Thank you so much in advance (5 points) Let X and Y, respectively, be p- and q-d

Question

Thank you so much in advance

(5 points) Let X and Y, respectively, be p- and q-dimensional random vectors. By the singular value decomposition we can write cov(X,Y)- jujy, where u, are orthonormal eigenvectors of cov(x, Y) cov(Y, X) (left-singular vectors), v, are those of cov(Y, X) . cov(X, Y) (right-singular vectors) and 12 ?···20 are termed singular values (a) Prove that cov(X, Y) cov(Y, X) and cov(Y, X) cov(X, Y) have shared eigen- values, 2 20 (b) Prove where the maximum is attained at u u, and v-vi, the normalized eigenvectors of cov(X, Y) cov(Y, X) and cov(Y, X) cov(X, Y), respectively, corresponding to the first shared eigenvalue

Explanation / Answer

[ Here I use some results of linear algebra. If you have any doubts about any steps feel free to ask me ]

a)

let, cov(x,y)=A1

then cov(y,x)=tranpose(A1)=A2, say

let an another matrix

B=A1.A2 - A2.A1

then the eigenvalues of B is necessarily zero because A1 and transpose(A1) has the same set of eigenvalues (result 1)

let the order of A1.A2 is n and the order of A2A1 is m and n<m

then the first n eigenvalues of A2A1 will be non zero and the remaining (m-n) are equal to zero.

b)

cov(u'x,v'y)= u'cov(x,y)v=u'(A1)v [ ' denotes transpose]

now

u'(A1)v<=(u'A1A1'u) / (v'v) [ from CS inequality]

u'A1A1'u=u'A1A2u=u'(sigma1)u= sigma1. u'u

since u is the eigenvector of A1A2 with corresponding eigen value sigma1 [from principle component analysis]

now v'v=u'u=1 [since they are orthonormal vectors]

which shows that

u'A1v <= sigma1

i.e the maximum value is sigma 1

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