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Unformatted text preview: MASZlZ Linear Algebra I Exercises 11 (with solutions) © Francis J. Wright, 2006 Do not submit your solutions to this ﬁnal coursework, which is not assessed and will not be marked.
Solutions are available from the course web site. 1. For each of the following matrices A, ﬁnd an orthogonal matrixP and a diagonal matrix A such
that P A PT = A. 022
12 b 2—2 202
a , ,c .
()21()_25 ()
220
VSolution
(a) eigenvalues := —1, 3 —1 1
ei envectors := ,
g 1 1 anew
Normalizing the eigenvectors gives PT = 1 1 and henceP=
35 35
L5 i5
2 2
_i5 iﬁ I
2 2
A is the diagonal matrix of eigenvalues
3 0
_ 0 —1
Check:
in in iﬁ _iﬁ
2 2 2 2 1 0
PPT= . =
_i5 L5 iﬁ i5 0 1
2 2 2 2
1 1 1 1
252512 2‘/7_2‘/7 30
PAPT: 1 1 ' 21 ' 1 1 = 0 1
‘3” 35 3” 3J7 (b) eigenvalues := 1, 6
2
1 ’ eigenvectors := [
1 %ﬁ%5
Normalizing the eigenvectors gives PT = 1 2 and hence P =
— J? — J?
5 5
2 J? i J?
5 5
—% J? %J? A is the diagonal matrix of eigenvalues 1 0
A:
0 6
Check:
35 iv? 15 Av?
S 5 5 5 1 0
PPT: 1 2 ' 1 2 = 01
——J? —¢? —¢? —¢?
5 5 5 5
2 1 2 1
T 5 J? 5 J? 2 —2 5 J? 5 J? l 0
PAP — . =
15 £5 ‘2 5 i5 £5 0 6
5 5 5 5
(C)
eigenvalues := 4, —2, —2
1 —l —1
eigenvectors := l , l , 0
1 0 1 Call the eigenvectors x1, x2, x3. The last two eigenvalues are identical and the corresponding
eigenvectors x2, x3 are linearly independent but not orthogonal, so replace the last eigenvector x3
by a linear combination of the last two eigenvectors x2, x3 that is orthogonal to )52 using x3 'x2 x3 := x3— x2 to give: x2 .x2 J? and hence P
i
3 _L
6 0 5&2
J? 1_31_3 __ w e .W . g m n a m _ o _
13 13 m a n 6 e m 1_31_21_6 g _ .m m E E 6 a m 1_31_21_6 O _ _ N A is the diagonal matrix of eigenvalues 0—20 400
00 A: —2 Check: 100 J?
Hl01o 1 1
3 2 2 6 L
3 J? L
3 J? L
3 J? L
3 _L
6 1
322 i
3 001 0 J? i
3 J?
PAPT 11
66JF3 _i
6 > Maple Solution 2. (a) Find an orthogonal matrix whose ﬁrst row is the vector [ L, i
J? J?
1 3 — — 10] and check it.
m, J and check it. (b) Find an orthogonal matrix whose ﬁrst column is the vector [ V Solution
An orthogonal matrix has orthonormal rows and columns. (a) We need a vector in the direction orthogonal to the direction of the given vector, namely
(1, 2). The orthogonal direction must be (2, —1) or (—2, 1). Normalizing this vector and making 1 2 1 2
it the second row of a matrix gives P1 = L or P2 = L .
J? 2 —l J? —2 1
2
1 l 2 1 0 T 1 1 2 1 —2 1 0
Check:PPT=P2=— = ,PP =— = .
11152—1 01225—21 21 01 (b) We need a vector in the direction orthogonal to the direction of the given vector, namely
(1, 3). The orthogonal direction must be (3, —1) or (—3, 1). Normalizing this vector and making 1 3 1 —3
1 orP2=L
3 —1 J10 J10 3 1 2
10 T1 1—3 13 10
_ ,P2P2=—[ :01 0 1 10 3 1 —3 1
3. LetA and B be symmetric matrices. Give a necessary and sufﬁcient condition forA B to be
symmetric. it the second column of a matrix gives P1 = 13
3—1 CheckzP1P1T=P12 = 11—0 > Maple Solution Solution IfA B is symmetric then (A B)T=A B. But (A B) T=BTAT =B A sinceA andB are symmetric.
Hence, a necessary and sufﬁcient condition forA B to be symmetric is that B A =A B, i.e. thatA
and B are square matrices of the same size that commute. 4. Show that Q QT and QT Q are both symmetric matrices, where Q is any m x n matrix. Solution
l(QQT)T=(QT)TQT=QQTand(QTQ)T=QT(QT)T=QTQ. 5. Prove that (A1A2An)T =AZ:A5_1 "A; where A1, A2,. . ., A" are any matrices such that the
productA1A2An is deﬁned. (State clearly any results you use.) Solution Suppose the result is true for n. Then 11(A1 A2 ”A" An +1)T =An +1 (A1 A2 A n)T Tbecause (A B)T =BT AT as proved in Chapter 11
=An + 1 ATA_1A1T by using the result for 11. Hence, the result is true for n + 1. It is clearly true for n = 1 so by induction it is true for all
positive integer n. 6. (a) Let Q be any m X n real matrix andx any vector in IR". Prove that the scalar xT QT Qx Z 0.
(b) LetZ be any m x n complex matrix andz be any vector in 0'. Let 2+ denote the transposed complex conjugate of 2, Le. 2*T, and  2“ = \I z z. Prove that the real scalar z+Z+Zz 2 0 and that if
Z is a square, nonsingular n x 11 matrix andz at 0 then z+Z+ Z 2 > 0. Solution (a) Lety = Qx, which is a vector in R". ThenyTy = My  2 and My  2 2 0 by the norm axioms.
Substituting the deﬁnition ofy gives 0 s yT y = ( Qx) T ( Qx) =xT QT Qx. (b) Lety =Zz, which is a vector in 0". Theny+y = My  2 and My  2 Z 0 by the norm axioms.
Substituting the deﬁnition of y gives 0 S y+y = (Z z) + (Z z) =z+Z+Zz. If Z is a square, non
singular n x 11 matrix andz at 0 then y at 0, because if y = 0 thenz =Z_1 y => 2 = 0, which is a
contradiction. If y at 0 then H y H 2 > 0 by the norm axioms, hence z+Z+ Z z > 0. ...
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 Spring '08
 McAllister
 Math

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