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determinant C1 for that permutation matrix. More than that, all other P that permute a,
b and separately c , d , e will come out with the correct sign when the 2 by 2 determinant
for columns a; b multiplies the 3 by 3 determinant for columns c; d; e .
41 The CauchyBinet formula gives the determinant of a square matrix AB (and AAT in particular) when the factors A, B are rectangular. For (2 by 3) times (3 by 2) there are
3 products of 2 by 2 determinants from A and B (printed in boldface):
"g j #
"g j #
abc
hk
hk
def
i`
i`
#
"
11
14 30
AB D
BD 2 4
Check
30 66
37
CauchyBinet: .4 2/.4 2/ C .7 3/.7 3/ C .14 12/.14 12/ D 24
.14/.66/ .30/.30/ D 24 a
d b
e "g j #
hk
i`
123
AD
147
c
f a
d b
e c
f Problem Set 6.1, page 293
1 The eigenvalues are 1 and 0:5 for A, 1 and 0:25 for A2 , 1 and 0 for A1 . Exchanging the rows of A changes the eigenvalues to 1 and 0:5 (the trace is now 0:2 C 0:3/.
Singular matrices stay singular during elimination, so D 0 does not change. 2 A has 1 D 1 and 2 D 5 with eigenvectors x1 D . 2 ; 1/ and x2 D .1; 1/. The
matrix A C I has the same eigenvectors, with eigenvalues increased by 1 to 0 and 6.
That zero eigenvalue correctly indicates that A C I is singular. 3 A has 1 D 2 and 2 D x 2 D .2; 1/. A 1 1 (check trace and determinant) with x 1 D .1; 1/ and
has the same eigenvectors, with eigenvalues 1= D 1 and 1.
2 Solutions to Exercises 59 4 A has 1 D 3 and 2 D 2 (check trace D 1 and determinant D 6) with x1 D
.3; 2/ and x2 D .1; 1/. A2 has the same eigenvectors as A, with eigenvalues 2 D 9
1
and 2 D 4.
2 5 A and B have eigenvalues 1 and 3. A C B has 1 D 3, 2 D 5. Eigenvalues of A C B are not equal to eigenvalues of A plus eigenvalues of B . 6 A and B have 1 D 1 and 2 D 1. AB and BA have D 2 ˙ p 3. Eigenvalues of AB
are not equal to eigenvalues of A times eigenvalues of B . Eigenvalues of AB and BA
are equal (this is proved in section 6.6, Problems 1819). 7 The eigenvalues of U (on its diagonal) are the pivots of A. The eigenvalues of L (on its diagonal) are all 1’s. The eigenvalues of A are not the same as the pivots.
8 (a) Multiply Ax to see x which reveals (b) Solve .A I /x D 0 to ﬁnd x . 9 (a) Multiply by A: A.Ax / D A.x / D Ax gives A x D 2 x
2 A 1 : x D A 1 Ax D A
.A C I /x D . C 1/x . 1 x D A 1 x gives A 1 xD 1 x (b) Multiply by
(c) Add I x D x : 10 A has 1 D 1 and 2 D :4 with x 1 D .1; 2/ and x 2 D .1; 1/. A1 has 1 D 1 and 2 D 0 (same eigenvectors). A100 has 1 D 1 and 2 D .:4/100 which is near zero.
So A100 is very near A1 : same eigenvectors and close eigenvalues. 11 Columns of A 1 I are in the nullspace of A 2 I because M D .A 2 I /.A 1 I / D zero matrix [this is the CayleyHamilton Theorem in Problem 6.2.32]. Notice that
M has zero eigenvalues .1 2 /.1 1 / D 0 and .2 2 /.2 1 / D 0. 12 The projection matrix P has D 1; 0; 1 with eigenvectors .1; 2; 0/, .2; 1; 0/, .0; 0; 1/. Add the ﬁrst and last vectors: .1; 2; 1/ also has D 1. Not...
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 Spring '12
 Minki
 Mass

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