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11
ing 2 1 D 0 as the CayleyHamilton Theorem predicts.
When A D SƒS 1 is diagonalizable, the matrix A j I D S.ƒ j I /S 1 will have
0 in the j; j diagonal entry of ƒ j I . In the product p.A/ D .A 1 I / .A n I /,
each inside S 1 cancels S . This leaves S times (product of diagonal matrices ƒ j I )
times S 1 . That product is the zero matrix because the factors produce a zero in each
diagonal position. Then p.A/ D zero matrix, which is the CayleyHamilton Theorem.
(If A is not diagonalizable, one proof is to take a sequence of diagonalizable matrices
approaching A.)
Comment I have also seen this reasoning but I am not convinced:
Apply the formula AC T D .det A/I from Section 5.3 to A I with variable . Its
cofactor matrix C will be a polynomial in , since cofactors are determinants:
.A I / cof .A I /T D det.A I /I D p./I: “For ﬁxed A, this is an identity between two matrix polynomials.” Set D A to ﬁnd
the zero matrix on the left, so p.A/ D zero matrix on the right—which is the CayleyHamilton Theorem.
I am not certain about the key step of substituting a matrix for . If other matrices
B are substituted, does the identity remain true? If AB ¤ BA, even the order of
multiplication seems unclear : : :
33 D 2; 1; 0 are in ƒ and the eigenvectors are in S (below). Ak D S ƒk S " 2
1
1 1
1
1 # 0
1
1 ƒk
6
1 " 2
2
0 1
2
3 # 1
2k
2D
6
3 " 4
2
2 # 22
. 1 /k
11C
3
11 " 1
1
1 1 1
1
1 is
1
1
1 # Check k D 4. The .2; 2/ entry of A4 is 24 =6 C . 1/4 =3 D 18=6 D 3. The 4step paths
that begin and end at node 2 are 2 to 1 to 1 to 1 to 2, 2 to 1 to 2 to 1 to 2, and 2 to 1 to
3 to 1 to 2. Much harder to ﬁnd the eleven 4step paths that start and end at node 1. Solutions to Exercises 64 34 If AB D BA, then B has the same eigenvectors .1; 0/ and .0; 1/ as A. So B is also diagonal b D D 0. The nullspace forthe following equation is 2dimensional:
c
10 ab
ab
10
0
b
00
AB BA D
D
D
. The
02 cd
cd
02
c
0
00
coefﬁcient matrix has rank 4 2 D 2.
p
35 B has D i and i , so B 4 has 4 D 1 and 1 and B 4 D I . C has D .1 ˙ 3i /=2.
This is exp.˙ i=3/ so 3 D 1 and 1. Then C 3 D I and C 1024 D C .
cos
sin
36 The eigenvalues of A D
are D e i and e i (trace 2 cos and
sin
cos
det D 1). Their eigenvectors are .1; i / and .1; i /:
1 1 e i n
i
1
An D Sƒn S 1 D
=2i
ii
i
1
e i n
i n
.e
C e i n /=2
cos n
sin n
D
D
:
sin n
cos n
.e i n e i n /=2i
Geometrically, n rotations by give one rotation by n .
37 Columns of S times rows of ƒS 1 will give r rank1 matrices .r D rank of A/. 38 Note that ones.n/ ones.n/ D n ones.n/. This leads to C D 1=.n C 1/. AA 1 D .eye.n/ C ones.n// .eye.n/ C C ones.n// D eye.n/ C .1 C C C C n/ ones.n/ D eye.n/: Problem Set 6.3, page 325
1 u1 D e 4t
1
1
1
t
4t 1
t
, u2 D e
. If u.0/ D .5; 2/, then u.t / D 3e
C 2e
.
0
1
0
1 2 z.t / D 2e t ; then dy=dt D 4y Problem 1. 6e t with y.0/ D 5 gives y.t / D 3e 4t C 2e...
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This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.
 Spring '12
 Minki
 Mass

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