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Unformatted text preview: B U Department of Mathematics
Math 201 Matrix Theory
Summer 2003 Final Exam
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it, without nonproﬁt purpose may result in severe civil and criminal penalties. 1. For what values of a does the system :
ax + y = 1
4x + ay = 2
have (i) a unique solution (ii) inﬁnitely many solutions (iii) no solution? Find also the rank of
the coeﬃcient matrix in each case.
Solution:
To have a unique solution the coeﬃcient matrix A= a1
4a must be nonsingular since the system is square.
detA = a2 − 4 = 0 ⇒ a = 2, a = −2
(i) unique solution if a = ±2 , which is :
x
y = A− 1 1
2 In this case rankA = 2
(ii) if a = 2 :
2x + 1 = 1
4x + 2y = 2 Aug = 211
422 −−−> 211
000 2x + y = 1. One equation two unknowns ⇒inﬁnitely many solutions.
In this case rankA = 1
−2 1  1
4 −2  2
sistent and hence no solution. (iii) if a = −2 : Aug = In this case 00
3 2
2. Let A=
41 −−− > −2 1  1
0 04 rankA = 1 1
2 .
3 (a) Find QRdecomposition of A, where Q is an orthogonal matrix.
Solution: . Incon Q is the orthogonal matrix obtained from A by GramSchmidt process
α1 =< 0, 3, 4 >, α2 =< 0, 2, 1 > , α3 =< 1, 2, 3 > labelling the columns
x1 = α1
αT x
2
x2 = α2 − x1 1 x1 =< 0, 4 , −3 >
2
55
αT x αT x 3
3
x3 = α3 − x1 1 x1 − x2 2 x2 =< 1, 0, 0 >
2
2
x1 , x2 , x3 is an orthogonal set. Then q1 = q2 = 34
x1
=< 0, , >
x1 
55 x2
4 −3
=< 0, ,
>
x2 
55 q3 = x3
=< 1, 0, 0 >
x3  form an orthonormal set. 0
0
1
Q = [q1 q2 q3 ] = 3/5 4/5 0 .
4/5 −3 0 /5 T
T
T
5 2 18/5
q1 α1 q1 α2 q1 α3
T
T
q2 α2 q2 α3 = 0 1 −1/5 .
R= T
q3 α3
00
1
So that A = QR.
(b) Find the inverse of Q, if it exists.
Solution:
Since Q is orthogonal, we have QT Q = QQT = I . Hence Q−1 = QT .
3. (a) Show that if A is similar to B , then Ak is similar to B k Solution:
Given that there exists an invertible M such that M −1 AM = B or A = M BM −1 ,
compute Ak
Ak = (M BM −1 )(M BM −1 )...(M BM −1 )[ktimes] ⇒ Ak = M B k M −1 .
Hence Ak is similar to B k , via the same matrix M .
(b) Let A be an m × n matrix. Prove that if tr(AT A) = 0 then A = 0.
Solution: Let A = [q1 q2 ...q3 ]mxn
T
T
T
q1 q1 q1 q2 ... q1 qn
T q T q q T q ... q2 qn
Hence, AT A = 2 1 2 2
:
:
T
T
T
qn q1 qn q2 ... qn qn T ⇒ (AT A)ii = qi qi = qi 2 T
T
T
tr(AT A) = q1 q1 + q2 q2 + ... + qn qn = q1 2 + ... + qn 2 tr(AT A) = 0 ⇒ qi = 0 for every i = 1, 2, ..., n
So each column of A is zero. Therefore A = 0. 2 −1 3
−1
6 has an eigenvector v1 = −2 . It is also known that
4. The matrix A = 0 0
0 3 −7
3
λ = 2 is an eigenvalue of A.
(a) Using the information, diagonalize A.
Solution:
If v1 is an eigenvector, then there is an eigenvalue λ such that 2 −1 3
−1
−1
9
−1
6 −2 = λ −2 ⇒ 18 = λ −2 Av1 = λv1 ⇒ 0 0
0 3 −7
3
3
−27
3
Therefore another eigenvalue is λ = −9, with an eigenvector v1
λ = 2 is an eigenvalue(given). Let us ﬁnd 0 −1 3
0
A − 2I = 0 −2 6 − −− > 0
0 3 −9
0 the eigenvector(s) for λ = 2 : −1 3
0 0 ⇒ −x2 + 3x3 = 0
00 x1
1
0 3x3 = x1 0 + x3 3 x1 is free.⇒ (A − 2I )x = 0 is satisﬁed if x =
x3
0
1
There are eigenvectors for λ = 2. Taking these two eigenvectors to be : two 1
0
v2 = 0 and v3 3 , We can diagonalize A
0
1 
−1 1 0
−9 2
S −1 AS = Λ where S = v1 v2 v3 = −2 0 3 and = 
3 01
2
(b) Find A2003 . (Leave it as a product.)
Solution:
A = S ΛS −1 ⇒ A2003 = S Λ2003 S −1 − 1
−1 1 0
−92003
−1 1 0 −2 0 3 22003
= −2 0 3 2003
3 01
2
3 01 ⇒ A2003 5. Let T : R2 − > R2 be a linear transformation satisfying T (< 1, 0 >) =< −4, 3 > and
T (< 1, 1 >) =< −10, 8 > . Let A be a matrix of T in the standart basis. (a) Find A. Solution:
We need T (< 0, 1 >). But < 0, 1 >=< 1, 1 > − < 1, 0 >
⇒ T (< 0, 1 >) = T (< 1, 1 >) − T (< 1, 0 >)
⇒ T (< 0, 1 >) =< −10, 8 > − < −4, 3 >=< −6, 5 >
−4 −6
⇒ A = [T e1 T e2 ] =
3
5
(b) What is the matrix B representing T in the basis that consist of eigenvectors of A ?
Solution:
Eigenvalues of A : A − λI  = −4 − λ −6
3
5−λ = (5 − λ)(−4 − λ) + 18 = 0 ⇒ λ = 2, λ = −1
λ = 2 : A − 2λ = −6 −6
3
3 ⇒ x1 = −x2 ⇒ an eigenvector is p1 = λ = −1 : A + λ = −3 −6
3
6 ⇒ x1 = −2x2 ⇒ an eigenvector is p2 = T (p1 ) = −4 −6
3
5 1
−1 = 2
−2 −4 −6
3
5 −2
1 = 2
−1 −2
1 = 2p1 + 0p2 T (p2 ) = 1
−1 = 0p1 − 1p2 B = [T p1 T p2 ] = 2
2
−2 −1 (c) Solve the system of diﬀerential equations
−1
u0 =
−2 du
dt = Au with the initial condition Solution:
du
dt = Au has the general solution : u = SeΛt S −1 u0 S= 1 −2
−1 1 e2t
e−t (found above, eigenvectors of A.) Λ = 2
−1 ⇒ eΛt = S −1 = −1 −2
−1 −1 , c = S −1 u0 = ⇒ u = c1 eλ1 t p1 + c2 eλ2 t p2
−2
1
u = −e2t
− e−t
−1
1 −1 −2
−1 −1 1
0 = −1
−1 = c1
c2 u1
u2 . Let u = u1 = −e2t + 2e−t
u2 = +e2t − e−t
6. Let A be square matrix with eigenvalues λ1 = 2, λ2 = 1 and λ3 = 5 with multiplicities 3, 2 and
2 respectively. Let Ei be the eigenspace associated with the eigenvalue λi , i = 1, 2, 3. Assume
that dimE1 = 2, dimE2 = 1 and dimE3 = 2
(a) Is A diagonalizable? Explain.
Solution:
Counting the multiplicities, we understand that A is 7x7.
dimE1 + dimE2 + dimE3 = 2 + 1 + 2 = 5 = 7. Hence A is not diagonalizable.
(b) Give a Jordan form J of A.
Solution:
λ = (0), (0, 0) 2 eigenvectors.
λ = (1, 1) 1 eigenvector.
λ = (5), (5) 2 eigenvectors.
J contains 5 blocks. 01 0 [0] 11
J = 1 [5]
[5] 7 x7 (c) Write down the characteristic polynomial of A.
Solution:
Using the fact that eigenvalues are roots of the characteristic polynomial
p(λ) = −λ3 (λ − 1)2 (λ − 5)2 (the front minus sign comes from the order of A)
(d) Is A invertible?Justify your answer fully.
Solution: To be invertible, detA = 0 has to be satisﬁed. But A has λ = 0 as an eigenvalue and
detA = λ1 λ2 ...λ7 which yields that detA = 0. The answer is no.
(e) Find the trace of A.
Solution:
trA = λ1 + λ2 + ... + λ7 = 0 + 0 + 0 + 1 + 1 + 5 + 5 = 12 ...
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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
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