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I have a problem in understanding the exact meaning of degenerate eigenvalue. I have some database and I calculate the covariance matrix among it. the obtained eigenvalues are same ( all of them =5000) and the obtained eigenvectors are different each other and orthogonal. what does that tell? I know in this case covariance matrix is symmetric and diagonal. eigenspace forms a sphere. But is there any advantage of this eigenspace ( any special property)?

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Well, that can be answered only depending on the application you are looking for. Covariance matrices are always positive semi-definite and symmetric. Thus, their eigenvalues are always non-negative and also their eigenvectors form a orthogonal space. That happens because every symmetric matrix is a normal matrix. Other than that, you will need to specify the application –  dineshdileep Feb 21 '13 at 7:16
    
@dineshdileep The application is object recognition. –  amr Feb 21 '13 at 13:56
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Let your covariance matrix be $X^TX$, where $X$ is the data matrix. Put your eigenvectors together to form an invertible matrix $P$. Your computation result shows that $X^TXP=5000P$, i.e. $X^TX=5000I$.

Since every set of $n$ linearly independent vectors (be they orthogonal or not) form an eigenbasis of a scalar matrix in $\mathbb{R}^n$, I don't think there is anything special about the eigenbasis you obtained. What is special, however, is the data matrix itself. The fact that $X^TX=5000I$ indicates that, without a change of basis, the random variables in question are already uncorrelated and have identical variances ($5000$). If these random variables are jointly normal, we can further conclude that they are mutually independent.

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you are right because the database is orthogonal too. But think with me a bit, you can not ignore or remove any eigenvector here because they form a spherical cloud, right? I think if we employ this space to recognize some object, the performance will be enhanced! what do you think? –  amr Feb 21 '13 at 13:51
    
Another point, as you mentioned the first eigenvector is [1 0 0 .... 0] and the second is [0 1 0 ......0]...etc. But the eigenvectors I obtained are different and have no uniform pattern! May be due to the noise or program. –  amr Feb 22 '13 at 1:44
    
@amr Hmm, I didn't say that the eigenvectors you obtain are the canonical basis vectors. What I said was: every set of $n$ linearly independent vectors is a set of eigenvectors of $5000I$. This includes, of course, the canonical basis vectors, but it can also be any arbitrary basis of $\mathbb{R}^n$. Since all eigenvalues/singular values of $5000I$ are identical, if you use PCA, dimension reduction is impossible regardless of the basis you choose. (Actually, for real data, it's practically impossible you would get $5000I$ as the covariance matrix. Did you use synthetic data?) –  user1551 Feb 22 '13 at 12:05
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