06-240/Term Test

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The Test

Front Page

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Math 240 Algebra I - Term Test


University of Toronto, October 24, 2006

Solve the 5 problems on the other side of this page.

Each of the problems is worth 20 points.

You have an hour and 45 minutes.

Notes.

  • No outside material other than stationary and a basic calculator is allowed.
  • We will have an extra hour of class time in our regular class room on Thursday, replacing the first tutorial hour.
  • The final exam date was posted by the faculty - it will take place on Wednesday December 13 from 2PM until 5PM at room 3 of the Clara Benson Building, 320 Huron Street (south west of Harbord cross Huron, home of the Faculty of Physical Education and Health).
Good Luck!

Questions Page

Solve the following 5 problems. Each of the problems is worth 20 points. You have an hour and 45 minutes.

Problem 1. Let F be a field with zero element 0_F, let V be a vector space with zero element 0_V and let v\in V be some vector. Using only the axioms of fields and vector spaces, prove that 0_F\cdot v=0_V.

Problem 2.

  1. In the field {\mathbb C} of complex numbers, compute
    \frac{1}{2+3i}+\frac{1}{2-3i}      and     \frac{1}{2+3i}-\frac{1}{2-3i}.
  2. Working in the field {\mathbb Z}/7 of integers modulo 7, make a table showing the values of a^{-1} for every a\neq 0.

Problem 3. Let V be a vector space and let W_1 and W_2 be subspaces of V. Prove that W_1\cup W_2 is a subspace of V iff W_1\subset W_2 or W_2\subset W_1.

Problem 4. In the vector space M_{2\times 2}({\mathbb Q}), decide if the matrix \begin{pmatrix}1&2\\-3&4\end{pmatrix} is a linear combination of the elements of S=\left\{\begin{pmatrix}1&0\\-1&0\end{pmatrix},\ \begin{pmatrix}0&1\\0&1\end{pmatrix},\ \begin{pmatrix}1&1\\0&0\end{pmatrix}\right\}.

Problem 5. Let V be a finite dimensional vector space and let W_1 and W_2 be subspaces of V for which W_1\cap W_2=\{0\}. Denote the linear span of W_1\cup W_2 by W_1+W_2. Prove that \dim(W_1+W_2)=\dim W_1 + \dim W_2.

Good Luck!
[print]

The Results

Excluding a few late exams and before appeals, 63 students took the exam; The average grade was 67.46 and the standard deviation was 24.52.

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 and 2 were graded by Dmitry Donin, problem 3 was graded by Dror and problems 4 and 5 were graded by Paul Lee.

The deadline to start the appeal process is Thursday November 2 at 4PM.

Solution Set

Students are most welcome to post a solution set here.

WARNING: The solution set below, written for students and by students, is provided "as is", with absolutely no warranty. It can not be assumed to be complete, correct, reliable or relevant. If you don't like it, don't read it. Visit this pages' history tab to see who added what and when.

Problem 1. 0_F\cdot v=(0_F+0_F)\cdot v (by F3)

(0_F+0_F)\cdot v=0_F\cdot v+0_F\cdot v (by VS8)

By VS4, \exists\ (0_F\cdot v)' s.t. (0_F\cdot v)+(0_F\cdot v)'=0_V

Add (0_F\cdot v)' to both sides of 0_F\cdot v=0_F\cdot v+0_F\cdot v

(0_F\cdot v)'+(0_F\cdot v)=[(0_F\cdot v)'+0_F\cdot v]+0_F\cdot v

0_V=0_V+0_F\cdot v (by construction)

0_V=0_F\cdot v (by VS3)

Problem 2. 1) \frac{1}{2+3i}+\frac{1}{2-3i}=\frac{(2-3i)+(2+3i)}{(2+3i)(2-3i)}=\frac{4}{2^2+3^3}=\frac{4}{13}

\frac{1}{2+3i}-\frac{1}{2-3i}=\frac{(2-3i)-(2+3i)}{(2+3i)(2-3i)}=\frac{-6i}{2^2+3^3}=-\frac{6}{13}i

2) 1^{-1}=1

2^{-1}=4

3^{-1}=5

4^{-1}=2

5^{-1}=3

6^{-1}=6

Problem 3. First suppose W_1\subset W_2 or W_2\subset W_1, then W_1\cup W_2 equals W_2 or W_1. Either case W_1\cup W_2 is a subspace of V.

Now suppose W_1\cup W_2 is a subspace of V and neither W_1\subset W_2 nor W_2\subset W_1. It must follow that \exists\ x\in W_1 s.t. x is not in W_2 and \exists\ y\in W_2 s.t. y is not in W_1. Since x,y\in W_1\cup W_2, x+y\in W_1\cup W_2. If x+y\in W_1, then \exists\ -x\in W_1 s.t. (-x)+x+y\in W_1 and (-x)+x=0, so y\in W_1, a contradiction. If x+y\in W_2, then \exists\ -y\in W_2 s.t. x+y+(-y)\in W_2 and (-y)+y=0, so x\in W_2, a contradiction. Therefore either W_1\subset W_2 or W_2\subset W_1.

Problem 4. Suppose \begin{pmatrix}1&2\\-3&4\end{pmatrix} is a linear combination of \begin{pmatrix}1&0\\-1&0\end{pmatrix}, \begin{pmatrix}0&1\\0&1\end{pmatrix}, and \begin{pmatrix}1&1\\0&0\end{pmatrix}, then \exists\ a_1, a_2, a_3\in F s.t. \begin{pmatrix}1&2\\-3&4\end{pmatrix}=a_1\begin{pmatrix}1&0\\-1&0\end{pmatrix}+a_2\begin{pmatrix}0&1\\0&1\end{pmatrix}+a_3\begin{pmatrix}1&1\\0&0\end{pmatrix}.

\mbox{Need to solve}\begin{cases}
1=a_1^{}+a_3^{}\\
2=a_2^{}+a_3^{}\\
-3=-a_1^{}\\
4=a_2^{}\end{cases}

Solving the equations yields a_1=3, a_2=4, a_3=-2, so \begin{pmatrix}1&2\\-3&4\end{pmatrix} is a linear combination of \begin{pmatrix}1&0\\-1&0\end{pmatrix}, \begin{pmatrix}0&1\\0&1\end{pmatrix}, and \begin{pmatrix}1&1\\0&0\end{pmatrix}. Specifically, \begin{pmatrix}1&2\\-3&4\end{pmatrix}=3\begin{pmatrix}1&0\\-1&0\end{pmatrix}+4\begin{pmatrix}0&1\\0&1\end{pmatrix}-2\begin{pmatrix}1&1\\0&0\end{pmatrix}.

Problem 5. Since V is finite-dimensional, then so are W_1 and W_2 and their basis. Let {a_1, a_2,..., a_m} be the basis of W_1 and {b_1, b_2,..., b_n} be the basis of W_2 so \dim W_1=m and \dim W_2=n.

We know {a_1, a_2,..., a_m} and {b_1, b_2,..., b_n} are linearly independent and clearly {a_1, a_2,..., a_m, b_1, b_2,..., b_n} spans W_1+W_2. If {a_1, a_2,..., a_m, b_1, b_2,..., b_n} is linearly dependent, then there exist not all zero coefficients c_1, c_2,..., c_{m+n}\in F s.t. c_1 a_1+c_2 a_2+...+c_m a_m+c_{m+1} b_1+c_{m+2} b_2+...+c_{m+n} b_n=0. Then some linear combinations of {a_1, a_2,..., a_m} with not all zero coefficients can be expressed in linear combinations of {b_1, b_2,..., b_n}, but this would imply W_1\cap W_2\neq \{0\}, a contradiction. Therefore {a_1, a_2,..., a_m, b_1, b_2,..., b_n} is a linearly independent set that spans W_1+W_2, it's the basis of W_1+W_2. We have \dim (W_1+W_2)=m+n=\dim W_1+\dim W_2.