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Material Type: Exam; Class: Elementary Linear Algebra; Subject: Math; University: Weber State University; Term: Spring 2007;
Typology: Exams
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Contents Overview 1 Problem 1 2 Problem 2 3 Problem 3 3 Problem 4 3 Problem 5 4 Problem 6 5 Problem 7 6 Problem 8 7
Overview Scores on this exam were a bit lower than last time, and not especially good in general. Of the 21 people who took the exam at or before the scheduled time, the average score was 17. Here is the score distribution: Score Number of Students 24 2 22 4 21 1 20 1 19 1 18 2 17 2 16 1 15 5 14 1 13 1 12 1 11 1
1
I do not curve exams. I occasionally curve overall class scores. If we follow the preset scale outlined on the syllabus and only counted the midterm, then the grade scale would be as follows. Score Grade 23-25 A+ , A, A- 19-22 B+, B, B- 17, 18 C+, C 15, 16 C- 12-14 D+, D, D- 0-11 E If the overall class scores were like this, I would curve a little, bump- ing up your grade by one-third. The next table shows how many people attempted each problem. Problem Number who attempted 1 0 2 20 3 12 4 13 5 17 6 16 7 9 8 16 Not everyone attempted five problems which is why the sum of at- tempted problems is 103 rather than 105.
Problem 1 Write down an injective linear transformation from P 4 to F^6. (You need not prove that it is linear or injective.) Does there exist a surjec- tive linear transformation from P 4 to F^6? Prove your claim.
Solution 1. Let p 1 (t) = 1, p 2 (t) = t, p 3 (t) = t^2 , p 4 (t) = t^3 , and p 5 (t) = t^4. Then {p 1 , p 2 , p 3 , p 4 , p 5 } is the standard basis for P 4. Let {e 1 , e 2 , e 3 , e 4 , e 5 , e 6 } be the standard basis for F^6. For each q ∈ P 4 , there exist scalars c 1 , c 2 , c 3 , c 4 , c 5 (depending on q)
such that q =
k=
ckpk. Define T : P 4 → F^6 by T (q) =
k=
ckek. It
should be reasonably clear that T is linear and injective, so we will not prove it. Now if T : P 4 → F^6 is linear, then the corresponding coordinate matrix, A, is 6 × 5. By the rank matrix theorem,
(1) rankA + dim NulA = 5.
Solution 7. This statement is false. Note that if A and B are similar, they must be square. Let A =
and B = 2A. Then B and A are row equivalent, but not square, hence not similar.
Remark 8. I meant to add the hypothesis that A and B were square. However, even if A and B are square, they need not be similar just because they are row equivalent. The statement fails even in the 1 × 1 case, since, for example,
is not similar to
even though the two matrices are row equivalent. I was disappointed that several people who claimed that the state- ment was true on past homework assignments still clung to their claim, despite careful correction.
Problem 5
Let A =
. Let T : R^2 → R^2 be the linear transformation
given by[ T x = Ax, for every x ∈ R^2. Let B = {b 1 , b 2 } where b 1 = 1 − 1
, and b 2 =
. Find the B-matrix for T.
Solution 9. Let D be the B-matrix for T. Let P =
b 1 b 2
. By a theorem, D = P −^1 AP. We compute:
Remark 10. The relevant result is Theorem 8 in 5.4. Note that the comments at the bottom of page 331 observe that the hypothesis that D is diagonal is unnecessary. This was one of those problems that almost every one misinterpreted. I certainly was surprised by this, although with hindsight, perhaps I should not have been. I graded this problem relatively generously.
Problem 6
Let A =
.^ Find the eigenvalues of^ A^ and two
linearly independent eigenvectors.
Solution 11. The characteristic polynomial is
det(A − λI) =
1 − λ − 2 0 0 − 1 1 − λ 0 0 0 0 − 1 − λ 1 0 0 − 3 3 − λ
= (1 − λ)
1 − λ 0 0 0 − 1 − λ 1 0 − 3 3 − λ
0 − 1 − λ 1 0 − 3 3 − λ
= (1 − λ)^2
− 1 − λ 1 − 3 3 − λ
− 1 − λ 1 − 3 3 − λ
= (1 − λ)^2 [(− 1 − λ)(3 − λ) + 3] − 2[(− 1 − λ)(3 − λ) + 3] = (1 − λ)^2 (λ^2 − 2 λ) − 2(λ^2 − 2 λ) = [(1 − λ)^2 − 2](λ^2 − 2 λ) = (λ^2 − 2 λ − 1)(λ^2 − 2 λ) = (λ^2 − 2 λ − 1)λ(λ − 2).
The eigenvalues are the roots of the characteristic polynomial. Hence, two of the eigenvalues are 0 and 2. The other two are the roots of λ^2 − 2 λ − 1 = 0. In this case, λ = 2 ±
√4+ 2 = 1^ ±
v 1 v 2 v 3 v 4
Hence, the dot product is commutative. Further,
(8) u · (v + w) =
∑^ n
i=
ui(vi + wi) =
∑^ n
i=
uivi +
∑^ n
i=
uiwi = u · v + u · w,
and so the dot product distributes over vector addition. Since u ∈ Rn, each component of u is real, and hence the square of each component is real and non-negative. Thus,
(9) u · u =
∑^ n
i=
uiui =
∑^ n
i=
u^2 i ≥ 0.
If equality held, then each ui = 0, and so equality holds if and only if u = 0.
Remark 14. Most people did this problem well.
Problem 8
Let u =
(^) ∈ R^3. Find the orthogonal projection of u onto
v =
Solution 15. We wish to compute
(10) projvu =
u · v v · v
v =
Remark 16. Several people transposed u and v in the formula, which, while easy to do, doesn’t make sense. We are projecting u onto v, and so the result must be a scalar multiple of v; not u.
