06-240/Term Test - Revision history

C8sd: /* Solution Set */ grammar

2009-10-22T03:49:46Z

Solution Set: grammar

← Older revision		Revision as of 23:49, 21 October 2009
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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>.		'''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>.

	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 ~~follows~~ <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>.		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>.

	'''Problem 4.''' Suppose <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}</math> is a linear combination of <math>\begin{pmatrix}1&0\\-1&0\end{pmatrix}</math>, <math>\begin{pmatrix}0&1\\0&1\end{pmatrix}</math>, and <math>\begin{pmatrix}1&1\\0&0\end{pmatrix}</math>, then <math>\exists\ a_1, a_2, a_3\in F</math> s.t. <math>\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}</math>.		'''Problem 4.''' Suppose <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}</math> is a linear combination of <math>\begin{pmatrix}1&0\\-1&0\end{pmatrix}</math>, <math>\begin{pmatrix}0&1\\0&1\end{pmatrix}</math>, and <math>\begin{pmatrix}1&1\\0&0\end{pmatrix}</math>, then <math>\exists\ a_1, a_2, a_3\in F</math> s.t. <math>\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}</math>.

Drorbn: /* Solution Set */

2009-10-21T16:43:32Z

Solution Set

← Older revision		Revision as of 12:43, 21 October 2009
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	==Solution Set==		==Solution Set==
	{{09-240/Class Notes Warning}}

	Students are most welcome to post a solution set here.		Students are most welcome to post a solution set here.

			<div style="color: red;"><span style="font-size: 150%;">WARNING: </span>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.
	Solutions posted by me, feel free to make any changes:

			Visit this pages' history tab to see who added what and when.</div>

	'''Problem 1.''' <math>0_F\cdot v=(0_F+0_F)\cdot v</math> (by F3)		'''Problem 1.''' <math>0_F\cdot v=(0_F+0_F)\cdot v</math> (by F3)

Drorbn at 16:33, 21 October 2009

2009-10-21T16:33:44Z

← Older revision		Revision as of 12:33, 21 October 2009
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	{{06-240/Results of the Term Test}}		{{06-240/Results of the Term Test}}

⚫	{{09-240/Class Notes Warning}}
	==Solution Set==		==Solution Set==
		⚫	{{09-240/Class Notes Warning}}

	Students are most welcome to post a solution set here.		Students are most welcome to post a solution set here.

Drorbn at 16:33, 21 October 2009

2009-10-21T16:33:08Z

← Older revision		Revision as of 12:33, 21 October 2009
Line 50:		Line 50:
	{{06-240/Results of the Term Test}}		{{06-240/Results of the Term Test}}

			{{09-240/Class Notes Warning}}
	==Solution Set==		==Solution Set==

Drorbn: Reverted edit of 218.74.122.100, changed back to last version by Drorbn

2007-06-15T23:43:14Z

Reverted edit of 218.74.122.100, changed back to last version by Drorbn

← Older revision		Revision as of 19:43, 15 June 2007
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	'''Problem 2.'''		'''Problem 2.'''
	# In the field <math>{\mathbb C}</math> of complex numbers, compute <center><math>\frac{1}{2 3i} \frac{1}{2-3i}</math>		# In the field <math>{\mathbb C}</math> of complex numbers, compute <center><math>\frac{1}{2+3i}+\frac{1}{2-3i}</math>      and     <math>\frac{1}{2+3i}-\frac{1}{2-3i}</math>.</center>
			# Working in the field <math>{\mathbb Z}/7</math> of integers modulo 7, make a table showing the values of <math>a^{-1}</math> for every <math>a\neq 0</math>.

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

			'''Problem 4.''' In the vector space <math>M_{2\times 2}({\mathbb Q})</math>, decide if the matrix <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}</math> is a linear combination of the elements of <math>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\}</math>.

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

			<center> '''Good Luck!''' </center>

			{{06-240/Results of the Term Test}}

			==Solution Set==

			Students are most welcome to post a solution set here.

			Solutions posted by me, feel free to make any changes:

			'''Problem 1.''' <math>0_F\cdot v=(0_F+0_F)\cdot v</math> (by F3)

			<math>(0_F+0_F)\cdot v=0_F\cdot v+0_F\cdot v</math> (by VS8)

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

			Add <math>(0_F\cdot v)'</math> to both sides of <math>0_F\cdot v=0_F\cdot v+0_F\cdot v</math>

			<math>(0_F\cdot v)'+(0_F\cdot v)=[(0_F\cdot v)'+0_F\cdot v]+0_F\cdot v</math>

			<math>0_V=0_V+0_F\cdot v</math> (by construction)

			<math>0_V=0_F\cdot v</math> (by VS3)

			'''Problem 2.''' 1) <math>\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}</math>

			<math>\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</math>

			2) <math>1^{-1}=1</math>

			<math>2^{-1}=4</math>

			<math>3^{-1}=5</math>

			<math>4^{-1}=2</math>

			<math>5^{-1}=3</math>

			<math>6^{-1}=6</math>

			'''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>.

			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 follows <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>.

			'''Problem 4.''' Suppose <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}</math> is a linear combination of <math>\begin{pmatrix}1&0\\-1&0\end{pmatrix}</math>, <math>\begin{pmatrix}0&1\\0&1\end{pmatrix}</math>, and <math>\begin{pmatrix}1&1\\0&0\end{pmatrix}</math>, then <math>\exists\ a_1, a_2, a_3\in F</math> s.t. <math>\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}</math>.

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

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

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

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

218.74.122.100 at 18:28, 15 June 2007

2007-06-15T18:28:31Z

← Older revision		Revision as of 14:28, 15 June 2007
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	'''Problem 2.'''		'''Problem 2.'''
	# In the field <math>{\mathbb C}</math> of complex numbers, compute <center><math>\frac{1}{2+3i}~~+\frac{1}{2-3i}</math>~~ ~~     and     <math>\frac{1}{2+3i}-~~\frac{1}{2-3i}</math~~>.</center~~>		# In the field <math>{\mathbb C}</math> of complex numbers, compute <center><math>\frac{1}{2 3i} \frac{1}{2-3i}</math>
	# Working in the field <math>{\mathbb Z}/7</math> of integers modulo 7, make a table showing the values of <math>a^{-1}</math> for every <math>a\neq 0</math>.

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

	'''Problem 4.''' In the vector space <math>M_{2\times 2}({\mathbb Q})</math>, decide if the matrix <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}</math> is a linear combination of the elements of <math>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\}</math>.

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

	<center> '''Good Luck!''' </center>

	{{06-240/Results of the Term Test}}

	==Solution Set==

	Students are most welcome to post a solution set here.

	Solutions posted by me, feel free to make any changes:

	'''Problem 1.''' <math>0_F\cdot v=(0_F+0_F)\cdot v</math> (by F3)

	<math>(0_F+0_F)\cdot v=0_F\cdot v+0_F\cdot v</math> (by VS8)

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

	Add <math>(0_F\cdot v)'</math> to both sides of <math>0_F\cdot v=0_F\cdot v+0_F\cdot v</math>

	<math>(0_F\cdot v)'+(0_F\cdot v)=[(0_F\cdot v)'+0_F\cdot v]+0_F\cdot v</math>

	<math>0_V=0_V+0_F\cdot v</math> (by construction)

	<math>0_V=0_F\cdot v</math> (by VS3)

	'''Problem 2.''' 1) <math>\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}</math>

	<math>\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</math>

	2) <math>1^{-1}=1</math>

	<math>2^{-1}=4</math>

	<math>3^{-1}=5</math>

	<math>4^{-1}=2</math>

	<math>5^{-1}=3</math>

	<math>6^{-1}=6</math>

	'''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>.

	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 follows <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>.

	'''Problem 4.''' Suppose <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}</math> is a linear combination of <math>\begin{pmatrix}1&0\\-1&0\end{pmatrix}</math>, <math>\begin{pmatrix}0&1\\0&1\end{pmatrix}</math>, and <math>\begin{pmatrix}1&1\\0&0\end{pmatrix}</math>, then <math>\exists\ a_1, a_2, a_3\in F</math> s.t. <math>\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}</math>.

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

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

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

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

Drorbn: Reverted edit of 87.249.225.110, changed back to last version by Drorbn

2007-05-28T14:51:08Z

Reverted edit of 87.249.225.110, changed back to last version by Drorbn

← Older revision		Revision as of 10:51, 28 May 2007
Line 37:		Line 37:

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

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

			'''Problem 4.''' In the vector space <math>M_{2\times 2}({\mathbb Q})</math>, decide if the matrix <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}</math> is a linear combination of the elements of <math>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\}</math>.

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

			<center> '''Good Luck!''' </center>

			{{06-240/Results of the Term Test}}

			==Solution Set==

			Students are most welcome to post a solution set here.

			Solutions posted by me, feel free to make any changes:

			'''Problem 1.''' <math>0_F\cdot v=(0_F+0_F)\cdot v</math> (by F3)

			<math>(0_F+0_F)\cdot v=0_F\cdot v+0_F\cdot v</math> (by VS8)

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

			Add <math>(0_F\cdot v)'</math> to both sides of <math>0_F\cdot v=0_F\cdot v+0_F\cdot v</math>

			<math>(0_F\cdot v)'+(0_F\cdot v)=[(0_F\cdot v)'+0_F\cdot v]+0_F\cdot v</math>

			<math>0_V=0_V+0_F\cdot v</math> (by construction)

			<math>0_V=0_F\cdot v</math> (by VS3)

			'''Problem 2.''' 1) <math>\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}</math>

			<math>\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</math>

			2) <math>1^{-1}=1</math>

			<math>2^{-1}=4</math>

			<math>3^{-1}=5</math>

			<math>4^{-1}=2</math>

			<math>5^{-1}=3</math>

			<math>6^{-1}=6</math>

			'''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>.

			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 follows <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>.

			'''Problem 4.''' Suppose <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}</math> is a linear combination of <math>\begin{pmatrix}1&0\\-1&0\end{pmatrix}</math>, <math>\begin{pmatrix}0&1\\0&1\end{pmatrix}</math>, and <math>\begin{pmatrix}1&1\\0&0\end{pmatrix}</math>, then <math>\exists\ a_1, a_2, a_3\in F</math> s.t. <math>\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}</math>.

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

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

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

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

87.249.225.110 at 06:40, 28 May 2007

2007-05-28T06:40:47Z

← Older revision		Revision as of 02:40, 28 May 2007
Line 37:		Line 37:

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

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

	'''Problem 4.''' In the vector space <math>M_{2\times 2}({\mathbb Q})</math>, decide if the matrix <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}</math> is a linear combination of the elements of <math>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\}</math>.

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

	<center> '''Good Luck!''' </center>

	{{06-240/Results of the Term Test}}

	==Solution Set==

	Students are most welcome to post a solution set here.

	Solutions posted by me, feel free to make any changes:

	'''Problem 1.''' <math>0_F\cdot v=(0_F+0_F)\cdot v</math> (by F3)

	<math>(0_F+0_F)\cdot v=0_F\cdot v+0_F\cdot v</math> (by VS8)

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

	Add <math>(0_F\cdot v)'</math> to both sides of <math>0_F\cdot v=0_F\cdot v+0_F\cdot v</math>

	<math>(0_F\cdot v)'+(0_F\cdot v)=[(0_F\cdot v)'+0_F\cdot v]+0_F\cdot v</math>

	<math>0_V=0_V+0_F\cdot v</math> (by construction)

	<math>0_V=0_F\cdot v</math> (by VS3)

	'''Problem 2.''' 1) <math>\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}</math>

	<math>\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</math>

	2) <math>1^{-1}=1</math>

	<math>2^{-1}=4</math>

	<math>3^{-1}=5</math>

	<math>4^{-1}=2</math>

	<math>5^{-1}=3</math>

	<math>6^{-1}=6</math>

	'''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>.

	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 follows <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>.

	'''Problem 4.''' Suppose <math>\begin{pmatrix}1&2\\-3&4\end{pmatrix}</math> is a linear combination of <math>\begin{pmatrix}1&0\\-1&0\end{pmatrix}</math>, <math>\begin{pmatrix}0&1\\0&1\end{pmatrix}</math>, and <math>\begin{pmatrix}1&1\\0&0\end{pmatrix}</math>, then <math>\exists\ a_1, a_2, a_3\in F</math> s.t. <math>\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}</math>.

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

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

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

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

Drorbn: Reverted edit of 221.203.53.51, changed back to last version by Drorbn

2007-05-18T20:23:55Z

Reverted edit of 221.203.53.51, changed back to last version by Drorbn

@@ Line 1: / Line 1: @@
 ==The Test==
@@ Line 37: / Line 37: @@
 '''Problem 2.'''
-# In the field <math>{\mathbb C}</math> of complex numbers, compute <center><math>\frac{1}{2 3i} \frac{1}{2-3i}</math>
+# In the field <math>{\mathbb C}</math> of complex numbers, compute <center><math>\frac{1}{2+3i}+\frac{1}{2-3i}</math> &nbsp;&nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp; <math>\frac{1}{2+3i}-\frac{1}{2-3i}</math>.</center>

221.203.53.51 at 19:45, 18 May 2007

2007-05-18T19:45:38Z

Show changes