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Math Exam 2 Answers

1. Explain in detail how you would find the t curve using techniques of this
course. (Warning: To check your answer, you might want to make up four data
points and work out the solution explicitly .)

Answer: We want to solve

Let A denote that big N ·4 matrix made from the xi. From our discussion of least squares we
that if A has rank 4, then AT>A is invertible and the least squares solution
However, our A is not of rank 4, because the set of functions is not
linearly independent ! For example, So let's throw out the cos 2x term.
Now we want to solve

The procedure is to let A be that big N ·3 matrix (which is of rank 3, as needed) and compute
the least squares solution as

This gives us the coefficients a , b, c to use in the t curve This function
satisfies the requirements of the problem, because it is of Ms. Ogunmola's requested form with
d = 0.

[Remark: I expected that most students would not realize that the four functions are not
independent. That is why I suggested working an example; in any example it becomes clear
that AT>A is not invertible (although it may not be clear how to x this).]

[Remark: Instead of throwing out cos 2x we could throw out 1 or cos2 x. We could not throw
out cos x without truly making our wind speed model less expressive.]

2. Show that if is perpendicular to a given polygon, then is perpen-
dicular to the transformed polygon.

Answer: Let be any vector lying in the transformed polygon. We wish to show that
Toward that end, let and be the points at the head and tail of , so
that These points and are in the transformed polygon, so there must exist
points and in the original polygon such that and Because and
are points in the original polygon, is a vector lying in the original polygon, and so
Then, using the basic facts that and , we have

3. What happens in the special case when A is a rotation? Explain in detail.

Answer: If A is a rotation, then A preserves the length of any vector, so A is orthogonal.
This implies that , so that In this special case, normals transform by
A. This makes sense, because an orthogonal transformation such as a rotation preserves angles ;
if is perpendicular to a polygon, then after both are transformed orthogonally the results will
still be perpendicular.

4. Using only the definitions of trace and matrix multiplication, prove that for
any two matrices A and B,

tr (AB) = tr (BA):

Answer: For any n · n matrices A and B,

5. Using Problem 4, prove that for any matrix C and any invertible matrix S,

Answer: Let A = S and B = CS-1. Then, using Problem 4,

6. Prove that B is really a basis for .
Answer: First we show that is linearly independent . Suppose that
Apply this function to v1:

So c1 = 0. Similarly, applying the function to any vj shows that cj = 0. Thus
This shows that B is linearly independent. To show that it spans , let f : be an arbitrary
linear transformation. Let I claim that To
see this, let vj be any element of B. Then

So the functions and f agree on every element of B, and hence on all of V .

7. What is the relationship between and ?

Answer: In order to simplify the notation, let and These mean that

We now compute in two different ways . On the one hand,

(because only the k = j term survives ). On the other hand, using the fact that fi is linear we

(because only the k = i term survives). Thus We conclude that

[Remark: Some students were skeptical of this problem; they seemed to suspect that I made
it up just to irritate them! I did not; the dual space (I substituted the term \mirror space" to
throw o potential cheaters) is a foundational concept used throughout linear algebra and its
applications. For example, you can't do general relativity without it.]

[Remark: Remember that matrices are used to represent linear transformations (among other
things). Multiplying matrices corresponds to composing transformations; that's why matrix
multiplication exists . Adding matrices corresponds to adding transformations. So to what does
transposing matrices correspond? Now you know: dualizing transformations.]

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