09-240/Term Test: Difference between revisions
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'''Problem 1''' |
'''Problem 1''' |
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If <math> |
If <math>c \cdot v = 0</math> and <math>c \neq 0,</math> |
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then <math> |
then <math>c^{-1} \cdot (c \cdot v) = 0;</math> |
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thus <math>( |
thus <math>(c^{-1} \cdot c) \cdot v = 0;</math> |
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thus <math>1 \cdot |
thus <math>1 \cdot v = 0;</math> |
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thus <math> |
thus <math>v = 0.</math> |
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It should be remembered that "or" means "one or the other, or both." |
It should be remembered that "or" means "one or the other, or both." |
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'''Problem 3.''' First suppose <math>W_1\subset W_2</math> or <math>W_2\subset W_1</math>, then <math>W_1\cup W_2</math> equals <math>W_2</math> or <math>W_1</math>. Either case <math>W_1\cup W_2</math> is a subspace of <math>V</math>. |
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Now suppose <math>W_1\cup W_2</math> is a subspace of <math>V</math> and neither <math>W_1\subset W_2</math> nor <math>W_2\subset W_1</math>. It must follow that <math>\exists\ x\in W_1</math> s.t. <math>x</math> is not in <math>W_2</math> and <math>\exists\ y\in W_2</math> s.t. <math>y</math> is not in <math>W_1</math>. Since <math>x,y\in W_1\cup W_2</math>, <math>x+y\in W_1\cup W_2</math>. If <math>x+y\in W_1</math>, then <math>\exists\ -x\in W_1</math> s.t. <math>(-x)+x+y\in W_1</math> and <math>(-x)+x=0</math>, so <math>y\in W_1</math>, a contradiction. If <math>x+y\in W_2</math>, then <math>\exists\ -y\in W_2</math> s.t. <math>x+y+(-y)\in W_2</math> and <math>(-y)+y=0</math>, so <math>x\in W_2</math>, a contradiction. Therefore either <math>W_1\subset W_2</math> or <math>W_2\subset W_1</math>. |
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Let <math>S_1 = \{ u_1, \ldots, u_n \}</math> and <math>S_2 = \operatorname{span}(S_1)</math>. |
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Assume <math>S_1</math> is linearly independent. Then <math>S_1</math> is a basis, and by basis properties, <math>\forall x \in S_2, \exists</math> unique <math>a_1, \ldots, a_n \in F</math> such that <math>a_1 u_1 + \ldots + a_n u_n = x</math>. <math>\forall a \in F</math>, <math>a = 0</math> or <math>a = 1</math>, so there are two possibilities for every <math>a</math>. Therefore, <math>|S_2| = \underbrace{2 \times \ldots \times 2}_n = 2^n</math>, so <math>S_1</math> is linearly independent <math>\Rightarrow |S_2| = 2^n</math>. |
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'''Problem 4''' |
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Assume <math>S_1</math> is linearly dependent. Then <math>\exists</math> non-trivial <math>a_i, \ldots, a_n</math> such that <math>a_1 u_1 + \ldots + a_n u_n = 0</math>. Since 0 has at least two representations, <math>|S_2| < 2^n</math>. Hence, <math>S_1</math> is linearly dependent <math>\Rightarrow |S_2| \ne 2^n</math>. By contrapositive, <math>|S_2| = 2^n \Rightarrow S_1</math> is linearly independent. |
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Hence, <math>S_1</math> is linearly independent <math>\iff |S_2| = 2^n</math>. |
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'''Problem 5''' |
'''Problem 5''' |
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Therefore, |
Therefore, |
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<math>g(x) = 5 \cdot -\frac{1}{6}(x^3 - 3x^2 + 2x) + 4 \cdot \frac{1}{2}(x^3 - 2x^2 - x + 2) + 3 \cdot -\frac{1}{2}(x^3 - x^2 - 2x) + 8 \cdot \frac{1}{6}(x^3 - x)</math> |
<math>g(x) = 5 \cdot \left( -\frac{1}{6}(x^3 - 3x^2 + 2x) \right) + 4 \cdot \frac{1}{2}(x^3 - 2x^2 - x + 2) + 3 \cdot \left( -\frac{1}{2}(x^3 - x^2 - 2x) \right) + 8 \cdot \frac{1}{6}(x^3 - x)</math> |
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Infinitely many eraser bits later... |
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⚫ |
Latest revision as of 02:53, 7 December 2009
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Announcement
Our one and only Term Test is coming up. It will take place in class on Thursday October 22 2009, starting promptly at 1:10PM and ending at 3:00PM sharp, in our normal classroom, MP103. It will consist of 4-5 questions (each may have several parts) on everything that will be covered in class by October 16: the axiomatic definition of fields and some basic properties of fields, and other examples, a tiny bit on the field with elements , the axiomatic definition of vector spaces, basic properties and examples of vector spaces, spans, linear combinations and linear equations, linear dependence and independence, bases, the replacement lemma and its consequences, a bit about linear transformations and a few smaller topics that we touched but that do not deserve their own headers.
Note that there may be some computations, but nothing that will require a calculator. Note also that I may include some questions from the homework assignments verbatim or nearly verbatim.
- Will there be "proof questions"?
- Sure. What else have we done so far?
- Do we need to know the proofs from class?
- Sure. There's a reason why these proofs are in class to start with; if they weren't valuable, we wouldn't have covered them.
No electronic devices capable of displaying text or sounding speech will be allowed.
In style and spirit this exam will not be very different of the one I gave 3 years ago. See 06-240/Term Test.
The Test
Front Page
Do not turn this page until instructed.
Math 240 Algebra I - Term Test
University of Toronto, October 22, 2009
Solve the 5 problems on the other side of this page.
Each of the problems is worth 20 points.
You have an hour and 50 minutes.
Notes.
- No outside material other than stationary and a basic calculator is allowed.
- We will have an hour of discussion time right after this test.
- The final exam date was posted by the faculty --- it will take place on Wednesday December 16 from 9AM until noon at room BN2S of the Clara Benson Building, 320 Huron Street (south west of Harbord cross Huron, home of the Faculty of Physical Education and Health).
Questions Page
Solve the following 5 problems. Each of the problems is worth 20 points. You have an hour and 50 minutes.
Problem 1. Let be a vector space over a field , let and let . Prove that if , then either or .
Problem 2.
- In the field of complex numbers, compute
(To be precise, "compute" means "write in the form , where ").
- In the field of complex numbers, find an element so that .
- In the 11-element field of remainders modulo 11, find
all solutions of the equation .
Problem 3. Let be a vector space and let and be subspaces of . Prove that is a subspace of iff or .
Problem 4. Let be vectors in a vector space over the field with two element . Show that the number of elements in the set is equal to if and only if are linearly independent..
Problem 5. Find a polynomial that satisfies , , , and .
The Results
99 students took the term test. Before appeals, the average grade was 64.41 and the standard deviation was 23.79.
The results are quite similar to what I expected them to be. The easiest questions (on average) were the computational ones, the hardest were the ones involving proofs.
How should you read your grade?
- If you got 100 you should pat yourself on your shoulder and feel good.
- If you got something like 95, you're doing great. You made a few relatively minor mistakes; find out what they are and try to avoid them next time.
- If you got something like 80 you're doing fine but you did miss something significant, probably more than just a minor thing. Figure out what it was and make a plan to fix the problem for next time.
- If you got something like 60 you should be concerned. You are still in position to improve greatly and get an excellent grade at the end, but what you missed is quite significant and you are at the risk of finding yourself far behind. You must analyze what happened - perhaps it was a minor mishap, but more likely you misunderstood something major or something major is missing in your background. Find out what it is and try to come up with a realistic strategy to overcome the difficulty!
- If you got something like 35, most likely you are not gaining much from this class and you should consider dropping it, unless you are convinced that you fully understand the cause of your difficulty (you were very sick, you really couldn't study at all for the two weeks before the exam because of some unusual circumstances, something like that) and you feel confident you have a fix for next time. If you do decide to drop the class, don't feel too bad about it. It is the hardest first year algebra class at UofT and of the thousands of students taking math here, very few come with sufficient preparation to do well in it.
Note that problems with writing are problems, period. Perhaps you got a low grade but you feel you know the material enough for a high grade only you didn't write everything you know or you didn't it write well enough or the silly graders simply didn't get what you wrote (and it isn't a simple misunderstanding - see "appeals" below). If this describes you, don't underestimate your problem. If you don't process and resolve it, it is likely to recur.
Appeals.
Remember! Grading is a difficult process and mistakes always happen - solutions get misread, parts are forgotten, grades are not added up correctly. You must read your exam and make sure that you understand how it was graded. If you disagree with anything, don't hesitate to complain! Your first stop should be the person who graded the problem in question, and only if you can't agree with him you should appeal to Dror.
Problem 1 was graded by Dror problems 2 and 3 were graded by Alan Lai and problems 4 and 5 were graded by Nevena Francetic.
The deadline to start the appeal process is Thursday November 5 at 4PM.
Solution Set
Students are most welcome to post a solution set here.
Problem 1
If and
then
thus
thus
thus
It should be remembered that "or" means "one or the other, or both."
Problem 2
(1)
(2)
If then
(3)
Problem 3. First suppose or , then equals or . Either case is a subspace of .
Now suppose is a subspace of and neither nor . It must follow that s.t. is not in and s.t. is not in . Since , . If , then s.t. and , so , a contradiction. If , then s.t. and , so , a contradiction. Therefore either or .
Problem 4
Let and .
Assume is linearly independent. Then is a basis, and by basis properties, unique such that . , or , so there are two possibilities for every . Therefore, , so is linearly independent .
Assume is linearly dependent. Then non-trivial such that . Since 0 has at least two representations, . Hence, is linearly dependent . By contrapositive, is linearly independent.
Hence, is linearly independent .
Problem 5
Recall that and that
After tedious computation with the addition of eraser bits covering your paper, you should obtain the following:
Therefore,