Why I get two different covariance matrix?

Hi all,

Prof. Andrew Ng in his ML class says that we can calculate covariance matrix as: 1/m*X'*X .

Where;

examples are in rows of X,

X' is transpose of X,

and, m is number of examples.

For example:

X=randi(12,[6,2]);
cov1=1/size(X,1)*X'*X

cov1 = 2×2

92.5000 42.0000 42.0000 21.8333

And, covariance with cov function is:

cov2=cov(X)

cov2 = 2×2

2.7000 -0.9000 -0.9000 1.9000

As you can see, cov1 is different from cov2 !!!

What is the reasan for that? Do you have any idea?

Thanks

Answer 1

[Paul]

Hi Ali,

Perhaps Prof. Ng has some additional assumptions about the data that aren't included in your question. To compute the covariance we have to subtract off the mean. As for whether or not the outer product should be scaled by 1/m or 1/(m-1) depends on assumptions about the underlying data. IIRC, if we know the data is drawn from a Normal distribution then divide by m (perhaps also for other distributions as well?), but typically we don't assume that and so divide by m-1 for unbiased estimation. As can be seen below, cov subtracts the mean and divides by m-1.

rng(100);
X=randi(12,[6,2])

X = 6×2

7 9 4 10 6 2 11 7 1 11 2 3

cov1=1/(size(X,1)-1)*(X-mean(X))'*(X - mean(X))

cov1 = 2×2

13.3667 -1.6000 -1.6000 14.0000

cov(X)

ans = 2×2

13.3667 -1.6000 -1.6000 14.0000

As for the second question

sigma=[6 2;2 3]; % cov matrix
[a1,v]   = eig(sigma)

a1 = 2×2

0.4472 -0.8944 -0.8944 -0.4472

v = 2×2

2.0000 0 0 7.0000

[a2,s,~] = svd(sigma)

a2 = 2×2

-0.8944 -0.4472 -0.4472 0.8944

s = 2×2

7 0 0 2

we see that eig and svd just have a different order for the results.

[ali yaman]

Hi @Paul thanks, very appreciated for your help.

You are right. I checked the data set, and I saw that, as you indicated, all feature has 0 mean and 1 standard deviation. And for that type of data set COV and 1/m*X'*X gives same result. Thanks.

Probably, Prof. NG forgot to say that assumption or maybe he said but ı did not catch.

About 2nd part:

Yes, they just have diffrent order. So, how can i know which one should i use? or does not matter, both of them give me same result if i use?

[Paul]

Which one of eig() or svd() to use? it would never occur to me to use svd() to get the eigenvectors of a symmetric matrix. Don't know if eig() or svd() is better for that special case. Asking that as a new question is more likely to get the attention of knowledgeable people that can answer.

[ali yaman]

@Paul Ok, Thanks Bro.

Answer 2

[ali yaman]

What about the second question!!

Just like covarince matrixe we can get two different eigenvectors.

For example;

sigma=[6 2;2 3]; % cov matrix
[a,~]=eig(sigma)

a = 2×2

0.4472 -0.8944 -0.8944 -0.4472

And, lets calculate eigenvectors with svd, singular value decomposition function:

[a,~,~]=svd(sigma)

a = 2×2

-0.8944 -0.4472 -0.4472 0.8944

As you can see we are getting two different eigenvectors value.

What is the reason?

Thanks, in advance.

[John D'Errico]

PLEASE DON"T ASK a new question as an answer to your own question! In fact, I won't answer a question posed as you have done, and I could probably do so.

[ali yaman]

@John D'Errico Ok. sorry for that. I did not know it. I will be more careful from now on

[John D'Errico]

Since this is probably of some general interest, I'll actually post a question of my own, then answer it myself, discussing the relative issues between eig and svd.

https://www.mathworks.com/matlabcentral/answers/1760195-should-i-use-eig-or-svd-to-compute-the-eigenvalues-of-a-symmetric-matrix

[ali yaman]

OK. that would be nice. thanks