Dear Allstats,
I came across this problem :
Let A be a p X p symmetric matrix. Show that there exists a matrix C of
order p X p such that C'AC = I; where I is the identity matrix.
I was able to solve the problem partially by using eigen values and eigen
vectors as follows :-
Let x(1), x(2), .... ,x(p) be eigen vectors of A corresponding to the p
eigen values k(1), k(2),...,k(p) respectively, which are so taken to make
them mutually orthogonal and of unit length. We then have,
1) A x( i ) = k( i ) x( i ) for i=1,2,...p
2) x'( i ) x( j ) = 0 if i <>j (i, j = 1,2,...n); and
3) x'( i ) x( i ) = 1 i= 1,2,...,p
Consequently,
Q'AQ = [x'(1) x'(2) ..... x'(p)] A ( x(1) x(2).... x(p) )
= [x'(1) x'(2) ..... x'(p)] ( Ax(1) Ax(2) ..... Ax(p) )
= [x'(1) x'(2) ..... x'(p)] ( k(1)x(1) k(2)x(2) .....k(p)x(p) )
= ( k(1)x'(1)x(1) k(2)x'(1)x(2) ..... k(p)x'(1)x(p) )
( k(1)x'(2)x(1) k(2)x'(2)x(2) ..... k(p)x'(2)x(p) )
( ..... ..... .... ..... )
(k(1)x'(p)x(1) ..... .... k(p)x'(p)x(p) )
= diag[ k(1) k(2) .... k(p) ]
= D (say)
Now if we consider another matrix say B which is also a diagonal matrix with
the diagonal element 1/(k(i))^(1/2) for the ith row or column, and if we
post multiply this matrix with D and premultiply by B' , we get the Identity
matrix I. But this will hold good only if the diagonal elements of matrix D,
that is the eigen values are non zero. So the last part of the proof is
incorrect. I was told this problem could be solved by using quadratic forms
only. Could anybody possibly give me a hint on how to proceed?
Thanks in advance,
______________________
Indrajit SenGupta
Department Of Statistics
St. Xavier's College
Calcutta University
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______________________
EC- 195
Salt Lake City, Sector -1
Calcutta 700064
West Bengal
India
Phone #337-5424
______________________
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