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Quiz 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
answer. Answers with no justification will not be graded.
Question 1: Find a parametric representation of the line in R4 that passes through P =
(1, 2, −1, −2) in the direction of u = (2, 1, 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 = (2t, t, 2t, t).
The parametric representation of the line is
x1 = 1 + 2t,
x2 = 2 + 2t,
x3 = −1 + 2t,
x4 = −2 + t.
or
L = {(1 + 2t, 2 + 2t, −1 + 2t, −2 + t) ∈ R4 , t ∈ R}.
Question 2: Let u = (1 + 2i, 2 − i) and v = (1, 1 − i) be two complex vectors in C2 . Compute
kuk`2 , kvk`2 , hu, vi and hv, ui.
(a) By definition
kuk2`2 =< u, u >= u · u = (1 + 2i)(1 − 2i) + (2 − i)(2 + i) = 1 − 4i2 + 4 − i2 = 10,
√
i.e., kuk`2 = 10.
kvk2 =< v, v >= v · v = 1 + (1 − i)(1 + i) = 1 + 1 − i2 = 3,
i.e., kvk`2 =
√
3.
hu, vi = (1 + 2i) + (2 − i)(1 + i) = 1 + 2i + 2 + 2i − i + 1 = 4 + 3i.
The definition of te bracket product implies that < v, u >= hu, vi = 4 − 3i. Let us check it!
hv, ui = 1 − 2i + (1 − i)(2 + i) = 1 − 2i + 2 + i − 2i + 1 = 4 − 3i.
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Quiz 1, September 9, 2014
Question 3: Find an equation of the hyperplane in R3 that passes through P = (−1, 0, 1) and
is orthogonal to the vector n = (1, 2, 1).
By definition H is the collection of points X = (x1 , x2 , x3 ) so that (X − P )·n = 0. This means
0 =(X − P )·(1, 2, 1) = (x1 + 1, x2 , x3 − 1)·(1, 2, 1) =
= x1 + 1 + 2x2 + x3 − 1 = 3x1 + 2x2 + x3 .
The equation of the hyperplane in question is
x1 + 2x2 + x3 = 0,
or
H{(x1 , x2 , x3 ) ∈ R3 , x1 + 2x2 + x3 = 0}.
Question 4: 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 · v̄) = Re(v · 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 · ū + tu · v̄ + tv · ū + v · v̄)
= Re(t2 kuk`2 + tu · v̄ + tv · ū + kvk`2 )
= t2 kuk`2 + tRe(u · v̄ + v · ū) + kvk`2
= t2 kuk`2 + 2tRe(u · v̄) + kvk`2 .
The polynomial t2 kuk2`2 + 2tRe(u · v̄) + kvk2`2 is non-negative for all values of t. This means that
the discriminant is non-positive:
4Re(u · v̄) − 4kuk`2 kvk`2 ≤ 0.
In conclusion
Re(u · v̄) ≤ kuk`2 kvk`2 ,
∀u, v ∈ Cn .
The same inequality obviously holds by replacing v by −v:
−Re(u · v̄)| ≤ kuk`2 kvk`2 ,
∀u, v ∈ Cn .
|Re(u · v̄)| ≤ kuk`2 kvk`2 ,
∀u, v ∈ Cn .
In conclusion