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Unformatted text preview: B U Department of Mathematics
Math 201 Matrix Theory
Spring 2006 Second Midterm
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it, without nonproﬁt purpose may result in severe civil and criminal penalties. 1. Let T : R3 → R3 be a linear transformation satisfying;
T (1, 0, 0) = (3 − k, −1, 0)
T (0, 1, 0) = (−1, 2 − k, −1)
T (0, 0, 1) = (0, −1, 3 − k ), where k is a real number.
What should k be so that the dimension of the range of T is 2? (15 points)
Solution:
⎤
3 − k −1
0
For the matrix A = ⎣ −1 2 − k −1 ⎦ of the given transformation, R(T), the range of T is
0
−1 3 − k
spanned by the column vectors of T. Since its dimension is 2, the column vectors are linearly dependent and so det(A) = 0.
Using cofactor expansion along the ﬁrst row gives
⎡ detA = (3 − k ) 2 − k −1
−1 3 − k +1 −1 −1
0 3−k = (3 − k )(k 2 − 5k + 5) − (3 − k )
= (3 − k )(k 2 − 5k + 4)
= (3 − k )(k − 4)(k − 1)
and detA = 0 if k = 1, 3, or 4.
⎡
⎤
101
2. Let A = ⎣ 0 1 1 ⎦.
11b
a. Find b so that A has zero as an eigenvalue;(6 points)
For λ = 0 to be an eigenvalue det(A − λI ) = detA = 0 must hold. Expanding along the ﬁrst column
we get
11
01
detA =
+
= b − 1 − 1 = b − 2 = 0 if b = 2
1b
11
b. Find the other eigenvalues and all the corresponding eigenvectors when b has the value found in
part (a) above. (15 points)
1−λ
0
1
1
0
1−λ
For b = 2, det(A − λI ) =
1
1
2−λ
if (1 − λ)[(1 − λ)(2 − λ) − 1] − (1 − λ) = 0
if (1 − λ)(λ2 − 3λ) = λ(1 − λ)(λ − 3) = 0
So λ = 0, 1 or 3 are eigenvalues. =0 i For λ =⎤ : ⎡ ⎤ ⎡
0
⎤⎡
⎤⎡
⎡
x1
101
0
x1
A ⎣ x2 ⎦=⎣ 0 ⎦ if ⎣ 0 1 1 ⎦ ⎣ x2 ⎦=⎣
000
0
x3
x3
x + x3 = 0
if 1
x2 + x3 = 0
if x1 = x2 = −x⎤
3
⎡
−1
⇒ x = a ⎣ −1 ⎦ , a ∈ R, a = 0 (eigenvectors
1 ⎤
0
0⎦
0 corresponding to λ = 0). For λ = 1: ⎡ ⎤⎡
⎤⎡ ⎤
x1
0
001
(A − I )x = ⎣ 0 0 1 ⎦ ⎣ x2 ⎦=⎣ 0 ⎦ if x3 = 0 and x1 + x2 = 0
0
1⎤ 1 1
x3
⎡
−1
⎣ 1 ⎦ , b ∈ R, b = 0 (eigenvectors corresponding to λ = 1).
⇒x=b
0
⎤⎡
−2 0
1
(A − 3I )x = ⎣ 0 −2 1 ⎦ ⎣
1
1 −1
⎡
⎤
⎡
−2 0
1
−2 0
⎣ 0 −2 1 ⎦ → ⎣ 0 0
if
0
1 −1
01
2
−2x1 + x3 = 0
if
x2 − 1 x3 = 0
2
if x1 = 1 x3 = ⎤ 2
x
2⎡
⎡ For λ = 3: ⇒ x = c⎣ 1
2
1
2 ⎤⎡
x1
x2 ⎦=⎣
x3
⎤⎡
1
0 ⎦⎣
−1
2 ⎤
0
0⎦
0
⎤⎡ ⎤
x1
0
⎦=⎣ 0 ⎦
x2
x3
0 ⎦ , c ∈ R, c = 0 (eigenvectors corresponding to λ = 3). 1 3. Decide whether the followings are TRUE or FALSE. If true prove; if false, give a counter example
or explain.(15 points)
i. For A and B be both invertible nxn matrices (A + B )−1 = A−1 + B −1 ;
10
−1 0
,B=
, A and B are invertible since detA = detB = 1 = 0
01
0 −1
00
But A + B =
is not invertible since det(A + B ) = 0.
00
So (A + B )−1 is meaningless since A+B is not invertible.
FALSE: For A = ii. Transformation T : R3 → R3 deﬁned by T (x, y, z ) = (x, 2) is linear.
FALSE: Since T (0, 0, 0) = (0, 2), ie T (0) = 0, hence T is not linear. ii ⎡ ⎤
⎡
⎤
1
1
iii. The orthogonal projection of u = ⎣ 1 ⎦ on the subspace of R3 spanned by v = ⎣ −1 ⎦ is u.
−1
0
FALSE: Since uT v = 1 − 1 + 0 = 0 we have u⊥v . So the orthogonal projection of u on the subspace
of R3 spanned by v (ie on v line) must be the zero vector. 4. Let A = [aij ] be an nxn matrix such that a1,n = 1, a2,n−1 = 1, . . . , an,1 = 1 and ai,j = 0 otherwise;
i. Write down the matrix A.
⎤
⎡
0 ···
1
⎥
⎢.
.
⎥
⎢.
A=⎢
.⎥
.⎦
⎣
.
1
1
··· 0
ii. Evaluate the determinant of A in the case that n=7.
detA = (−1)(−1)(−1)detI
= −1
since we exchange r1 ↔ r7 , r2 ↔ r6 , r3 ↔ r5 in order to obtain the identity matrix I from A.
iii. Find a formula for the determinant of A in case that n is an arbitrary positive integer.
Consider the number of row exchanges to reduce A to I.
For n > 2, if n is even, then n row exchanges are needed. If n is odd, then n−1 row exchanges are
2
2
needed.
If the number of row exchanges is even then detA=1. If it is odd, then detA=1.
So for k ∈ Z+
if n = 4k then n = 42k = 2k even ⇒ detA = 1
2
k
if n = 4k + 1, then n−1 = 42 = 2k even ⇒ detA = 1
2
if n = 4k + 2, then n = 2k + 1 odd ⇒ detA = −1
2
if n = 4k + 3, then n−1 = 2k + 1 odd ⇒ detA = −1
2
In fact, if n ≡ 0 or 1 (mod4) ⇒ detA = 1 and if n ≡ 2 or 3 (mod4) ⇒ detA = −1. ⎤
x1
⎢x ⎥
5. Let W be the subspace of R4 containing all vectors ⎢ 2 ⎥ satisfying x1 + x2 + x3 + x4 = 0. Find
⎣ x3 ⎦
x4
⊥
a basis for W , the orthogonal complement of W. (16 points)
⎡ Solution: iii ⎡ −x2 − x3 − x4
⎢
x2
W = {⎢
⎣
x3
x
⎤ ⎡4 ⎤
⎡
−1
−1
⎢1⎥⎢0⎥
⎥
⎥⎢
= S(⎢
⎣ 0 ⎦, ⎣ 1 ⎦,
0
0 ⎤
⎥
⎥ : x2 , x3 , x4 ∈ R}
⎦
⎤
⎡
−1
⎢
⎢0⎥
⎥= R(A), columnspaceof A = ⎢
⎢
⎣
⎣0⎦
1
⎡ ⎤
−1 −1 −1
1
0
0⎥
⎥
0
1
1⎦
0
0
1 We know that R(A)⊥ = N(A⊥) .
⎡
⎤
⎡
⎤
−1 1 0 0 − − − − − − − − − − − − − − −
1 −1 0 0
− − − − − − − − − − − − − −→
AT = ⎣ −1 0 1 0 ⎦ (−1)r1 → r1 , r1 + r2 → r2 , r1 + r3 → r3 ⎣ 0 −1 1 0 ⎦
−1 0 0 1
0 −1 0 1
⎤
⎡
− − − − − − − − − → 1 −1 0 0
−−−−−−−−−
(−1)r2 → r2 , r2 + r3 → r3 ⎣ 0 1 −1 0 ⎦
0 0 −1 1
⎡
⎡
⎤
⎤⎡ ⎤
⎤ x1
⎡
x1
0
1 −1 0 0 ⎢
⎢ x2 ⎥
⎥ ⎢0⎥
T
⎥
⎦ ⎢ x2 ⎥ ⎢ ⎥
⎣
So x = ⎢
⎣ x3 ⎦ ∈ N(A ) if 0 1 −1 0 ⎣ x3 ⎦=⎣ 0 ⎦
0 0 −1 1
x4
x4
0
if x1 − x2 = x2 − x3 = −x3 + x4 = 0
if x3 = x2 = x1 = x4
⎤
⎡
1
⎢1⎥
⎢
Hence N(AT ) = {x4 ⎢ ⎥ : x4 ∈ R} = S(⎢
⎣1⎦
⎣
1
⎡ 6. ⎡ 01
⎣1 0
i. Find QRdecomposition of A=
00
triangular matrix.(17 points)
Solution: ⎤
1
1⎥
⎥)= W ⊥ .
1⎦
1 1
√
2 ⎤ 0 ⎦, where Q is an orthogonal matrix and R is an upper 1
√
2 ⎤
⎡⎤
⎡1⎤
√
0
1
2
The vectors a= ⎣ 1 ⎦, b= ⎣ 0 ⎦, c= ⎣ 0 ⎦ are linearly independent.
1
√
0
0
2
⎡ Take q1 = a and q2 = b, since a and b have unit lengths and a⊥b. iv c T
T
= c − (q1 c)q1 − (q2 c)q2
= c − (aT c)a − (bT c)b
⎡⎤
⎡1⎤
√
1
2
1
= ⎣ 0 ⎦ − √2 ⎣ 0 ⎦
1
√
0
⎡ 2⎤
0
⎣ 0 ⎦ sinceaT c = 0, bT c =
=
1
√
2 ⇒ q3 = c
c 1
√
2 ⎡ ⎤⎡⎤
0
0
√
= 2⎣ 0 ⎦ = ⎣ 0 ⎦
1
√
1
2 T
T
T
q1 a = 1 q1 b = 0 q1 c = 0
1
T
T
q2 b = 1 q2 c = √2
Now
1
T
q3 c = √2 ⎡ 01
⎣1 0
Hence A=
00 ⎤⎡
10
010
⎦= ⎣ 1 0 0 ⎦ ⎣ 0 1
0
1
√
00
001
2
1
√
2 ⎤ ⎡ 0 1
√
2
1
√
2 ⎤
⎦= QR ii. Find the inverse of the matrix Q found in part i) above. (4 points)
Solution:
Q−1 = QT ⇒ Q−1 ⎡ ⎤
010
=⎣ 1 0 0 ⎦=Q
001 v ...
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This note was uploaded on 11/16/2011 for the course MATH 201 taught by Professor Soysal during the Fall '08 term at Boğaziçi University.
 Fall '08
 SOYSAL
 Math

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