| dbp:proof
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- Case: All are independent
Fix some , define , and diagonalize by an orthogonal transform . Then consider . It is diagonalized as well.
Let , then it is also standard Gaussian. Then we have
Inspect their diagonal entries, to see that implies that their nonzero diagonal entries are disjoint.
Thus all eigenvalues of are 0, 1, so is a dist with degrees of freedom.
Case: Each is a distribution.
Fix any , diagonalize it by orthogonal transform , and reindex, so that . Then for some , a spherical rotation of .
Since , we get all . So all , and have eigenvalues .
So diagonalize them simultaneously, add them up, to find .
Case: .
We first show that the matrices B can be simultaneously diagonalized by an orthogonal matrix and that their non-zero eigenvalues are all equal to +1. Once that's shown, take this orthogonal transform to this simultaneous eigenbasis, in which the random vector becomes , but all are still independent and standard Gaussian. Then the result follows.
Each of the matrices B has rank r'i and thus r'i non-zero eigenvalues. For each i, the sum has at most rank . Since , it follows that C has exactly rank N − r'i.
Therefore B and C can be simultaneously diagonalized. This can be shown by first diagonalizing B, by the spectral theorem. In this basis, it is of the form:
:
Thus the lower rows are zero. Since , it follows that these rows in C in this basis contain a right block which is a unit matrix, with zeros in the rest of these rows. But since C has rank N − r'i, it must be zero elsewhere. Thus it is diagonal in this basis as well. It follows that all the non-zero eigenvalues of both B and C are +1. This argument applies for all i, thus all B are positive semidefinite.
Moreover, the above analysis can be repeated in the diagonal basis for . In this basis is the identity of an vector space, so it follows that both B and are simultaneously diagonalizable in this vector space . By iteration it follows that all B-s are simultaneously diagonalizable.
Thus there exists an orthogonal matrix such that for all , is diagonal, where any entry with indices , , is equal to 1, while any entry with other indices is equal to 0. (en)
- . Since is symmetric, and , by the previous claim, has the same eigenvalues as 0. (en)
- Fix i, and consider the eigenvectors v of such that . Then we have , so all . Thus we obtain a split of into , such that V is the 1-eigenspace of , and in the 0-eigenspaces of all other . Now induct by moving into . (en)
- Let the eigenvalues of be , then calculate the characteristic function of . It comes out to be
For and to be equal, their characteristic functions must be equal, so have the same eigenvalues . (en)
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