Unformatted text preview: indeﬁnite ( < 0 and
> 0) because the determinant ac b 2 is negative.
If c > 9 the graph of z is a bowl, if c < 9 the graph has a saddle point. When c D 9
the graph of z D .2x C 3y/2 is a “trough” staying at zero along the line 2x C 3y D 0.
Orthogonal matrices, exponentials e At , matrices with det D 1 are groups. Examples
of subgroups are orthogonal matrices with det D 1, exponentials e An for integer n.
Another subgroup: lower triangular elimination matrices E with diagonal 1’s.
A product AB of symmetric positive deﬁnite matrices comes into many applications.
The “generalized” eigenvalue problem K x D M x has AB D M 1 K . (often we use
eig.K; M / without actually inverting M .) All eigenvalues are positive:
AB x D x gives .B x /T AB x D .B x /T x: Then D x T B T AB x =x T B x > 0: p
p
3; 2 1; 2; 2 C 1; 2 C 3. The
2 cos k D 2
6
product of those eigenvalues is 6 D det K .
35 Put parentheses in x T AT CAx D .Ax /T C.Ax /. Since C is assumed positive deﬁnite,
this energy can drop to zero only when Ax D 0. Sine A is assumed to have independent
columns, Ax D 0 only happens when x D 0. Thus AT CA has positive energy and is
positive deﬁnite. 34 The ﬁve eigenvalues of K are 2 My textbooks Computational Science and Engineering and Introduction to Applied Mathematics start with many examples of AT CA in a wide range of applications.
I believe this is a unifying concept from linear algebra. Problem Set 6.6, page 360
so M D F G 1 . C similar to A and B ) A similar to B .
10
30
01
1
2 AD
is similar to B D
D M AM with M D
.
03
01
10 1 B D GC G 1 D GF 1 AF G 1 Solutions to Exercises 72
1
3 BD
0
1
BD
1
4
BD
2
1
0
10
10 10
D
DM
0
11
10 11
1
1
1
0
11 10
D
;
1
0
1
11 0
1
1
3
01
12 01
D
.
1
10
34 10 4 A has no repeated so it can be diagonalized: S 1
5
0
1
0
1
00
10
0
,
,
,
0
11
10
0
0
0
is by itself and also
1
1 1 AM ; 1 AS D ƒ makes A similar to ƒ.
1
are similar (they all have eigenvalues 1 and 0).
1
1
is by itself with eigenvalues 1 and 1.
0
6 Eight families of similar matrices: six matrices have D 0, 1 (one family); three matrices have D 1, 1 and three have D 0, 0 (twop
families each!); one has D
1
1, 1; one has D 2, 0; two matrices have D 2 .1 ˙ 5/ (they are in one family). 1
AM /.M 1 x / D M 1 .Ax / D M 1 0 D 0
(b) The nullspaces of A and
of M 1 AM have the same dimension. Differentvectors and different bases.
Same ƒ
01
0 2 have the same line of eigenvectors
But A D
and B D
Same S
00
0 0 and the same eigenvalues D 0; 0.
12
13
1k
10
11
A2 D
, A3 D
, every Ak D
. A0 D
and A 1 D
.
01
01
01
01
01
2
k
1
c
2c
c
kc k 1
c
c2
2
k
0
1
JD
and J D
; J D I and J D
.
0 c2
0
c1
0
ck
du
dv
5
v.0/
1
u.0/ D
D
. The equation
D
u has
D v C w and
2
w.0/
0
dt
dt
dw
D w . Then w.t / D 2e t and v.t / must include 2t e t (this comes from the
dt
repeated ). To match v.0/ D 5, the solution is v.t / D 2t e t...
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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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