Unformatted text preview: 17,3 14. 24. 28. FURMﬁ. E {W U 1 U
A1:0,A1:A3:1,eigenba5is:r 1 , —5 , 2 0 a b 0 a 0
20. F01" A1 = 1, E1 = her 0 0 c = ker 0 ﬂ .1 so ifa = (J then E1 is 2dimeneional,
0 O 1 0 0 0
otherwise it is 1dimensional.
1 —a. —b
For A2 : 2, E3 = ker 0 1. —r: so E2 is 1dimensional.
O 0 0 Hence, there is an eigeubasis if a = 0. LetA: 0‘ b.Firstwewant “ b 2 : 2 ,or2a+b=2,‘2c+d=1. This
c of c d 1 1
condition is satisﬁed by all matrices of the form A = 2' i : g: . Next, we want there to be no other eigenvalue, besides 1I so that 1 must have an algebraic multiplicity of 2. We want the characteristic polynomial to be {A — 1)2 = A2 — 2A + 1, so that the trace
must be 2, and u + (1 i 20) = 21 or, a = l + 2c. Thus we want a. matrix of the form _ 1 + 2c —4c
A # l c 1  2c '
Finally, we have to make sure the E; = span 3 instead of E1 = R2. This means that we must exclude the case A = I2. In order to ensure this, we state simply that A: [l—i—Zc —4c 0 1 2c] , where c is any nunzem cumtaut Since Jaw) is triangular, its eigenvalues are its diagonal entries, hence its only eigenvalue
is 36:. Moreover, 0 I U 0
U 0 1 '
E]; = ker(.]ﬂ (k) — kIn) = leer ' Q = span (El).
: : f 1
0 0 U 0 The geometric multiplicity of k: is 1 while its algebraic multiplicity is n. ...
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 Spring '08
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 Linear Algebra, Algebra

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