*Solution:*
We assume that the x-axis is the
domain and the y-axis is the
codomain
of the function that is to be defined by the given curve.
According to Definition 2.1.1,
a subset of the plane defines a function
if for each element x in the domain
there is a unique element y in the codomain
such that (x, y) belongs to the subset of the plane.
If a vertical line intersects the curve in two distinct points,
then there will be points (x_{1}, y_{1}) and
(x_{2}, y_{2}) on the curve
with x_{1} = x_{2} and
y_{1} y_{2}.
Thus if we apply Definition 2.1.1
to the given curve,
the uniqueness part of the definition translates directly
into the ``vertical line test''.

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*Solution:*
If a horizontal line intersects the graph of the function more than once,
then the points of intersection represent points

(x_{1},y_{1}) and (x_{2},y_{2})
for which x_{1}
x_{2} but y_{1} = y_{2}.
According to Definition 2.1.4,
a function is one-to-one if

f(x_{1}) = f(x_{2}) implies
x_{1} = x_{2}.
Equivalently, if (x_{1}, y_{1}) and
(x_{2}, y_{2}) line on its graph,
then we cannot have

y_{1} = y_{2} while
x_{1} x_{2}.
In this context, the ``horizontal line test''
is exactly the same as the condition given in
Definition 2.1.4.

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*Solution:*
We first note that the reflection of a point (a,b)
in the line y = x is the point (b,a).
This can be seen by observing that the line segment
joining (a,b) and (b,a) has slope -1,
which makes it perpendicular to the line y = x,
and that this line segment intersects the line y = x
at the midpoint ((a+b)/2,(a+b)/2) of the segment.

If f : **R** -> **R** has an inverse,
and the point (x,y) lies on the graph of f, then y = f(x), and so

f^{ -1}(y) = f^{ -1}(f(x)) = x.
This shows that the point (x,y) lies on the graph of f^{ -1}.
Conversely, if

(x,y) lies on the graph of f^{ -1}, then

x = f^{ -1}(y), and therefore
y = f(f^{ -1}(y)) = f(x),
which shows that (y,x) lies on the graph of f.

On the other hand,
suppose that the graph of the function g is defined
by reflecting the graph of f in the line y = x.
For any real number x, if

y = f(x) then we have g(f(x)) = g(y) = x and
for any real number y we have

f(g(y)) = f(x) = y, where x = g(y).
This shows that g = f^{ -1},
and so f has an inverse.

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(a) Show that L is an invertible function if and only if det(A) 0.

(b) Show that if L is either one-to-one or onto, then it is invertible.

*Solution:*
(a) We need to assume that you know that a square matrix A
is invertible if and only if
det(A) 0.

First, if L has an inverse, then it can also be described by multiplication by a matrix B, which must satisfy the conditions BA = I, and AB = I, where I is the n × n identity matrix. Thus A is an invertible matrix, and so det(A) 0.

On the other hand, if
det(A) 0,
then A is invertible, and so L has an inverse, defined by

L^{-1}(**x**) = A^{-1}**x**,
for all **x** in **R**^{n}.

(b) The rank of the matrix A is the dimension of the column space of A, and this is the image of the transformation L, so L is onto if and only if A has rank n.

On the other hand, the nullity
of A is the dimension of the solution space of the equation

A**x** = **0**,
and L is one-to-one if and only if the nullity of A is zero, since

A**x**_{1} = A**x**_{2} if and only if
A(**x**_{1} - **x**_{2}) = **0**.

To prove part (b) we need to use the Rank-Nullity Theorem, which states that if A is an n × n matrix, then the rank of A plus the nullity of A is n. Since the matrix A is invertible if and only if it has rank n, it follows that L is invertible if and only if L is onto, and then the Rank-Nullity Theorem shows that this happens if and only if L is one-to-one.

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L :

det(A

*Solution:*
If det(A^{T}A) 0,
then A^{T}A is an invertible matrix.
If we define

K : **R**^{m} -> **R**^{n} by
K(**x**) = (A^{T}A)^{ -1}A^{T}**x**,
for all **x** in **R**^{m},
then KL is the identity function on **R**^{m}.
It then follows from Exercise 2.1.17 that L is one-to-one.

*Comment:*
There is a stronger result that depends on
knowing a little more linear algebra.
In some linear algebra courses it is proved that
det(A^{T}A) gives the n-dimensional ``content''
of the parallepiped defined by the column vectors of A.
This content is nonzero if and only if the vectors are
linearly independent, and so
det(A^{T}A) 0
if and only if the column vectors of A are linearly independent.
According to the Rank-Nullity Theorem,
this happens if and only if the nullity of A is zero.
In other words, L is a one-to-one linear transformation if and only if
det(A^{T}A) 0.

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*Note:*
A vector **x** is called an eigenvector of A
if it is nonzero and there exists a scalar
such that
A**x** = **x**.

*Solution:*
As noted in the solution to Problem 2.1.23,
A**x**_{1} = A**x**_{2} if and only if
A(**x**_{1}-**x**_{2}) = **0**,
and so L is one-to-one if and only if
A**x** **0** for
all nonzero vectors **x**.
This is equivalent to the statement that
there is no nonzero vector **x** for which
A**x** = 0 · **x**,
which translates into the given statement about eigenvalues of A.

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*Comment:*
The problem is that there can be two names
for the same element of **Z**_{6}.
For example,

[1]_{6} = [7]_{6}.
The given formula looks like it might depend on

the particular name of an element, since
f([x]_{6}) = [5x]_{15}.
For example, we have

f([1]_{6}) = [5 · 1]_{15} = [5]_{15}.
On the other hand,

f([7]_{6}) = [5 · 7]_{15} = [35]_{15}.
Fortunately, this doesn't cause any problems because

[35]_{15} = [5]_{15}.
We need to prove that the function behaves properly
for *all* choices of a representative for *all*
of the congruence classes in **Z**_{6}.

*Solution:*
Suppose that [a]_{6} = [b]_{6}. Then

a b (mod 6),
so a = b + 6k, for some integer k.
Multiplying both sides of this equation by 5 gives us

5a = 5b + 30k, so
5a 5b (mod 30),
which means that

[5a]_{30} = [5b]_{30}.
This shows that the formula for f
will have the same effect on a and b, so that

f ([a]_{6}) = [5a]_{30}
= [5b]_{30} = f ([b]_{6}).
This allows us to conclude that f is a well-defined function.

Here are the values of the function f:

f ([0]_{6}) = [5 · 0]_{30} = [0]_{30} ;

f ([1]_{6}) = [5 · 1]_{30} = [5]_{30} ;

f ([2]_{6}) = [5 · 2]_{30} = [10]_{30} ;

f ([3]_{6}) = [5 · 3]_{30} = [15]_{30} ;

f ([4]_{6}) = [5 · 4]_{30} = [20]_{30} ;

f ([5]_{6}) = [5 · 5]_{30} = [25]_{30} .

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*Solution:*
First, if x is relatively prime to 24,
then so is 5x, and therefore 5x is relatively prime to 12.
This shows that the formula does allow any input in
**Z**_{24}^{×},
and will produce an output in
**Z**_{12}^{×}.
That is, the formula does work on the given domain and codomain.

To show that f is well-defined,
suppose that [a]_{24} = [b]_{24}. Then

5a 5b (mod 24),
since multiplication is well-defined in

**Z**_{24}^{×},
and it follows that
5a 5b (mod 12), so

f ([a]_{24}) = [5a]_{12}
= [5b]_{12} = f ([b]_{24}).

This allows us to conclude that f is a well-defined function.

The elements of
**Z**_{24}^{×} are
{ 1, 5, 7, 11, 13, 17, 19, 23 }.

It may be easier to use
{ ±1, ±5, ±7, ±11 }.

Here are the values of the function f:

f ([1]_{24}) = [5 · 1]_{12} = [5]_{12} ;

f ([-1]_{24}) = [5 · (-1)]_{12}
= [-5]_{12} = [7]_{12} ;

f ([5]_{24}) = [5 · 5]_{12}
f ([25]_{12}) = [1]_{12} ;

f ([-5]_{24}) = [5 · (-5)]_{12}
f ([-25]_{12}) = [11]_{12} ;

f ([7]_{24}) = [5 · 7]_{12}
f ([35]_{12}) = [11]_{12} ;

f ([-7]_{24}) = [5 · (-7)]_{12}
f ([-35]_{12}) = [1]_{12} ;

f ([11]_{24}) = [5 · 11]_{12}
f ([55]_{12}) = [7]_{12} ;

f ([-11]_{24}) = [5 · (-11)]_{12}
f ([-55]_{12}) = [5]_{12} .

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*Solution:*
Remember that the set **Z**_{n}^{×}
is defined to be
the set of all congruence classes modulo n
that are represented by an integer relatively prime to n.
Since 17 is prime, this is the set of nonzero congruence classes modulo 17.
By Proposition 1.4.5,
this is precisely the set of congruence classes [k]_{17}
for which there is a multiplicative inverse in
**Z**_{17}^{×}.
Therefore we can define

g : **Z**_{17}^{×} ->
**Z**_{17}^{×}
by g([x]) = [a]^{ -1}[x],
for all [x] in **Z**_{17}^{×}.
Then

g(f([x]))
= g([ax])
= [a]^{ -1}([a][x])
= ([a]^{ -1}[a])[x]
= [x] and

f(g([x]))
= f([a]^{ -1}[x])
= [a]([a]^{ -1}[x])
= ([a][a]^{ -1})[x]
= [x],
which shows that g = f^{ -1}.

This implies that f is both one-to-one and onto.

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