eigenvectors of the covariance matrix sigma.
I have the following questions:
I need to select the first k = 800 eigenvectors corresponding to the eigenvalues with the largest magnitude to rank the selected features. The final matrix named Aq. How can I do this in MATLAB?
What is the meaning of these selected eigenvectors?
It seems the size of the final matrix Aq is 1000*800 double once I calculate Aq. The time points/observation information of 500 has disappeared. For the final matrix Aq, what does the value 1000 in matrix Aq represent now? Also, what does the value 800 in matrix Aq represent now?
I'm assuming you determined the eigenvectors from the eig function. What I would recommend to you in the future is to use the eigs function. This not only computes the eigenvalues and eigenvectors for you, but it will compute the k largest eigenvalues with their associated eigenvectors for you. This may save computational overhead where you don't have to compute all of the eigenvalues and associated eigenvectors of your matrix as you only want a subset. You simply supply the covariance matrix of your data to eigs and it returns the k largest eigenvalues and eigenvectors for you.
Now, back to your problem, what you are describing is ultimately Principal Component Analysis. The mechanics behind this would be to compute the covariance matrix of your data and find the eigenvalues and eigenvectors of the computed result. It has been known that doing it this way is not recommended due to numerical instability with computing the eigenvalues and eigenvectors for large matrices. The most canonical way to do this now is via Singular Value Decomposition. Concretely, the columns of the V matrix give you the eigenvectors of the covariance matrix, or the principal components, and the associated eigenvalues are the square root of the singular values produced in the diagonals of the matrix S.
See this informative post on Cross Validated as to why this is preferred:
https://stats.stackexchange.com/questions/79043/why-pca-of-data-by-means-of-svd-of-the-data
I'll throw in another link as well that talks about the theory behind why the Singular Value Decomposition is used in Principal Component Analysis:
https://stats.stackexchange.com/questions/134282/relationship-between-svd-and-pca-how-to-use-svd-to-perform-pca
Now let's answer your question one at a time.
Question #1
MATLAB generates the eigenvalues and the corresponding ordering of the eigenvectors in such a way where they are unsorted. If you wish to select out the largest k eigenvalues and associated eigenvectors given the output of eig (800 in your example), you'll need to sort the eigenvalues in descending order, then rearrange the columns of the eigenvector matrix produced from eig then select out the first k values.
I should also note that using eigs will not guarantee sorted order, so you will have to explicitly sort these too when it comes down to it.
In MATLAB, doing what we described above would look something like this:
sigma = cov(B);
[A,D] = eig(sigma);
vals = diag(D);
[~,ind] = sort(abs(vals), 'descend');
Asort = A(:,ind);
It's a good thing to note that you do the sorting on the absolute value of the eigenvalues because scaled eigenvalues are also eigenvalues themselves. These scales also include negatives. This means that if we had a component whose eigenvalue was, say -10000, this is a very good indication that this component has some significant meaning to your data, and if we sorted purely on the numbers themselves, this gets placed near the lower ranks.
The first line of code finds the covariance matrix of B, even though you said it's already stored in sigma, but let's make this reproducible. Next, we find the eigenvalues of your covariance matrix and the associated eigenvectors. Take note that each column of the eigenvector matrix A represents one eigenvector. Specifically, the ith column / eigenvector of A corresponds to the ith eigenvalue seen in D.
However, the eigenvalues are in a diagonal matrix, so we extract out the diagonals with the diag command, sort them and figure out their ordering, then rearrange A to respect this ordering. I use the second output of sort because it tells you the position of where each value in the unsorted result would appear in the sorted result. This is the ordering we need to rearrange the columns of the eigenvector matrix A. It's imperative that you choose 'descend' as the flag so that the largest eigenvalue and associated eigenvector appear first, just like we talked about before.
You can then pluck out the first k largest vectors and values via:
k = 800;
Aq = Asort(:,1:k);
Question #2
It's a well known fact that the eigenvectors of the covariance matrix are equal to the principal components. Concretely, the first principal component (i.e. the largest eigenvector and associated largest eigenvalue) gives you the direction of the maximum variability in your data. Each principal component after that gives you variability of a decreasing nature. It's also good to note that each principal component is orthogonal to each other.
Here's a good example from Wikipedia for two dimensional data:
I pulled the above image from the Wikipedia article on Principal Component Analysis, which I linked you to above. This is a scatter plot of samples that are distributed according to a bivariate Gaussian distribution centred at (1,3) with a standard deviation of 3 in roughly the (0.878, 0.478) direction and of 1 in the orthogonal direction. The component with a standard deviation of 3 is the first principal component while the one that is orthogonal is the second component. The vectors shown are the eigenvectors of the covariance matrix scaled by the square root of the corresponding eigenvalue, and shifted so their tails are at the mean.
Now let's get back to your question. The reason why we take a look at the k largest eigenvalues is a way of performing dimensionality reduction. Essentially, you would be performing a data compression where you would take your higher dimensional data and project them onto a lower dimensional space. The more principal components you include in your projection, the more it will resemble the original data. It actually begins to taper off at a certain point, but the first few principal components allow you to faithfully reconstruct your data for the most part.
A great visual example of performing PCA (or SVD rather) and data reconstruction is found by this great Quora post I stumbled upon in the past.
http://qr.ae/RAEU8a
Question #3
You would use this matrix to reproject your higher dimensional data onto a lower dimensional space. The number of rows being 1000 is still there, which means that there were originally 1000 features in your dataset. The 800 is what the reduced dimensionality of your data would be. Consider this matrix as a transformation from the original dimensionality of a feature (1000) down to its reduced dimensionality (800).
You would then use this matrix in conjunction with reconstructing what the original data was. Concretely, this would give you an approximation of what the original data looked like with the least amount of error. In this case, you don't need to use all of the principal components (i.e. just the k largest vectors) and you can create an approximation of your data with less information than what you had before.
How you reconstruct your data is very simple. Let's talk about the forward and reverse operations first with the full data. The forward operation is to take your original data and reproject it but instead of the lower dimensionality, we will use all of the components. You first need to have your original data but mean subtracted:
Bm = bsxfun(#minus, B, mean(B,1));
Bm will produce a matrix where each feature of every sample is mean subtracted. bsxfun allows the subtraction of two matrices in unequal dimension provided that you can broadcast the dimensions so that they can both match up. Specifically, what will happen in this case is that the mean of each column / feature of B will be computed and a temporary replicated matrix will be produced that is as large as B. When you subtract your original data with this replicated matrix, the effect will subtract every data point with their respective feature means, thus decentralizing your data so that the mean of each feature is 0.
Once you do this, the operation to project is simply:
Bproject = Bm*Asort;
The above operation is quite simple. What you are doing is expressing each sample's feature as a linear combination of principal components. For example, given the first sample or first row of the decentralized data, the first sample's feature in the projected domain is a dot product of the row vector that pertains to the entire sample and the first principal component which is a column vector.. The first sample's second feature in the projected domain is a weighted sum of the entire sample and the second component. You would repeat this for all samples and all principal components. In effect, you are reprojecting the data so that it is with respect to the principal components - which are orthogonal basis vectors that transform your data from one representation to another.
A better description of what I just talked about can be found here. Look at Amro's answer:
Matlab Principal Component Analysis (eigenvalues order)
Now to go backwards, you simply do the inverse operation, but a special property with the eigenvector matrix is that if you transpose this, you get the inverse. To get the original data back, you undo the operation above and add the means back to the problem:
out = bsxfun(#plus, Bproject*Asort.', mean(B, 1));
You want to get the original data back, so you're solving for Bm with respect to the previous operation that I did. However, the inverse of Asort is just the transpose here. What's happening after you perform this operation is that you are getting the original data back, but the data is still decentralized. To get the original data back, you must add the means of each feature back into the data matrix to get the final result. That's why we're using another bsxfun call here so that you can do this for each sample's feature values.
You should be able to go back and forth from the original domain and projected domain with the above two lines of code. Now where the dimensionality reduction (or the approximation of the original data) comes into play is the reverse operation. What you need to do first is project the data onto the bases of the principal components (i.e. the forward operation), but now to go back to the original domain where we are trying to reconstruct the data with a reduced number of principal components, you simply replace Asort in the above code with Aq and also reduce the amount of features you're using in Bproject. Concretely:
out = bsxfun(#plus, Bproject(:,1:k)*Aq.', mean(B, 1));
Doing Bproject(:,1:k) selects out the k features in the projected domain of your data, corresponding to the k largest eigenvectors. Interestingly, if you just want the representation of the data with regards to a reduced dimensionality, you can just use Bproject(:,1:k) and that'll be enough. However, if you want to go forward and compute an approximation of the original data, we need to compute the reverse step. The above code is simply what we had before with the full dimensionality of your data, but we use Aq as well as selecting out the k features in Bproject. This will give you the original data that is represented by the k largest eigenvectors / eigenvalues in your matrix.
If you'd like to see an awesome example, I'll mimic the Quora post that I linked to you but using another image. Consider doing this with a grayscale image where each row is a "sample" and each column is a feature. Let's take the cameraman image that's part of the image processing toolbox:
im = imread('camerman.tif');
imshow(im); %// Using the image processing toolbox
We get this image:
This is a 256 x 256 image, which means that we have 256 data points and each point has 256 features. What I'm going to do is convert the image to double for precision in computing the covariance matrix. Now what I'm going to do is repeat the above code, but incrementally increasing k at each go from 3, 11, 15, 25, 45, 65 and 125. Therefore, for each k, we are introducing more principal components and we should slowly start to get a reconstruction of our data.
Here's some runnable code that illustrates my point:
%%%%%%%// Pre-processing stage
clear all;
close all;
%// Read in image - make sure we cast to double
B = double(imread('cameraman.tif'));
%// Calculate covariance matrix
sigma = cov(B);
%// Find eigenvalues and eigenvectors of the covariance matrix
[A,D] = eig(sigma);
vals = diag(D);
%// Sort their eigenvalues
[~,ind] = sort(abs(vals), 'descend');
%// Rearrange eigenvectors
Asort = A(:,ind);
%// Find mean subtracted data
Bm = bsxfun(#minus, B, mean(B,1));
%// Reproject data onto principal components
Bproject = Bm*Asort;
%%%%%%%// Begin reconstruction logic
figure;
counter = 1;
for k = [3 11 15 25 45 65 125 155]
%// Extract out highest k eigenvectors
Aq = Asort(:,1:k);
%// Project back onto original domain
out = bsxfun(#plus, Bproject(:,1:k)*Aq.', mean(B, 1));
%// Place projection onto right slot and show the image
subplot(4, 2, counter);
counter = counter + 1;
imshow(out,[]);
title(['k = ' num2str(k)]);
end
As you can see, the majority of the code is the same from what we have seen. What's different is that I loop over all values of k, project back onto the original space (i.e. computing the approximation) with the k highest eigenvectors, then show the image.
We get this nice figure:
As you can see, starting with k=3 doesn't really do us any favours... we can see some general structure, but it wouldn't hurt to add more in. As we start increasing the number of components, we start to get a clearer picture of what the original data looks like. At k=25, we actually can see what the cameraman looks like perfectly, and we don't need components 26 and beyond to see what's happening. This is what I was talking about with regards to data compression where you don't need to work on all of the principal components to get a clear picture of what's going on.
I'd like to end this note by referring you to Chris Taylor's wonderful exposition on the topic of Principal Components Analysis, with code, graphs and a great explanation to boot! This is where I got started on PCA, but the Quora post is what solidified my knowledge.
Matlab - PCA analysis and reconstruction of multi dimensional data
I am trying to understand principal component analysis in Matlab,
There seems to be at least 3 different functions that do it.
I have some questions re the code below:
Am I creating approximate x values using only one eigenvector (the one corresponding to the largest eigenvalue) correctly? I think so??
Why are PC and V which are both meant to be the loadings for (x'x) presented differently? The column order is reversed because eig does not order the eigenvalues with the largest value first but why are they the negative of each other?
Why are the eig values not in ordered with the eigenvector corresponding to the largest eigenvalue in the first column?
Using the code below I get back to the input matrix x when using svd and eig, but the results from princomp seem to be totally different? What so I have to do to make princomp match the other two functions?
Code:
x=[1 2;3 4;5 6;7 8 ]
econFlag=0;
[U,sigma,V] = svd(x,econFlag);%[U,sigma,coeff] = svd(z,econFlag);
U1=U(:,1);
V1=V(:,1);
sigma_partial=sigma(1,1);
score1=U*sigma;
test1=score1*V';
score_partial=U1*sigma_partial;
test1_partial=score_partial*V1';
[PC, D] = eig(x'*x)
score2=x*PC;
test2=score2*PC';
PC1=PC(:,2);
score2_partial=x*PC1;
test2_partial=score2_partial*PC1';
[o1 o2 o3]=princomp(x);
Yes. According to the documentation of svd, diagonal elements of the output S are in decreasing order. There is no such guarantee for the the output D of eig though.
Eigenvectors and singular vectors have no defined sign. If a is an eigenvector, so is -a.
I've often wondered the same. Laziness on the part of TMW? Optimization, because sorting would be an additional step and not everybody needs 'em sorted?
princomp centers the input data before computing the principal components. This makes sense as normally the PCA is computed with respect to the covariance matrix, and the eigenvectors of x' * x are only identical to those of the covariance matrix if x is mean-free.
I would compute the PCA by transforming to the basis of the eigenvectors of the covariance matrix (centered data), but apply this transform to the original (uncentered) data. This allows to capture a maximum of variance with as few principal components as possible, but still to recover the orginal data from all of them:
[V, D] = eig(cov(x));
score = x * V;
test = score * V';
test is identical to x, up to numerical error.
In order to easily pick the components with the most variance, let's fix that lack of sorting ourselves:
[V, D] = eig(cov(x));
[D, ind] = sort(diag(D), 'descend');
V = V(:, ind);
score = x * V;
test = score * V';
Reconstruct the signal using the strongest principal component only:
test_partial = score(:, 1) * V(:, 1)';
In response to Amro's comments: It is of course also possible to first remove the means from the input data, and transform these "centered" data. In that case, for perfect reconstruction of the original data it would be necessary to add the means again. The way to compute the PCA given above is the one described by Neil H. Timm, Applied Multivariate Analysis, Springer 2002, page 446:
Given an observation vector Y with mean mu and covariance matrix Sigma of full rank p, the goal of PCA is to create a new set of variables called principal components (PCs) or principal variates. The principal components are linear combinations of the variables of the vector Y that are uncorrelated such that the variance of the jth component is maximal.
Timm later defines "standardized components" as those which have been computed from centered data and are then divided by the square root of the eigenvalues (i.e. variances), i.e. "standardized principal components" have mean 0 and variance 1.
I get this error when I call iwishrnd() function:
??? Error using ==> iwishrnd at 41
Covariance matrix must be symmetric and positive definite.
as I explained in my previous question:
Why does eig(A) function (in which A is a positive semidefinite function) returns negative doubles?
I like to know, why matlab calls this parameter a Covariance matrix ?
I know that, this matrix is used as the conjugate prior for the covariance matrix of a multivariate normal distribution, but it is proportional to mean of that covariance matrix (as you can see in http://en.wikipedia.org/wiki/Inverse-Wishart_distribution). So, isn't it better to call this the mean of the distribution, instead of the covariance matrix?
This is a nomenclature issue; it's relatively common to call the A parameter in the Wishart distribution a 'covariance matrix', since it (1) has to have all the properties of a covariance matrix, (2) the output of the Wishart distribution is almost always used as a covariance matrix (e.g., the Wishart is the conjugate prior for a Gaussian, see link below) and (3) A represents the mean of the expected covariance matrix output from the Wishart distribution.
A does not represent the covariance of the Wishart distribution, if that's what you're wondering.
http://en.wikipedia.org/wiki/Conjugate_prior#Table_of_conjugate_distributions
I am using LAPACK's ssteqr function to calculate eigenvalues/eigenvectors. The documentation for ssteqr says that the eigenvalues are sorted "in ascending order". Is it reasonable to assume that the list of eigenvectors is also sorted in ascending order?
Yes, it is reasonable to assume that the eigenvectors are ordered so that the i-th eigenvector corresponds to the i-th eigenvalue.
Nevertheless, if I were you, I would check for each eigenvalue the result of the multiplication of the eigenvector by the matrix. This way you are sure that you interpret the output right, and you see explicitly the accuracy of the calculations.
An old question, this, but I struggled with this recently, so am adding this for current and future readers.
The basic answer is that, yes, the eigenvectors are sorted such that the ith eigenvector corresponds to the ith eigenvalue. However, note that the eigenvectors thus obtained may not be the actual eigenvectors you want. This is so because of the following.
Since the ?steqr functions work only on tridiagonal matrices, one typically uses LAPACK's ?sytrd functions to first transform one's original symmetric matrix, call it M, to a tridiagonal form, call it T, such that M = QTQT where Q is an orthogonal matrix (and QT denotes its transpose). One then applies the ?steqr function on this tridiagonal matrix T to find its eigenvalues and eigenvectors. Now the eigenvalues thus obtained (of T) are exactly the same as the eigenvalues of M, so if one only wants the eigenvalues one can stop here. But if one is interested in the eigenvectors, like the OP, then one needs to bear in mind that the eigenvectors of T and M are different. To find the eigenvectors of the original matrix M, one needs to left-multiply the obtained eigenvectors of T by Q. This is very easily done by using the LAPACK functions orgtr or ormtr. See here for a clear explanation: https://software.intel.com/en-us/mkl-developer-reference-fortran-sytrd.