# Last name: name: 1 Quiz 1 (Notes, books, and calculators are not authorized)

```Last name:
name:
1
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
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 =&lt; u, u &gt;= u &middot; u = (1 + 2i)(1 − 2i) + (2 − i)(2 + i) = 1 − 4i2 + 4 − i2 = 10,
√
i.e., kuk`2 = 10.
kvk2 =&lt; v, v &gt;= v &middot; 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 &lt; v, u &gt;= hu, vi = 4 − 3i. Let us check it!
hv, ui = 1 − 2i + (1 − i)(2 + i) = 1 − 2i + 2 + i − 2i + 1 = 4 − 3i.
2
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 )&middot;n = 0. This means
0 =(X − P )&middot;(1, 2, 1) = (x1 + 1, x2 , x3 − 1)&middot;(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 &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
Re(u &middot; v̄) ≤ kuk`2 kvk`2 ,
∀u, v ∈ Cn .
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
```