1. Let V be a vector space over a field F. Let W be a subspace of V and U be a subspace of W. Prove that U is a subspace of V.

2. Let F be any field. Here is a system of m linear equations in n unknowns in F.

a_{11}x_{1} + a_{12}x_{2}
+ ... + a_{1n}x_{n} = b_{1},

a_{21}x_{1} + a_{22}x_{2}
+ ... + a_{2n}x_{n} = b_{2},

......

a_{m1}x_{1} + a_{m2}x_{2}
+ ... + a_{mn}x_{n} = b_{m}.

All of the coefficients a_{ij } and
all of the constants b_{i} are in F.
A solution of the system is an n-tuple (r_{1}, r_{2},
... , r_{n}) in F^{n} which satisfies all of
the equations when each r_{i} is substituted for
the corresponding x_{i}. (Each r_{i}
is an element of F.)

The system of equations is called *homogeneous* if
every b_{i} = 0.

(a) Show that if every b_{i} = 0,
then the set of all solutions is a subspace of F^{n}.

(b) Show that if the set of all solutions is a subspace
of F^{n}, then every b_{i} = 0.

3. Let F be a field. We define
a nonempty subset E of F to be a subfield of
F if (i)

the additive identity element 0 of
F and the multiplicative identity element 1 of F
are in E, and (ii) under the operations of addition and
multiplication given by F, E itself is a field.
Examples are the real numbers form a subfield of the field of complex
numbers, the rational numbers form a subfield of the field of real
numbers.

If E is a subfield of F, we can make
F into a vector space over the field E by (i) declaring that
the vector addition in F is the addition that already exists
in F because it is a field,

(ii) declaring that if a is an element of
E (now playing the role of the field of scalars) and v is an
element of F (now playing the role of vector space) then the scalar
product av is the product given by multiplication in the field
F.

Prove that F is a vector space over E
by carefully checking all of the eight or so axioms for a vector space.

3. Here is an infinite sequence of polynomials in **R**[x]
which is a bit like the sequence of ordinary powers of x.

(x)_{0} = 1, (x)_{1} = x, (x)_{2}
= x(x - 1), (x)_{3} = x(x - 1)(x - 2),

..., (x)_{n} = x(x - 1) ...(x - n + 1), ...

Notice that each (x)_{n} has degree
n and leading coefficient 1.

(a) Viewing **R**[x] as a vector
space over the field **R **of real numbers, express each of the
powers x^{0}, x^{1},_{ }x^{2}, x^{3},
and x^{4}, explicitly_{ }as a linear combination
of (x)_{0}, (x)_{1},_{ }(x)_{2},
(x)_{3},_{ }and_{ }(x)_{4}. (This
gets to be a lot of work.)

(b) Let T be the set {(x)_{n}:
n greater than or equal to 0}. Prove that T is a linearly
independent set.

(c) Prove by induction on n that every
polynomial p(x) of degree n in **R**[x]
is a linear combination of (x)_{0}, (x)_{1}, (x)_{2},
... , (x)_{n}. (Notice that parts (b) and (c) together prove that
T is a basis for **R**[x].

(d)_{ }Prove, again by induction on
n, if the polynomial p(x) in part (c) has integer coefficients then
all of the coefficients in the linear combination in (c) are integers.

n

(e) From part (d) you now know that x^{n}
= S
S(n, r)(x)_{r}, where each coefficient

r = 0

S(n, r) is an integer. Show that

(i) for every n, S(n, n) = 1,

(ii) for every n > 0, S(n, 0) = 0,

(iii) if r > n, then the convention is that
S(n, r) = 0 (nothing to prove).

(f) Starting from the fact that (x)_{n
+ 1 }= (x)_{n}(x - n), show that for r less than
or equal to n and greater than 0,

n

(Hint: Start with x^{n} = S
S(n, j)(x)_{j}, and multiply each side by x. When you

j = 0

distribute the factor x through the sum on the
right, write each x as [(x - j) + j],

n + 1

according to the term that it is in. You also have
x^{n + 1} = S
S(n + 1, k)(x)_{k}. Now you

k = 0

can equate the coefficients of (x)_{r}
on the two sides of the equation because the set of polynomials T
is a linearly independent set.)

(Terminology: the numbers S(n, r) are
commonly called Stirling numbers of the second kind. The Stirling
numbers of the first kind are the coefficients that you get when you multiply
out each (x)_{n} = x(x - 1) ...(x - n + 1) to express
it as a sum of powers of x.)

(g) Use the results in (e) and
(f) to calculate recursively the coefficients 0, 1, 7, 6, 1
that you found for x^{4 }in part (a) above.