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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.


2 3 2
Question 1: Is A = 0 2 0 diagonalizable? Be very accurate when answering.
0 2 2
Solution 1: The characteristic polynomial is
PA (λ) = det(A − λI) =
2−λ
0
0
3
2
2−λ
0
= (2 − λ)3 .
2
2−λ
This means that λ = 2 is the only eigenvalue and it is of multiplicity three. Let v = (x, y, z)T be
an eigenvector, then
3y + 2z = 0
2y = 0
which implies that y = z = 0. The eigenvectors are of the following form v = (α, 0, 0)T , α ∈ R and
α 6= 0. This means that the eigenspace E1 is the one-dimensional line E1 := span{(1, 0, 0)T }, i.e.,
all the eigenvectors are parallel to (1, 0, 0)T . As a result, it is not possible to find three independent
eigenvectors. The matrix is not diagonalizable.
Solution 2: We deduce as above that 2 is the only eigenvalue of multiplicity three. If A was diagonal,
we would have A = P DP −1 where D = 2I. This would mean A = P 2IP −1 = P P −1 = 2I, i.e.,
A = 2I, which is obviously wrong. In conclusion A cannot be diagonalizable.
Question 2: The following linearly independent

0
and e3 := (1, 1, 1)T are eigenvectors of A = 1
1
diagonal matrix D so that A = P DP −1 .
T
T
vectors
 e1 := (−1, 0, 1) , e2 := (−1, 1, 0) ,
1 1
0 1. Find an invertible matrix P and a
1 0
The matrix is clearly diagonalizable since there exist three linearly independent eigenvectors. Let
us compute Ae1 , Ae2 and Ae3 to get the associated eigenvalues. (A second possibility would
be to compute the characteristic polynomial of A, but that would be too long and unnecessarily
complicated since we are already given the eigenvectors.)
Ae1 = (1, 0, −1)T = −e1
Ae2 = (1, −1, 0)T = −e2
Ae3 = (2, 2, 2)T = 2e3 .
This means that e1 and e2 are associated with the eigenvalue −1 (the multiplicity of this eigenvalue
is two) and e3 is associated with the eigenvalue 2. Let us now set




−1 0 0
−1 −1 1
1 1 .
D =  0 −1 0 , P =  0
0
0 2
1
0 1
Then A = P DP −1 .
2
Mid term exam 2, November 8, 2012
Question 3: Consider the matrix A =
4
2−i
. Find an invertible matrix P and a
2+i
0
diagonal matrix D so that A = P DP −1 .
PA (λ) := det(A − λI) =
4−λ
2+i
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
D=
−1
0
0
,
5
we have A = P DP −1 .
Question 4: Let u be a unit vector in Cn , i.e., ūT u = 1. (a) Prove that H = 1 − 2uūT is
Hermitian. (Recall that H is Hermitian iff H = H̄ T .)
The definition of H implies that
T
H̄ T = (I − 2uūT ) = (I − 2ūuT )T = (I − 2uūuT ) = H.
(b) Prove that H = I − 2uūT is unitary. (Recall that H is unitary iff H H̄ T = I.)
Using the fact that H is Hermitian, we infer that
H H̄ T = HH = (I − 2uūT )(I − 2uūT ) = I − 4uūT + 4uūT uūT .
Now using that u is a unit vector, ūT u = 1, we infer that
H H̄ T = I − 4uūT + 4uūT = I,
i.e., H is Hermitian.
(c) Prove that the determinant of any Hermitian matrix is real.
From class we know that Hermitian matrices are diagonalizable and the determinant of diagonalizable
matrices is the product of the eigenvalues. Since the eigenvalues of any Hermitian matrix are real,
its determinant is real.
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Question 5: Consider the matrix A from Question 2. (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−t v2 ,
and e2t v3 .
where
v1 = (−1, 0, 1)T ,
v2 = (−1, 1, 0)T ,
v3 = (1, 1, 1)T .
This is all I wanted you to write.
Recall that upon defining the matrix P := [v1 v2 v3 ], 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 e−2t
0
0
1
w(t) = α2 e−2t  = α1 e−t 0 + α2 e−t 1 + α3 e2t 0 ,
0
1
α3 e4t
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 e−2t P 0 + α2 e−2t P 1 + α3 e4t P
0
0
 
0
0 = α1 e−2t v1 + α1 e−2t v2 + α1 e4t v3 .
1
In conclusion, u(t) is a linear combination of the fundamental solutions e−2t v1 , e−2t v2 , e4t v3 .
(b) What is the solution of
du(t)
dt
= Au(t), with u(0) = (−2, 1, 1)?
We know that u(t) is a linear combination of the fundamental solutions:
u(t) = α1 e−t v1 + α2 e−t v2 + α3 e2t v3 .
The condition u(0) = (2, 1, −1) implies
(2, 1, −1) = α2 v1 + α2 v2 + α3 v3 .
The corresponding

−1 −1
0
1
1
0
augmented matrix is
 
 
1 −2
−1 −1 1 0
1
1 1 ∼ 0
1 1 1  ∼ 0
1 1
0 −1 2 −1
0
1
1
0
1
0
3
 
2
1
1 ∼ 0
0
0
The solution is (α1 , α2 , α3 ) = (1, 1, 0). This means
u(t) = e−t v1 + e−t v2 = e−t (−2, 1, 1)T .
0
1
0
0
0
1

1
1 .
0
4
Mid term exam 2, November 8, 2012
Question 6: Consider the following bases of R3 and R2 :
S 0 = {u1 , u2 } = {(1, 3), (2, 5)}.
S = {w1 , w2 , w3 } = {(1, 1, 1), (1, 1, 0), (1, 0, 0)},
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
−7
=
4
−33
19
−13
.
8
Question 7: 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
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Question 8: 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 9: Let F : P2 (t) −→ P2 (t) be the linear mapping defined by F (p) = p(0) + p(1)t +
p(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,
p(1) = a + b + c
p(2) = 4a + 2b + c
we infer that
0 = p(0) + p(1)t + p(2)t2 = c + (a + b + c)t + (4a + 2b + c)t2 .
This implies that c = 0, a + b + c = 0
The associated matrix is

4
1
0
and 4a + 2b + c = 0. This is a linear system for a, b and c.
 

2 1
4 2
1
1 1 ∼ 0 −2 −3
0 1
0 0
1
The rank is maximum. 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 8, 2012

1
Question 10: Let A = 2
0
2
1
0

0
0. (a) Compute the eigenvalues of A.
4
The characteristic polynomial is
(4 − λ)
1−λ
2
2
= (4 − λ)(1 − λ)2 − 4) = (4 − λ)(−1 − λ)(3 − λ).
1−λ
The eigenvalues are 4, −1 and 3.
(b) Let e1 and e2 be two unit eigenvectors associated with the two largest eigenvalues of A.
Compute he1 , e2 i.
The matrix is real-valued, symmetric, and the eigenvalues are different, as a result he1 , e2 i = 0.
There is no need to compute e1 and e2 . (It is a known result that eigenvectors of symmetric
real-valued matrices are orthogonal if they are associated with different eigenvalues.)
Question 11: Let M be a real-valued n×n matrix. (a)Prove that M T M + 2I is diagonalizable
M T M + 2I is diagonalizable since it is a symmetric matrix as shown be the following computation:
(M T M + 2I)T = (M T M )T + 2I T = M T (M T )T + 2I = M T M − 2I.
(b) Prove that the eigenvalues of M T M + 2I are larger that or equal to 2. (Hint v T M T M v =
kM vk2 )
Let (λ, v) an eigen-pair. Then (M T M + 2I)v = λv; this means v T (M T M + 2I)v = λv T v, which
can also be written kM vk2 + 2kvk2 = λkvk2 . This proves that 2kvk2 ≤ λkvk2 , from which we
conclude that 2 ≤ λ (Note that kvk =
6 0 since v is an eigenvector).
(c) Assume that the smallest eigenvalue of (M T + 2I)(M + 2I) is
eigenvalue of M .
1
π.
Prove that −2 is not an
Assume that −2 is an eignevalue of M . Let v be a nonzero vector of M associated with the
eigenvalue −2, i.e., (M + 2I)v = 0. This means that (M + 2I)T (M + 2I)v = 0v, which implies
that 0 is an eigenvalue of (M + 2I)T (M + 2I), which contradicts that π1 is the smallest eigenvalue.
In conclusion one cannot find a nonzero vector so that (M + 2I)v = 0. This implies that −2 cannot
be an eigenvalue of M .