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The Diagonalization of a Symmetric Matrix

Module by: Steven J. Cox. E-mail the author

Summary: (Blank Abstract)

By choosing an orthogonal basis q j , k 1knj q j , k 1 k nj for each RPj Pj and collecting the basis vectors in Qj=( q j , 1 q j , 2 q j , nj ) Qj q j , 1 q j , 2 q j , nj we find that Pj=QjQjT=k=1nj q j , k q j , k T Pj Qj Qj k 1 nj q j , k q j , k As a result, the spectral representation takes the form

B=j=1hλjQjQjT=j=1hλjk=1nj q j , k q j , k T B j 1 h λj Qj Qj j 1 h λj k 1 nj q j , k q j , k
(1)
This is the spectral representation in perhaps its most detailed dress. There exists, however, still another form! It is a form that you are likely to see in future engineering courses and is achieved by assembling the QjQj into a single nn-by-nn orthonormal matrix Q=( Q1Qh ) Q Q1 Qh Having orthonormal columns it follows that QTQ=I Q Q I . QQ being square, it follows in addition that QT=Q-1 Q Q . Now B q j , k =λj q j , k B q j , k λj q j , k may be encoded in matrix terms via
BQ=QΛ B Q Q Λ
(2)
where ΛΛ is the nn-by- nn diagonal matrix whose first n1n1 diagonal terms are λ1λ1, whose next n2n2 diagonal terms are λ2λ2, and so on. That is, each λjλj is repeated according to its multiplicity. Multiplying each side of Equation 2, from the right, by QT Q we arrive at
B=QΛQT B Q Λ Q
(3)
Because one may just as easily write
QTBQ=Λ Q B Q Λ
(4)
one says that QQ diagonalizes BB.

Let us return the our example B=( 111 111 111 ) B 1 1 1 1 1 1 1 1 1 of the last chapter. Recall that the eigenspace associated with λ1=0 λ1 0 had e 1 , 1 =( -1 1 0 ) e 1 , 1 -1 1 0 and e 1 , 2 =( -1 0 1 ) e 1 , 2 -1 0 1 for a basis. Via Gram-Schmidt we may replace this with q 1 , 1 =12( -1 1 0 ) q 1 , 1 1 2 -1 1 0 and q 1 , 2 =16( -1 -1 2 ) q 1 , 2 1 6 -1 -1 2 Normalizing the vector associated with λ2=3 λ2 3 we arrive at q 2 , 1 =13( 1 1 1 ) q 2 , 1 1 3 1 1 1 and hence Q=( q11q12q2 )=16( 3-12 3-12 022 ) Q q11 q12 q2 1 6 3 -1 2 3 -1 2 0 2 2 and Λ=( 000 000 003 ) Λ 0 0 0 0 0 0 0 0 3

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