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MATH 115A  Lecture 4 Midterm 1
STUDENT ID #:
This is a closedbook and closednote examination.
Calculators are not allowed .
Please show all your work.
Use only the paper provided. You may write on the back if you need more
space, but clearly indicate this on the front.
There are 5 problems for a total of 100 points.
POINTS:
1.
2.
3.
4.
5.
1. (10 points each) For the given subset W of the Fvector space V ,
prove or disprove that W is a subspace of V .
Clearly, the zero vector (0, 0, 0) is in W. Now suppose a = (a_{1}, a_{2}, a_{3}) and
b = (b_{1}, b_{2}, b_{3}) are in W. Then (a_{1}+b_{1})(a_{2}+b_{2}) = a_{1}a_{2}+b_{1}b_{2} = a_{3}+b_{3},
so a + b ∈W. If 2 R is a scalar, then
λa_{1}  λa_{2} = λ(a_{1} 
a_{2}) = a_{3}, so
a ∈W. That is, W is closed under addition and scalar
multiplication and
contains 0, hence, it is a subspace.
The two complex numbers x = 1 and
are both in W.
However,
is not rational , so that x + y is not in W. That is,
W is not closed under addition and is therefore not a subspace.
2. (20 points) Let
.
Is S a basis of
R^{3}? If so, prove it. If not, disprove it.
Because the dimension of R^{3} is 3, it suffices to show either that
S is linearly
independent, or that S is a generating set. Either is easily accomplished by
solving linear equations.
3. (20 points) The set of complex numbers C can be viewed as both an
Rvector space and a Cvector space. Give an example of a map T : C > C
that is Rlinear, but not Clinear.
For example, the map T : C > C defined by T(a + bi) = a is easily
checked to be Rlinear, but not Clinear. First, we check that T is additive.
Let x = a + bi and y = c + di be two complex numbers. Then T(x + y) =
T((a + c) + (b + d)i) = (a + c) = T(x) + T(y).
Next, let λ ∈R be a real
scalar , and x = a+bi a complex number. Then
T(λx) = T((λa) + (λb)i)
=λa =λT(x). Thus, T is
Rlinear.
Finally, we show T is not Clinear. Let x = i ∈C
(element of the vector
space) and = i ∈C ( complex scalar ). then T(λx)
=1, but λT(x) = 0.
4. (20 points) Let D : C^{1}(R) > C(R) be the map on continuously
differentiable functions given by the derivative, that is, D(f)(x) = f'(x).
Show that D is linear (you may assume what you know from calculus) and
give a basis for the Null space (that is, the kernel) of D.
Let f and g be elements of C^{1}(R), and let c ∈R
be a scalar. Then
D(cf + g) = D(cf) + D(g) = cD(f) + D(g), where the first equality holds
because the derivative of a sum is the sum of derivatives, and the second
because the derivate of a multiple is the multiple of the derivative. This
shows that D is a linear transformation .
The Null space of D is the set of all continuously differentiable functions f
such that f'(x) = 0 for all x ∈R.
Such function are the constant ones, that
is,
A basis for this space is any one constant nonzero
function, for example,
5. (20 points) Let V be the Rvector space of functions in one variable x
of the form
where a, b and c are arbitrary real numbers .
Define a map
Show that T is a
linear transformation and determine the dimension of its Null space N(T)
and its range R(T).
Let f and g be in V , and let c ∈R
be a scalar. Then T(f + g) =
((f +g)(0), (f +g)(1), (f +g)(2)) = (f(0)+g(0), f(1)+g(1), f(2)+g(2)) =
(f(0), f(1), f(2))+(g(0), g(1), g(2)) = T(f)+T(g) and T(cf) = ((cf)(0), (cf)(1),
(cf)(2)) =
(cf(0), cf(1), cf(2)) = c(f(0), f(1), f(2)) = cT (f). That is, T is a linear
transformation.
We compute the range of T. I claim T is onto. To show that, it suffices to
give elements f, g and h in V such thatis a basis.
Let f(x) = x(x1)/2, g(x) = x(x2) and h(x) = (x1)(x2)/2. Then
T(f) = e_{1}, T(g) = e_{2} and T(h) = e_{3}. Thus, T is onto and the dimension of
the range is 3.
Since the dimension of V is equal to the dimension of R^{3}, the dimension
theorem implies that the dimension of the Null space is 0.
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