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Mid term exam 1 (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
Question 1: Find the equation of the hyperplane in R3 that passes through P = (2, −1, −5)
and is parallel to the hyperplane H 0 determined by the equation 4x1 − 3x2 + 7x3 = 1.
The definition of H 0 implies that the vector n = (4, −3, 7) is normal to H 0 . Since H and H 0 are
parallel, n = (4, −3, 7) is also a normal vector to H. By definition H is the collection of points
X = (x1 , x2 , x3 ) in R3 so that (X − P )&middot;n = 0. This means that
0 =(X − P )&middot;(4, −3, 7) = (x1 − 2, x2 + 1, x3 + 5)&middot;(4, −3, 7) =
= 4x1 − 8 − 3x2 − 3 + 7x3 + 35 = 4x1 − 3x2 + 7x3 + 24.
The equation of the hyperplane in question is
4x1 − 3x2 + 7x3 = −24.
Question 2: Find a parametric representation of the line in R4 that passes through P =
(1, 2, 3, 4) in the direction of u = (4, 3, 2, 1).
The line in question consists of the points X = (x1 , x2 , x3 , x4 ) in R4 that are such that there is
t ∈ R so that (X − P ) = tu. This means that
X − P = tu = (4t, 3t, 2t, t).
The parametric representation of the line is
x1 = 1 + 4t,
x2 = 2 + 3t,
x3 = 3 + 2t,
x4 = 4 + t.
or
L = {(1 + 4t, 2 + 3t, 3 + 2t, 4 + t) ∈ R4 , t ∈ R}.
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Mid term exam 1, October 16, 2014
Question 3: Prove the complex Cauchy-Schwarz inequality: i.e., |Re(hu, vi)| ≤ kuk`2 kvk`2 , for
all u, v ∈ Cn and all positive integer n. (Hint: consider the polynomial Re(htu + v, tu + vi) with
t ∈ R, and use the observation that Re(u &middot; v̄) = Re(v &middot; ū)).
We follow the hint and consider the polynomial Re(htu + v, tu + vi) with t ∈ R. Then
0 ≤ ktu + vk`2 Re(htu + v, tu + vi) = Re(t2 u &middot; ū + tu &middot; v̄ + tv &middot; ū + v &middot; v̄)
= Re(t2 kuk`2 + tu &middot; v̄ + tv &middot; ū + kvk`2 )
= t2 kuk`2 + tRe(u &middot; v̄ + v &middot; ū) + kvk`2
= t2 kuk`2 + 2tRe(u &middot; v̄) + kvk`2 .
The polynomial t2 kuk2`2 + 2tRe(u &middot; v̄) + kvk2`2 is non-negative for all values of t. This means that
the discriminant is non-positive:
4Re(u &middot; v̄) − 4kuk`2 kvk`2 ≤ 0.
In conclusion
∀u, v ∈ Cn .
Re(u &middot; v̄) ≤ kuk`2 kvk`2 ,
The same inequality obviously holds by replacing v by −v:
−Re(u &middot; v̄)| ≤ kuk`2 kvk`2 ,
∀u, v ∈ Cn .
|Re(u &middot; v̄)| ≤ kuk`2 kvk`2 ,
∀u, v ∈ Cn .
In conclusion
Question 4: Find a 2&times;2 real-valued matrix P whose columns are unit orthogonal vectors and
whose second column is a multiple of (1, −2)T .
The second column of P is λ(1, −2) where λ ∈ R. The norm of this vector is λ2 (1 + 4) = 5λ2 .
Since this is a unit vector, we infer that λ = &plusmn; √15 . Let us choose λ = √15 . The first column of P
is (a, b)T , i.e.,
#
&quot;
√1
a
5
P =
b − √25
with a2 + b2 = 1 and a √15 − b √25 = 0. This means a = 2b and b2 (1 + 4) = 1, i.e., a = &plusmn; √25 and
b = &plusmn; √15 . In conclusion we can choose
1 2
P =√
5 1
1
.
−2
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
1 0
Question 5: Let E = 0 1
0 −2
R2 and R3 are row vectors. (a)
3

 
0
R1
0 and let A = R2  be an arbitrary 3 &times; 3 matrix, where R1 ,
1
R3
Compute EA
E is the matrix on the elementary row operation that consists of replacing the second row of a
matrix by itself minus three times the first row:


R1

R2
EA = 
−2R2 + R3
(b) Give the inverse of E
E
because

1
0
0
0
1
2


0
1
0 EA = 0
1
0
−1

1

= 0
0
0
1
2

0
0
1

 

0
R1
R1
=
 = A.
0 
R2
R2
1 −R2 + R3
2R2 − 2R2 + R3
0
1
2
Using A = I in the above equality means that

1 0
0 1
0 2

0
0 E = I
1
Question 6: Let u be a unit vector in Cn , i.e., kuk2`2 = 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.
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Mid term exam 1, October 16, 2014
Question 7: Are the following functions f1 (t) = 1 − 3t + 2 sin(t), f2 (t) = 2 − 4t − sin(t),
f3 (t) = 1 − 5t + 7 sin(t) linearly independent in V := span{1, t, sin(t)}?
Let X = (x1 , x2 , x3 ) ∈ R3 be so that x1 f1 + x2 f2 + x3 f3 = 0. Then
x1 f1 (t) + x2 f2 (t) + x3 f3 (t) = x1 + 2x2 + x3 + (−3x1 − 4x2 − 5x3 )t + (2x1 − x2 + 7x3 ) sin(t) = 0.
This means that X solves

1
−3
2
2
−4
−1

1
−5 X = 0.
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Let us reduce the matrix of the linear system in echelon form

 


1
2
1
1 2
1
1
−3 −4 −5 ∼ 0 2 −2 ∼= 0
2 −1 7
0 −5 5
0

2 1
2 −2 .
0 0
There are only two pivots. There is one free variable. This means that the solution set of the above
linear system is not {0}. There is some nonzero vector X so that x1 f1 + x2 f2 + x3 f3 = 0. This
means that the functions f1 , f2 , f3 are linearly dependent.


4 −3 2 −1
16 −9 6 −3

Question 8: Compute the LU factorization of A = 
12 3 0 0 . Give all the details.
8
3 6 −2



4 −3 2 −1
1 0 0 0
4 1 0 0 0 3 −2 1 


A=
3 4 1 0 0 0
2
1
0 0
0
1
2 3 4 1
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Question 9: Express the function v(t) = e2t + 4et − 3 as a linear combination of z1 (t) =
e2t − 2et + 5, z2 (t) = 2e2t − 3et , z3 (t) = et + 3.
We look for (x1 , x2 , x3 ) ∈ R3 so that v = x1 z1 + x2 z2 + x3 z3 . This means that the following holds
for all t ∈ R:
e2t + 4et − 3 = x1 (e2t − 2et + 5) + x2 (2e2t − 3et ) + x3 (et + 3)
= e2t (x1 + 2x2 ) + et (−2x1 − 3x2 + x3 ) + 1(5x1 + 3x3 ).
This implies
x1 + 2x2 = 1,
We can write this
form:

1
2
−2 −3
5
0
−2x1 − 3x2 + x3 = 4
5x1 + 3x3 = −3.
system in the form of an augmented matrix and compute the reduced echelon
 
0 1
1
1 4  ∼ 0
3 −3
0
2
1
−10
 
0 1
1
1 6  ∼ 0
3 −8
0
2
1
0
0
1
13
 
1
1
6  ∼ 0
52
0
0
1
0
0
0
1

−3
2
4
This means that v = −3z1 + 2z2 + 4z3 .
Question 10: (a) Let A be a n &times; n real-valued matrix. Assume that X T AX ≥ −5kXk2`2 for
all X ∈ Rn . Prove that 7I + A is invertible.
(b) Let A be a n &times; n complex-valued matrix. Assume that Re(X̄ T AX) ≥ −5kXk2`2 for all
X ∈ Cn . Prove that 7I + A is invertible.
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Mid term exam 1, October 16, 2014
Question 11: Consider the two-variate polynomials Q1 (x, y) = (1 − x)(1 − y), Q2 (x, y) =
x(1 − y), Q3 (x, y) = y(1 − x), Q4 (x, y) = xy. (a) Show that the set S = {Q1 , Q2 , Q3 , Q4 } is
linearly independent.
Let x1 , x2 , x3 , x3 such that
x1 Q1 + . . . + x4 Q4 = 0.
Then
0 = x1 (1 − x − y + xy) + x2 (x − xy) + x3 (y − xy) + x4 xy
= x1 + (−x1 + x2 )x + (−x1 + x3 )y + (x1 − x2 − x3 + x4 ).
This means that
x1 = 0,
−x1 + x2 = 0,
−x1 + x3 = 0,
x1 − x2 − x3 + x4 = 0.
Then clearly x1 = x2 = x3 = x4 = 0, i.e., the set S is linearly independent.
(b) Is the polynomial Q(x, y) = 1 in span(S)?
Let us try to find x1 , x2 , x3 , x3 such that
1 = x1 Q1 + . . . + x4 Q4
This implies that
1 = x1 (1 − x − y + xy) + x2 (x − xy) + x3 (y − xy) + x4 xy
= x1 + (−x1 + x2 )x + (−x1 + x3 )y + (x1 − x2 − x3 + x4 ).
Then
x1 = 1,
−x1 + x2 = 0,
−x1 + x3 = 0,
x1 − x2 − x3 + x4 = 0.
Then clearly x1 = x2 = x3 = x4 = 1. The means that 1 = Q1 (x, y)+Q2 (x, y)+Q3 (x, y)+Q4 (x, y),
i.e., 1 ∈ span(S)
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