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1
Mid term exam 2 (Notes, books, and calculators are not authorized). Show all your work in the
blank space you are given on the exam sheet. Always justify your answer. Answers with
no justification will not be graded.
Question 1: The following λ1 := 1, v1 := (1, 1, 0), λ2 := −2, v2 := (1, 0, −1), and λ3 := 3, v3 :=
(1, 2, 3) are eigenpairs of a matrix A. (a) What are the fundamental solutions of the ODE
system du(t)
dt = Au(t)?
The fundamental solutions have been defined in class and are
e t v1 ,
e−2t v2 ,
and e3t v3 .
This is all I wanted you to write.
Recall that upon defining the matrix P := [v1T v2T v3T ], we have A = P DP −1 where D is the diagonal
matrix composed of the eigenvalues. Then the ODE can be re-written as follows:
d(P −1 u)
= DP −1 u.
dt
= Dw by defining w(t) := P −1 u(t). Then w(t) is given by


 
 
 
α1 et
1
0
0
w(t) = α2 e−2t  = α1 et 0 + α2 e−2t 1 + α3 e3t 1 ,
α3 e3t
0
0
0
This can also be re-written
dw
dt
where α1 , α2 , α3 are any real numbers. As a result,
 
 
0
1
u(t) = P w(t) = α1 et P 0 + α2 e−2t P 1 + α3 e3t P
0
0
 
0
1 = α1 et v1 + α1 e−2t v2 + α1 e3t v3 .
0
In conclusion, u(t) is a linear combination of the fundamental solutions et v1 , e−2t v2 , e3t v3 .
(b) What is the solution of
du(t)
dt
= Au(t), with u(0) = (6, 5, 1)?
We know that u(t) is a linear combination of the fundamental solutions:
u(t) = α1 et v1 + α2 e−2t v2 + α3 e3t v3 .
The condition u(0) = (6, 5, 1) implies
(6, 5, 1) = α1 v1 + α2 v2 + α3 v3 .
The corresponding augmented

 
1 1 1 6
1 1
1 0 2 5 ∼ 0 −1
0 −1 3 1
0 −1
matrix is
 
1 6
1 1
1 −1 ∼ 0 1
3 1
0 0
1
−1
2
 
6
1
1 ∼ 0
2
0
The solution is (α1 , α2 , α3 ) = (3, 2, 1). This means
u(t) = 3et v1 + 2e−2t v2 + e3t v3 .
1
1
0
0
0
1
 
5
1
2 ∼ 0
1
0
0
1
0
0
0
1

3
2 .
1
2
Mid term exam 2, November 13, 2014
Question 2: Let A be a n×n complex-valued Hermitian matrix (i.e., AT = A), n ≥ 2. We
define the complex inner product hv, wi := v T w and associated norm kvk2 = v T v.
(a) Show that hv2 , Av1 i = hAv2 , v1 i, for all vectors v1 , v2 in Cn . Give all the details.
hv2 , Av1 i = v2T Av1 = v2T Av1 = v2T AT v1 = (Av2 )T v1 = hAv2 , v1 i.
(b) Prove that the eigenvalues of A are real. Be very accurate in your statements.
Let (v, λ) be an eigenpair. Then
hAv, vi = (Av)T v = λv T v = λkvk2
and
hv, Avi = v T Av = λv T λv = λkvk2 .
From (a) we know that hAv, vi = hv, Avi; as a result
λkvk2 = λkvk2 ,
which implies λ = λ since v 6= 0.
(c) Let λ1 and λ2 be two different eigenvalues of A and let v1 , v2 be two associated eigenvectors.
Prove that v1 and v2 are orthogonal with respect to the complex inner product hv, wi := v T w.
We have
hv2 , Av1 i = hv2 , λ1 v1 i = v2T (λ1 v1 ) = λ1 v2T v1 , = λ1 hv2 , v1 i,
hAv2 , v1 i = hλ2 v2 , v1 i = (λ2 v2 )T v1 = λ2 v2T v1 = λ2 hv2 , v1 i.
From (a) and (b) we deduce that
0 = (λ1 − λ2 )hv2 , v1 i = (λ1 − λ2 )hv2 , v1 i,
which proves that v1 and v2 are orthogonal since λ1 6= λ2 .
1 7
Question 3: Consider the matrix A =
. Is it possible to find a set of two independent
7 −3
eigenvectors of A? Be very accurate and explicit when answering.
Since the matrix is symmetric, it is diagonalizable, which implies that, yes indeed, one can find a
set of two independent eigenvectors of A.
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Question 4: Consider the dynamical system
du
dt
=
3
2
1
4
u. Is this dynamical system stable?
−1
The two eigenvalues of the matrix are the roots of
0 = (2 − λ)(−1 − λ) − 4 = λ2 − λ − 6.
The two roots are λ1 = 3 and λ2 = −2. The dynamical system is unstable since λ1 > 0.
Question 5: Let du
dt = F (u) be five-dimensional nonlinear dynamical system. Assume that
u0 ∈ R5 solves F (u0 ) = 0 and assume that the dynamics in the neighborhood of u0 is described
A is 5 × 5 real-valued matrix with eigenvalues −1 + 3i, −1 − 3i, − 31 + 7i,
by dz
dt = Az where
√
1
− 3 − 7i, and − 2. Is u0 a stable equilibrium solution? Justify your answer.
The real part of the five eigenvalues are negative. As a result u0 is a stable equilibrium point.
1 2
Question 6: Find the eigenvalues of
and say if the matrix is diagonalizable in R2 .
−2 1
The characteristic polynomial is
(1 − λ)2 + 4 = 0.
The two eigenvalues are complex conjugate λ1 = 1+2i, λ2 = 1−2i. The matrix is not diagonalizable
in R2 . It is diagonalizable in C though.
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Mid term exam 2, November 13, 2014
Question 7: Consider the following basis of P3 (t), S := {1, t, t2 , t3 }, and the following basis of
R3 , S 0 := {u1 , u2 , u3 } := {(1, 0, 0), (0, 1, 0), (0, 0, 1)}. Let F : P3 −→ R3 be the linear mapping
R +1
R +1
R +1
defined by F (p) = ( −1 p(t)dt, −1 p(t)tdt, −1 p(t)t2 dt).
(a) Find the matrix of F in the bases S and S 0 .
We compute the coordinate vectors of F (1), F (t), F (t2 ) and F (t3 ) relative to S 0 .
+1
Z
F (1) = (
tdt,
−1
+1
F (t) = (
−1
t2 dt,
−1
t3 dt,
−1
+1
t3 dt,
−1
This means that
Z
2
t3 dt) = (0, , 0)
3
+1
Z
−1
+1
t4 dt,
−1

[F ]SS 0
2
t2 dt) = (2, 0, )
3
+1
Z
+1
Z
−1
F (t3 ) = (
t2 dt,
−1
+1
Z
−1
+1
Z
tdt,
Z
F (t2 ) = (
+1
Z
dt,
−1
Z
+1
Z
2
= 0
2
3
0
2
3
0
Z
+1
2
2
t4 dt) = ( , 0, )
3
5
−1
2
t5 dt) = (0, , 0)
5
2
3
0
0
2
5

2
5 .
0
(b) Is the mapping F defined above bijective? Justify your answer.
The mapping F is not bijective since the dimension of the domain (dimP3 = 4) is not equal to the
dimension of the co-domain (dimR3 = 3). (Note that the rank of [F ]SS 0 is three. This means that
F is injective but is not surjective.)
√
π 0
Question 8: Find the eigenvalues of
and give two orthogonal eigenvectors.
0 e
√
T
T
The matrix is diagonal. The
√ eigenvalues are π and e. The vectors (1, 0) and (0, 1) are
eigenvectors associated with π and e, respectively; they are obviously orthogonal.
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5
Question 9: Consider the matrix A =
diagonal matrix D so that A = P DP −1 .
4 − λ
PA (λ) := det(A − λI) = 2+i
4
2−i
. Find an invertible matrix P and a
2+i
0
2 − i
= −λ(4 − λ) − 5 = λ2 − 4λ − 5
−λ i.e., PA (λ) = (λ − 5)(λ + 1). The eigenvalues of A are λ1 = −1 and λ1 = 5.
Let v1 = (a, b)T be an eigenvector associated with the eigenvalue −1. The definition (A + I)v1 = 0
implies that (2 + i)a + b = 0, i.e., v1 = a(1, −(2 + i))T .
Let v2 = (a, b)T be an eigenvector associated with the eigenvalue 5. The definition (A − 5I)v2 = 0
implies that −a + (2 − i)b = 0, i.e., v2 = b(2 − i, 1)T .
In conclusion, upon setting
P =
1
2−i
,
−2 − i
1
−1
D=
0
0
,
5
we have A = P DP −1 .
Question 10: Let M be a real-valued n×n matrix. (a) Is M T M + M T + M + I diagonalizable?
The matrix M T M + M T + M + I is diagonalizable since it is a symmetric matrix as shown by the
following computation:
(M T M +M T +M +I)T = (M T M )T +M +M T +I T = M T (M T )T +M T +M +I = M T M +M T +M +I.
(b) Prove that the eigenvalues of M T M + M T + M + I are non-negative. (Hint: try to factor
M T M + M T + M + I)
Observe first that M T M + M T + M + I = (M T + I)(M + I). Let (λ, v) an eigen-pair. Then
(M T M + M T + M + Iv = λv; this means v T (M T + I)(M + I)v = λv T v, which can also be
written k(M + I)vk2 = λkvk2 . This proves that λ ≥ 0.
(c) Assume that the smallest eigenvalue of M T M + M T + M + I is 0. Prove that −1 is an
eigenvalue of M .
The above computation shows that v T (M T + I)(M + I)v = λv T v for any eigenpairs (v, λ) of
M T M + M T + M + I. Let v be an eigenvector associated with the eigenvalue 0 of the matrix
M T M + M T + M + I. Then 0 = v T (M T + I)(M + I)v = k(M + I)vk2 ; this means that
(M + I)v = 0, thereby proving that (v, −1) is an eigenpair of M .
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Mid term exam 2, November 13, 2014
Question 11: Consider the following bases of R3 and R2 :
S = {w1 , w2 , w3 } = {(1, 1, 1), (1, 1, 0), (1, 0, 0)},
S 0 = {u1 , u2 } = {(1, 3), (2, 5)}.
Let F : R3 −→ R2 be the linear mapping defined by F (x, y, z) = (3x+2y −4z, x−5y +3z). Find
the matrix of F in the bases S and S 0 . (Hint use the fact that (a, b) = (−5a+2b)u1 +(3a−b)u2 .)
We compute the coordinate vectors of F (w1 ), F (w2 ), and F (w3 ) relative to S 0 .
F (w1 ) = (1, −1) = (−5 − 2)u1 + (3 + 1)u2 = −7u1 + 4u2 ,
F (w2 ) = (5, −4) = (−25 − 8)u1 + (15 + 4)u2 = −33u1 + 19u2 ,
F (w3 ) = (3, 1) = (−15 + 2)u1 + (9 − 1)u2 = −13u1 + 8u2 .
This means that
[F ]SS 0

3
Question 12: Is A = 0
0
7
3
5
−7
=
4
−33
19
−13
.
8

1
0 diagonalizable? Be very accurate when answering.
3
Solution 1: The characteristic polynomial is
3 − λ
PA (λ) = det(A − λI) = 0
0
7
1 3−λ
0 = (3 − λ)3 .
5
3 − λ
This means that λ = 3 is the only eigenvalue and it is of multiplicity three. If A was diagonal, we
would have A = P DP −1 where D = 3I. This would mean A = P 3IP −1 = P P −1 = 3I, i.e.,
A = 3I, which is obviously wrong. In conclusion A cannot be diagonalizable.
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7
Question 13: Let G : R3 −→ P2 be the linear mapping defined by G(e1 ) = 1 + t2 , G(e2 ) =
2 + t + t2 , G(e3 ) = −1 + t − 2t2 , where {e1 , e2 , e3 } is the canonical basis of R3 . Find the
dimension of Ker(G) and the dimension of Im(G).
Let us first characterize ker(G). Let X be a member of Ker(G). Then x + 2y − z + (y + z)t + (x +
y − 2z)t2 = 0. This means that x + 2y − z = 0, y + z = 0 and x + y − 2z = 0. This is a linear
system. The matrix associated with this system is

 
 
 
 

1 2 −1
1 2 −1
1 2 −1
1 3 0
1 0 −3
0 1 1  ∼ 0 1
1  ∼ 0 1 1  ∼ 0 1 1 ∼ 0 1 1 
1 1 −2
0 −1 −1
0 0 0
0 0 0
0 0 0
There is one free variable. This means that dim(ker(G)) = 1. Then the rank Theorem gives
dim(Im(G)) = 3 − dim(Ker(G)) = 2. Note in passing that X ∈ ker(G) iff
 
 
x1
3
x2  = x3 −1
x3
1
This means that ker(G) = span((3, −1, 1)).
Question 14: Let F : P2 (t) −→ P2 (t) be the linear mapping defined by F (p) = p(0) + p0 (1)t +
P 00 (2)t2 . Show that F is bijective.
Since the domain P2 (t) and the co-domain P2 (t) have the same dimension, it suffices to prove that
F is either injective or surjective (by the rank Theorem). Let us prove that F is injective. Let
p = at2 + bt + c be a member of P2 (t) such that F (p) = 0. Since
p(0) = c,
p0 (1) = 2a + b
p00 (2) = 2a
we infer that
0 = p(0) + p0 (1)t + p00 (2)t2 = c + (2a + b)t + (2a)t2 .
This means that a = b = c = 0. As a result F is injective. In conclusion F is bijective.
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Mid term exam 2, November 13, 2014
Question 15: Consider the subspace U of R4 spanned by v1 = (1, 1, 1, 1), v2 = (1, 1, 2, 4),
v3 = (1, 2, −4, −3). Find an orthogonal basis of U .
Apply the Gram-Schmidt process.
w1 = (1, 1, 1, 1)
w2 = (−1, −1, 0, 2)
1 3
w3 = ( , , −3, 1).
2 2