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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 meyer@ncsu.edu 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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