# MATH 115A - Lecture 4 Midterm 1

STUDENT ID #:

This is a closed-book and closed-note 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 F-vector 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

R-vector space and a C-vector space. Give an example of a map T : C -> C

that is R-linear, but not C-linear.

For example, the map T : C -> C defined by T(a + bi) = a is easily

checked to be R-linear, but not C-linear. 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
R-linear.

Finally, we show T is not C-linear. 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 non-zero

function, for example,

5. (20 points) Let V be the R-vector 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(x-1)/2, g(x) = -x(x-2) and h(x) = (x-1)(x-2)/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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