The Diagonalization of a Symmetric Matrix 2.3 2002/04/09 2002/08/05 00:00:00.003 GMT-5 Steven Cox cox@rice.edu Steven Cox cox@rice.edu Jacob Grabczewski jgrab@owlnet.rice.edu (Blank Abstract) By choosing an orthogonal basis q j , k 1 k nj for each Pj and collecting the basis vectors in Qj q j , 1 q j , 2 … q j , nj we find that Pj Qj Qj k 1 nj q j , k q j , k As a result, the spectral representation takes the form B j 1 h λj Qj Qj j 1 h λj k 1 nj q j , k q j , k 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 Qj into a single n-by-n orthonormal matrix Q Q1 … Qh Having orthonormal columns it follows that Q Q I . Q being square, it follows in addition that Q Q . Now B q j , k λj q j , k may be encoded in matrix terms via B Q Q Λ where Λ is the n-by- n diagonal matrix whose first n1 diagonal terms are λ1, whose next n2 diagonal terms are λ2, and so on. That is, each λj is repeated according to its multiplicity. Multiplying each side of , from the right, by Q we arrive at B Q Λ Q Because one may just as easily write Q B Q Λ one says that Q diagonalizes B. Let us return the our example B 1 1 1 1 1 1 1 1 1 of the last chapter. Recall that the eigenspace associated with λ1 0 had e 1 , 1 -1 1 0 and e 1 , 2 -1 0 1 for a basis. Via Gram-Schmidt we may replace this with q 1 , 1 1 2 -1 1 0 and q 1 , 2 1 6 -1 -1 2 Normalizing the vector associated with λ2 3 we arrive at q 2 , 1 1 3 1 1 1 and hence Q q11 q12 q2 1 6 3 -1 2 3 -1 2 0 2 2 and Λ 0 0 0 0 0 0 0 0 3