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Consider a random vector X e RP with mean EX is a p x p dimensional matrix. Denote the jth eigenvalue and jth eigenvector of

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Answer #1

The covariance matrix Cov X] E(X - E[X]) (X - EX .

It is given the EX0 . But EZ EpX= pE[X] = 0 (Since expectation is linear)

Where (1, 2,. is made up of orthonormal vectors of \Sigma.

Now = E[X X| z=Cov[Z] EZZ= E[pXX*

= \varphi^t \Sigma \varphi = \varphi^t \big( \lambda_1 \phi_1, \lambda_2 \phi_2, \dots , \lambda_p \phi_p \big)

0 0 2 0 0 Ap 0 0

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