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Unformatted text preview: C 5e t .
2
3
2
3
m21 m22 m23 m24
0 m12 m13 0
60
6 0 m22 m23 0 7
0
0
07
7
6
7
If M 1 JM D K then JM D6
4 m41 m42 m43 m44 5 D MKD4 0 m32 m33 0 5.
0
0
0
0
0 m42 m43 0
That means m21 D m22 D m23 D m24 D 0. M is not invertible, J not similar to K .
The ﬁve 4 by 4 Jordan forms with D 0; 0; 0; 0 are J1 D zero matrix and
2
3
2
3
0100
0100
60 0 0 07
60 0 1 07
J2 D 4
J3 D 4
0 0 0 05
0 0 0 05
0000
0000
2
3
2
3
0100
0100
60 0 0 07
60 0 1 07
J4 D 4
J5 D 4
0 0 0 15
0 0 0 15
0000
0000 7 (a) .M
8
9
10
11 12 13 Solutions to Exercises 73 Problem 12 showed that J3 and J4 are not similar, even with the same rank. Every
matrix with all D 0 is “nilpotent” (its nth power is An D zero matrix). You see
J 4 D 0 for these matrices. How many possible Jordan forms for n D 5 and all D 0?
1 Ji Mi D MiT in each block
(2) M0 has those diagonal blocks Mi to get M0 JM0 D J T . (3) AT D .M 1 /T J T M T
equals .M 1 /T M0 1 JM0 M T D .MM0 M T / 1 A.MM0 M T /, and AT is similar to A. 14 (1) Choose Mi D reverse diagonal matrix to get Mi
1 AM I / D det.M 1 AM M 1 IM /. This is det.M 1 .A I /M /.
By the product rule, the determinants of M and M 1 cancel to leave det.A I /.
ab
dc
ba
cd
16
is similar to
;
is similar to
. So two pairs of similar
cd
ba
dc
ab
10
01
matrices but
is not similar to
: different eigenvalues!
01
10
15 det.M 1 17 (a) False: Diagonalize a nonsymmetric A D SƒS 1 . Then ƒ is symmetric and similar
01
0
1
and
are similar
10
1
0
(d) True: Adding I increases all eigenvalues by 1 (b) True: A singular matrix has D 0. (c) False:
(they have D ˙1) 18 AB D B 1 .BA/B so AB is similar to BA. If AB x D x then BA.B x / D .B x /. 19 Diagonal blocks 6 by 6, 4 by 4; AB has the same eigenvalues as BA plus 6
20 (a) A D M to B 2 .
3
(c)
0
3
(d)
0 1 2 1 1 1 2 4 zeros. 2 BM ) A D .M BM /.M BM / D M B M . So A is similar
(b) A2 equals . A/2 but A may not be similar to B D A (it could be!).
30
1
because 1 ¤ 2 , sothesematrices are similar.
is diagonalizableto
04
4
1
has only one eigenvector, sonot diagonalizable (e) PAP T is similar to A.
3 21 J 2 has three 1’s down the second superdiagonal, and two independent eigenvectors for #
010
01
.
D 0. Its 5 by 5 Jordan form is
with J3 D 0 0 1 and J2 D
00
J2
000
Note to professors: An interesting question: Which matrices A have (complex) square
roots R2 D A? If A is invertible, no problem. But any Jordan blocks for D 0 must
have sizes n1 n2 : : : nk nk C1 D 0 that come in pairs like 3 and 2 in this
example: n1 D (n2 or n2 C 1) and n3 D (n4 or n4 C 1) and so on.
#"
#
"
a10
a00
A list of all 3 by 3 and 4 by 4 Jordan forms could be 0 b 0 , 0 a 0 ,
00c
00b
3
2
"
#
a1
a10
(for any numbers a; b; c )
a
6
7
0a1
5,
b
with 3; 2; 1 eigenvectors; diag.a; b; c; d / and 4
00a
c
3
32
32
2
a1
a1
a1
a1
a1
a
76
7
6
76
with 4; 3; 2; 1 eigenvectors.
,
5, 4
4
a 15
a
b 15 4
b
a
b
J3 " Solutions to Exercises 74 I / must be just n . The CayleyHamilton Theorem in Problem 6.2....
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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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