M341 F10  Homework 2.
With the exception of 1.4, problems 1 and 3 which are purely computational, expla
nations in English are given as they should. Mostly in section 1.3 even when there are a
computations (in Math there are almost always some), reasoning is more important than
computing. See the blend of mathematical statements and English language, see also how
you should always try and explain until what you have in mind seems (or should seem)
obvious to the reader (to you in fact).
Section 1.3.
12.
This problem is a bit diﬃcult. Do not hesitate to read the solution several times
if necessary
Let us assume that three mutually orthogonal non zero vectors exist in
R
2
,
say
x
= [
x
1
, x
2
]
,
y
= [
y
1
, y
2
]
,
z
= [
z
1
, z
2
]. Since [
x
1
, x
2
]
̸
= [0
,
0], one of its coordinates
is not 0.
We may assume, changing the numbering of the coordinates if necessary that
x
1
̸
= 0. If
y
1
=
z
1
= 0, [
y
1
, y
2
]
.
[
z
1
, z
2
] =
y
1
z
1
+
y
2
z
2
=
y
2
z
2
= 0, would bring that either
y
2
= 0 or
z
2
= 0 so that either
y
or
z
would be the zero vector, which is not true. So at
least one of
y
1
and
z
1
is not zero. Changing the names of
y
and
z
if necessary, we may
assume that
y
1
̸
= 0.
If both
x
1
and
y
1
are not zero, we may consider
a
=
x
2
/x
1
and
b
=
y
2
/y
1
(observe
that
a
or
b
might be 0).
[
x
1
, x
2
] =
x
1
[1
, a
] and [
y
1
, y
2
] =
y
2
[1
, b
], so that the orthogonality conditions being
preserved and [1
, a
]
,
[1
, b
], and [
z
1
, z
2
] are still mutually orthogonal:
[1
, a
]
.
[
z
1
, z
2
] =
z
1
+
az
2
= 0, and also [1
, b
]
.
[
z
1
, z
2
] =
z
1
+
bz
2
= 0.
From
z
1
+
az
2
= 0 and
z
1
+
bz
2
= 0, we get (
a
−
b
)
z
2
= 0, so that either
a
=
b
which is
not possible since it would mean that
x
and
y
are parallel instead of being orthogonal, or
z
2
= 0 which is the only possibility. But
z
2
= 0 and
z
1
+
az
2
=
z
1
+
bz
2
= 0, would mean
that
z
1
=
z
2
= 0 which is not possible since
z
̸
= 0.
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 Spring '08
 HIETMANN
 Trigraph, Diagonal matrix, x.y ̸=

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