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Unformatted text preview: 1 11 Y1 −1 P=⎝ 2 , 1 0 ⎠ = X1 | X2 , P = ⎝ −2 3 2⎠ = T Y2 −2 01 2 −2 −1 so ⎛ 1 T G1 = X1 Y1 = ⎝ 2 −2 −1 −2 2 ⎞ −1 −2 ⎠ , 2 T H R Y IG ⎛ 0 T G2 = X2 Y2 = ⎝ −2 2 ⎞ 1 1 3 2⎠. −2 −1 Check that these are correct by conﬁrming the validity of (7.2.7)–(7.2.10). 3. Since λ1 = 1 is a simple eigenvalue, (7.2.12) may be used to compute G1 from any pair of associated right-hand and left-hand eigenvectors x and yT . Of course, P and P−1 are not needed to determine such a pair, but since P T and P−1 have been computed above, we can use X1 and Y1 to make the point that any right-hand and left-hand eigenvectors associated with λ1 = 1 T will do the job because they are all of the form x = αX1 and yT = β Y1 for α = 0 = β. Consequently, ⎛ ⎞ 1 ⎞ α ⎝ 2 ⎠ β ( 1 −1 −1 ) ⎛ 1 −1 −1 T −2 xy G1 = T = = ⎝ 2 −2 −2 ⎠ . yx αβ −2 2 2 P O C Invoking (7.2.10) yields the other spectral projector as G2 = I − G1 . 4. An even easier solution is obtained from the spectral theorem by writing A − I = (1G1 − 3G2 ) − (G1 + G2 ) = −4G2 , A + 3I = (1G1 − 3G2 ) + 3 (G1 + G2 ) = 4G1 , Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 520 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 so that G1 = (A + 3I) 4 and G2 = −(A − I) . 4 Can you see how to make this rather ad hoc technique work in more general situations? 5. In fact, the technique above is really a special case of a completely general formula giving each Gi as a function A and λi as It is illegal to print, duplicate, or distribute this material Please report violations to [email protected] k Gi = (A − λj I) j =1 j =i k D E . (λi − λj ) j =1 j =i T H This “interpolation formula” is developed on p. 529. Below is a summary of the facts concerning diagonalizability. IG R Summary of Diagonalizability For an n × n matrix A with spectrum σ (A) = {λ1 , λ2 , . . . , λk } , the following sta...
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