Problem Set 1 Solutions
MATH 110: Linear Algebra
PART 1
The following problems are each worth 5 points..
1. Curtis p. 14 1: a) The base case n= 1 is trivial. Suppose the theorem
is true for some n−1. Then 1 + 3 + . . . + (2n−1) + (2n+ 1) = n2+ 2n+ 1 =
(2n+ 1)2. b) Again, the base case is trivial. The inductive step involves
a bit of algebra, namely 12+. . . +n2+ (n+ 1)2is equal, by induction, to
n(n+1)(2n+1)
6+ (n+ 1)2=(n+1)(n+2)(2n+3)
6.
2. Curtis p. 15 4: The closure axioms are satisfied because of the way
the operations are defined (using a table). The “inverse element” axiom is
satisfied because 1−1= 1. The operations are commutative because they
are symmetric about the diagonal. The rest of the properties simply involve
verification by examining the tables.
3. Curtis p. 25 3. First of all, notice that the process of checking for
Rnand Fninvolves only checking Fn, since Rnis a special case of Fn(where
the field is R). The important thing is that all the axioms for a vector space
behave the same on a tuple as on a singleton. The zero element in the vector
space is <0,0, . . . , 0>. The function space is a bit harder to check. It is
important to recognize that each function is an element in the vector space.
For example, showing that (f+g)(x)=(g+f)(x) involves understanding
that (f+g) and (g+f) are each functions themselves. Now it is a fact that for
every x, (f+g)(x) = f(x)+g(x) (this is by the very definition of the operation
“+”). f(x), g(x) are both some real numbers (depending on x). Because of
the field axioms for the reals, it follows that f(x) + g(x) = g(x) + f(x). Once
again, by the definition of “+” this is (g+f)(x).
4. Curtis p. 33 3. The answers for this exercise are all at the back of the
book. Notice that the answer for part (h) is that the solutions are a subspace
only if gis the zero function. This is because if it is not, then the sum of two
solutions, when plugged into the differential equation (LHS) equal 2g, thus
sums do not satisfy the solution.
5. Curtis p. 37 I will assume in questions such as this one that you know
your basic calculus (if you don’t, review!). So you should know that if the
derivative of a function exists and df/dt = 0 then the function is of the form
f(x) = cfor some constant c.In terms of vector spaces this is the same as
saying that fconsists of the subspace generated by the element f(x) = 1,
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