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Math 24

# Math 24 - 26 If*2 1-1 13 Ifa2x2 matrix A has...

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Unformatted text preview: 26. If *2 1 -1. 13. Ifa2x2 matrix A has asingleeigenvalueA1,tlientl12e ' A = 1 _3 0 ’ characteristic polynomial must be p(}k) = (it—A1) . 3 _.5 0 According to Cayley’s Theorem, any matrix must : then - ’ satisfy its characteristic equation, so (A -—}q I )2 = 0. Further, we know that We know that 170») = — det(A — A1) 611: : A11t+(A—A|I)r __2 ~_ A l __1 = — 1 —3 e A o = el1lte(A"‘)Ll)t 3 3 *5 _)\ = eA111[I + (A -— MDt Expanding down the third column, 1 2 2 ] : i 1 _3 ~ A +—A—A)t+---. .. A: ~2—A 1 2( 1 , ,p(> 13 ‘5 [+1 1 4% But, from the above, (A — A] N = 0, for k 2 2. ‘ = (4 + 3A) + M12 + 5A + 5) Hence, it ‘ = A3 - 5A2 ~ 8A ~ 4, em = ehir [I + (A —- A11 )t] . ‘ Zeros must be factors of the constant term, so ——1 is a possibility. Dividing by A + 1 leads to the following 14 Matrix facton'zation ' ~2 1 A = (—1 0)’ 17(1) = (A+ l)(}t+2)2 has characteristic polynomial pot) = ()t + 1)2 and and eigenvalues A1 :: -1 and 12 = “2. Because repeated eigenvalue A 2 —~1. We can write “1 1 ~1 1 0 2 em : et(—I+(A+I)) A + I = l _2 0 fig 0 1 1. ze—tlet(A+I) 3 _5 1 p . O O 0 , t2 2 =e”‘ I+t(A+1)+—2-7(A+I) +... . Matrix A must satisfy its characteristic polynomial, the geometric multiplicity of A1 = —1 is one, and so (A + I)2 = O and (A + I)" : 0 fork 2 2. Thus, anveigenvector is v1 = (_2, _1, 1)T, leading to the solution _ I’__2\ Y1(t)=€mV1 = e" ——l . 1 i the series truncates. N ext, TV} . ‘ ’e 1 4" f ’1 0 —i‘ ’1 11)] A+21= 1 -1 0 31> 01 —1 , ’1 3 —5 2 0 0 0 .. ev—t (1 " t t ) the geometric multiplicity of A2 =: —2 is one, and _ “t 1 +1: an eigenvector is v20) 2 (1, 1, HT, leading to the solution 1 3’20) = 6”sz = 6—2’ 1 1 Notice that 2 ~2 4 — [ref 1 —2 1 (A +21) = ~1 2 ——‘l —-—> 0 0 0 l ——2 l 0 0 0 has dimension two, equalling the algebraic multiplic- ity of A2. Thus, we can pick a vector in the nullspace of (A + 21)2 that is not in the nullspace of A + 2i. Choose v3 = (—1, 0, 1)T, which is not a multiple of ...
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