Drorbn - User contributions [en]

14-240/Tutorial-December 2

2014-12-09T16:55:00Z

Bug: /* Theorem */

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==Boris==

====Theorem====

Let <math>A</math> be a <math>n \times n</math> matrix and <math>B</math> be the matrix <math>A</math> with two rows interchanged. Then <math>det(A) = -det(B)</math>. Boris decided to prove the following lemma first:

=====Lemma 1=====

Let <math>A</math> be a <math>n \times n</math> matrix and <math>B</math> be the matrix <math>A</math> with two '''adjacent''' rows interchanged. Then <math>det(A) = -det(B)</math>.

All we need to show is that <math>det(A) + det(B) = 0</math>. Assume that <math>B</math> is the matrix <math>A</math> with rows <math>i, i + 1</math> of <math>A</math> interchanged. Since the determinant of a matrix with two identical rows is <math>0</math>, then:

:::::::<math>det(A) + det(B) =</math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} =</math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i\\...\end{pmatrix} + det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix}</math>.

Since the determinant is linear in each row, then we continue where we left off:

:::::::<math>det(A) + det(B) =</math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i\\...\end{pmatrix} + det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} = </math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i + A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i + A_{i + 1}\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i + A_{i + 1}\\A_i + A_{i + 1}\\...\end{pmatrix} = 0</math>.

Then <math>det(A) + det(B) = 0</math> and <math>det(A) = -det(B)</math>. The proof of the lemma is complete.

For the proof of the theorem, assume that <math>B</math> is the matrix <math>A</math> with rows <math>i, j</math> of <math>A</math> interchanged and <math>i \neq j</math>. By '''Lemma 1''', we have the following:

:::::::<math>det(A) =</math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\\A_j\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\\A_j\\...\end{pmatrix} = (-1)^{j - i}det\begin{pmatrix}...\\A_{i + 1}\\...\\A_j\\A_i\\...\end{pmatrix} =</math>

:::::::<math>(-1)^{j - i}(-1)^{j - i - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)^{2(j - i) - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>

:::::::<math>(-1)^{- 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>

:::::::<math>-det(B)</math>.

Then the proof of the theorem is complete.

==Nikita==
==Scanned Tutorial Notes by [[User Boyang.wu|Boyang.wu]]==
[[File:Tut.pdf]]

14-240/Tutorial-December 2

2014-12-07T21:01:41Z

Bug: /* Boris */

{{14-240/Navigation}}

==Boris==

====Theorem====

Let <math>A</math> be a <math>n \times n</math> matrix and <math>B</math> be the matrix <math>A</math> with two rows interchanged. Then <math>det(A) = -det(B)</math>. Boris decided to prove the following lemma first:

=====Lemma 1=====

Let <math>A</math> be a <math>n \times n</math> matrix and <math>B</math> be the matrix <math>A</math> with two '''adjacent''' rows interchanged. Then <math>det(A) = -det(B)</math>.

All we need to show is that <math>det(A) + det(B) = 0</math>. Assume that <math>B</math> is the matrix <math>A</math> with rows <math>i, i + 1</math> of <math>A</math> interchanged. Since the determinant of a matrix with two identical rows is <math>0</math>, then:

:::::::<math>det(A) + det(B) =</math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} =</math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i\\...\end{pmatrix} + det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix}</math>.

Since the determinant is linear in each row, then we continue where we left off:

:::::::<math>det(A) + det(B) =</math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i\\...\end{pmatrix} + det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} = </math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i + A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i + A_{i + 1}\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i + A_{i + 1}\\A_i + A_{i + 1}\\...\end{pmatrix} = 0</math>.

Then <math>det(A) + det(B) = 0</math> and <math>det(A) = -det(B)</math>. The proof of the lemma is complete.

For the proof of the theorem, assume that <math>B</math> is the matrix <math>A</math> with rows <math>i, j</math> of <math>A</math> interchanged and <math>i \neq j</math>. By '''Lemma 1''', we have the following:

:::::::<math>det(A) =</math>

:::::::<math>(-1)det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\\A_j\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\\A_j\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\\A_j\\...\end{pmatrix} =</math>

:::::::<math>(-1)^{j - i}det\begin{pmatrix}...\\A_{i + 1}\\...\\A_j\\A_i\\...\end{pmatrix} = (-1)^{j - i}(-1)^{j - i - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>

:::::::<math>(-1)^{2(j - i) - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)^{- 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>

:::::::<math>-det(B)</math>.

Then the proof of the theorem is complete.

==Nikita==

14-240/Tutorial-December 2

2014-12-07T20:56:22Z

Bug:

{{14-240/Navigation}}

==Boris==

====Theorem====

Let <math>A</math> be a <math>n \times n</math> matrix and <math>B</math> be the matrix <math>A</math> with two rows interchanged. Then <math>det(A) = -det(B)</math>. Boris decided to prove the following lemma first:

=====Lemma 1=====

Let <math>A</math> be a <math>n \times n</math> matrix and <math>B</math> be the matrix <math>A</math> with two '''adjacent''' rows interchanged. Then <math>det(A) = -det(B)</math>.

All we need to show is that <math>det(A) + det(B) = 0</math>. Assume that <math>B</math> is the matrix <math>A</math> with rows <math>i, i + 1</math> of <math>A</math> interchanged. Since the determinant of a matrix with two identical rows is <math>0</math>, then:

:::::::<math>det(A) + det(B) =</math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} =</math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i\\...\end{pmatrix} + det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix}</math>.

Since the determinant is linear in each row, then we continue where we left off:

:::::::<math>det(A) + det(B) =</math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i\\...\end{pmatrix} + det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} = </math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i + A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i + A_{i + 1}\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i + A_{i + 1}\\A_i + A_{i + 1}\\...\end{pmatrix} = 0</math>.

Then <math>det(A) + det(B) = 0</math> and <math>det(A) = -det(B)</math>. The proof of the lemma is complete.

For the proof of the theorem, assume that <math>B</math> is the matrix <math>A</math> with row <math>i</math> of <math>A</math> interchanged with row <math>j</math> of <math>A</math> and that <math>i \neq j</math>. By Lemma 1, we have the following:

:::::::<math>det(A) =</math>

:::::::<math>(-1)det\begin{pmatrix}...\\A_(i + 1)\\A_i\\...\\A_j\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\\A_j\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\\A_j\\...\end{pmatrix} =</math>

:::::::<math>(-1)^{j - i}det\begin{pmatrix}...\\A_{i + 1}\\...\\A_j\\A_i\\...\end{pmatrix} = (-1)^{j - i}(-1)^{j - i - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>

:::::::<math>(-1)^{2(j - i) - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)^{- 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>

:::::::<math>-det(B)</math>.

Then the proof of the theorem is complete.

==Nikita==

14-240/Tutorial-December 2

2014-12-07T20:54:50Z

Bug:

{{14-240/Navigation}}

==Boris==

====Theorem====

Let <math>A</math> be a <math>n \times n</math> matrix and <math>B</math> be the matrix <math>A</math> with two rows interchanged. Then <math>det(A) = -det(B)</math>. Boris decided to prove the following lemma first:

=====Lemma 1=====

Let <math>A</math> be a <math>n \times n</math> matrix and <math>B</math> be the matrix <math>A</math> with two '''adjacent''' rows interchanged. Then <math>det(A) = -det(B)</math>.

All we need to show is that <math>det(A) + det(B) = 0</math>. Assume that <math>B</math> is the matrix <math>A</math> with rows <math>i, i + 1</math> of <math>A</math> interchanged. Since the determinant of a matrix with two identical rows is <math>0</math>, then:

:::::::<math>det(A) + det(B) =</math>

:::::::<math>det(A) + det(B) = det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} =</math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i\\...\end{pmatrix} + det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix}</math>.

Since the determinant is linear in each row, then we continue where we left off:

:::::::<math>det(A) + det(B) =</math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i\\...\end{pmatrix} + det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} = </math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i + A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i + A_{i + 1}\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i + A_{i + 1}\\A_i + A_{i + 1}\\...\end{pmatrix} = 0</math>.

Then <math>det(A) + det(B) = 0</math> and <math>det(A) = -det(B)</math>. The proof of the lemma is complete.

For the proof of the theorem, assume that <math>B</math> is the matrix <math>A</math> with row <math>i</math> of <math>A</math> interchanged with row <math>j</math> of <math>A</math> and that <math>i \neq j</math>. By Lemma 1, we have the following:

:::::::<math>det(A) =</math>

:::::::<math>(-1)det\begin{pmatrix}...\\A_(i + 1)\\A_i\\...\\A_j\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\\A_j\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\\A_j\\...\end{pmatrix} =</math>

:::::::<math>(-1)^{j - i}det\begin{pmatrix}...\\A_{i + 1}\\...\\A_j\\A_i\\...\end{pmatrix} = (-1)^{j - i}(-1)^{j - i - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>

:::::::<math>(-1)^{2(j - i) - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)^{- 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>

:::::::<math>-det(B)</math>.

Then the proof of the theorem is complete.

==Nikita==

14-240/Tutorial-December 2

2014-12-07T20:51:18Z

Bug: /* Theorem */

{{14-240/Navigation}}

==Boris==

====Theorem====

Let <math>A</math> be a <math>n \times n</math> matrix and <math>B</math> be the matrix <math>A</math> with two rows interchanged. Then <math>det(A) = -det(B)</math>. Boris decided to prove the following lemma first:

=====Lemma 1=====

Let <math>A</math> be a <math>n \times n</math> matrix and <math>B</math> be the matrix <math>A</math> with two '''adjacent''' rows interchanged. Then <math>det(A) = -det(B)</math>.

All we need to show is that <math>det(A) + det(B) = 0</math>. Assume that <math>B</math> is the matrix <math>A</math> with rows <math>i, i + 1</math> of <math>A</math> interchanged. Since the determinant of a matrix with two identical rows is <math>0</math>, then:

:::::::<math>det(A) + det(B) =</math>

:::::::<math>det(A) + det(B) = det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} =</math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i\\...\end{pmatrix} + det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix}</math>.

Since the determinant is linear in each row, then we continue where we left off:

:::::::<math>det(A) + det(B) =</math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i\\...\end{pmatrix} + det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\end{pmatrix} = </math>

:::::::<math>det\begin{pmatrix}...\\A_i\\A_i + A_{i + 1}\\...\end{pmatrix} + det\begin{pmatrix}...\\A_{i + 1}\\A_i + A_{i + 1}\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i + A_{i + 1}\\A_i + A_{i + 1}\\...\end{pmatrix} = 0</math>.

Then <math>det(A) + det(B) = 0</math> and <math>det(A) = -det(B)</math>. The proof of the lemma is complete.

For the proof of the theorem, assume that <math>B</math> is the matrix <math>A</math> with row <math>i</math> of <math>A</math> interchanged with row <math>j</math> of <math>A</math> and that <math>i \neq j</math>. By Lemma 1, we have the following:

:::::::<math>det(A) =</math>

:::::::<math>(-1)det\begin{pmatrix}...\\A_(i + 1)\\A_i\\...\\A_j\\...\end{pmatrix} = det\begin{pmatrix}...\\A_i\\A_{i + 1}\\...\\A_j\\...\end{pmatrix} = (-1)det\begin{pmatrix}...\\A_{i + 1}\\A_i\\...\\A_j\\...\end{pmatrix} = (-1)^{j - i}det\begin{pmatrix}...\\A_{i + 1}\\...\\A_j\\A_i\\...\end{pmatrix} =</math>

:::::::<math>(-1)^{j - i}(-1)^{j - i - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)^{2(j - i) - 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = (-1)^{- 1}det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} =</math>

:::::::<math>(-1)det\begin{pmatrix}...\\A_j\\A_{i + 1}\\...\\A_i\\...\end{pmatrix} = -det(B)</math>.

Then the proof of the theorem is complete.

==Nikita==

14-240/Tutorial-December 2

2014-12-07T20:17:29Z

Bug: /* Lemma 1 */

14-240/Tutorial-December 2

2014-12-07T20:17:03Z

Bug: /* Theorem */

14-240/Tutorial-December 2

2014-12-07T20:14:49Z

Bug: /* Theorem */

14-240/Tutorial-December 2

2014-12-07T20:14:35Z

Bug: /* Theorem */

14-240/Tutorial-December 2

2014-12-07T20:14:25Z

Bug: /* Lemma 1 */

14-240/Tutorial-December 2

2014-12-07T20:13:50Z

Bug: /* Lemma 1 */

14-240/Tutorial-December 2

2014-12-07T20:11:12Z

Bug: /* Lemma 1 */

14-240/Tutorial-December 2

2014-12-07T20:10:07Z

Bug:

14-240/Tutorial-December 2

2014-12-07T20:08:08Z

Bug: /* Boris */

14-240/Tutorial-December 2

2014-12-07T19:15:06Z

Bug: Created page with "{{14-240/Navigation}} ==Boris== ==Nikita=="

{{14-240/Navigation}}

==Boris==

==Nikita==

14-240/Navigation

2014-12-07T18:36:49Z

Bug:

<noinclude>Back to [[14-240]].<br/></noinclude>
{| border="1px" cellpadding="1" cellspacing="0" width="100%" style="font-size: small; align: left"
|- align=center style="color: red;"
|colspan=3|'''Welcome to Math 240!'''<br/>
|- align=left
!#
!Week of...
!Notes and Links
|- align=left
|align=center|1
|Sep 8
|[[14-240/About This Class|About This Class]], What is this class about? ({{Pensieve link|Classes/14-240/one/What_is_This_Class_AboutQ.pdf|PDF}}, {{Pensieve link|Classes/14-240/What_is_This_Class_AboutQ.html|HTML}}), [[14-240/Classnotes for Monday September 8|Monday]], [[14-240/Classnotes for Wednesday September 10|Wednesday]]
|- align=left
|align=center|2
|Sep 15
|[[14-240/Homework Assignment 1|HW1]], [[14-240/Classnotes for Monday September 15|Monday]], [[14-240/Classnotes for Wednesday September 17|Wednesday]], {{Pensieve link|Classes/14-240/nb/TheComplexField.pdf|TheComplexField.pdf}} 
|- align=left
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|Sep 22
|[[14-240/Homework Assignment 2|HW2]], [[14-240/Class Photo|Class Photo]], [[14-240/Classnotes for Monday September 22|Monday]], [[14-240/Classnotes for Wednesday September 24|Wednesday]] 
|- align=left
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|Sep 29
|[[14-240/Homework Assignment 3|HW3]], [[14-240/Classnotes for Wednesday October 1|Wednesday]], [[14-240/Tutorial-Sep30|Tutorial]] 
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|Oct 6
|[[14-240/Homework Assignment 4|HW4]], [[14-240/Classnotes for Monday October 6|Monday]], [[14-240/Classnotes for Wednesday October 8|Wednesday]], [[14-240/Tutorial-October7|Tutorial]] 
|- align=left
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|Oct 13
|No Monday class (Thanksgiving), [[14-240/Classnotes for Wednesday October 15|Wednesday]], [[14-240/Tutorial-October14|Tutorial]]
|- align=left
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|Oct 20
|[[14-240/Homework Assignment 5|HW5]], [[14-240/Term Test|Term Test]] at tutorials on Tuesday, [[14-240/Classnotes for Wednesday October 22|Wednesday]]
|- align=left
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|Oct 27
|[[14-240/Homework Assignment 6|HW6]], [[14-240/Classnotes for Monday October 27|Monday]], [[14-240/Linear Algebra - Why We Care|Why LinAlg?]], [[14-240/Classnotes for Wednesday October 29|Wednesday]], [[14-240/Tutorial-October28|Tutorial]]
|- align=left
|align=center|9
|Nov 3
|Monday is the last day to drop this class, [[14-240/Homework Assignment 7|HW7]], [[14-240/Classnotes for Monday November 3|Monday]], [[14-240/Classnotes for Wednesday November 5|Wednesday]], [[14-240/Tutorial-November4|Tutorial]]
|- align=left
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|Nov 10
|[[14-240/Homework Assignment 8|HW8]], [[14-240/Classnotes for Monday November 10|Monday]], [[14-240/Tutorial-November11|Tutorial]]
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|Nov 17
|Monday-Tuesday is UofT November break
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|Nov 24
|[[14-240/Homework Assignment 9|HW9]]
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|Dec 1
|[[14-240/Classnotes for Wednesday December 3|Wednesday]] is a "makeup Monday"! [[14-240/End-of-Course Schedule|End-of-Course Schedule]], [[14-240/Tutorial-December 2|Tutorial]]
|- align=left
|align=center|F
|Dec 8
|[[14-240/The Final Exam|Our Final Exam]] will take place on Wednesday December 10, 2-5PM, at GB 404 for students with last names between A and Lo, and at GB 412 for the rest.
|- align=left
|colspan=3 align=center|[[14-240/Register of Good Deeds|Register of Good Deeds]]
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14-240/Tutorial-November4

2014-11-30T22:49:26Z

Bug: /* Cite Carefully */

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==Boris==

====Question 26 on Page 57 in Homework 5====

Let <math>a \in R</math> and <math>W = \{f \in P_n(R): f(a) = 0\}</math> be a subspace of <math>P_n(R)</math>. Find <math>dim(W)</math>.

First, let <math>f(x) \in W</math>. Then we can decompose <math>f(x)</math> since there is a <math>g(x) \in P_{n - 1}(R)</math> such that <math>f(x) = (x - a)g(x)</math>. From here, there are several approaches:

'''Approach 1: Use Isomorphisms'''

We show that <math>W</math> is isomorphic to <math>P_{n - 1}(R)</math>. Let <math>B = \{1, x, x^2, ..., x^{n - 1}\}</math> be the standard ordered basis of <math>P_{n - 1}(R)</math> and <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> be a subset of <math>W</math>. Then there is a unique linear transformation <math>T:P_{n - 1} \to W</math> such that <math>T(f(x)) = (x - a)f(x)</math> where <math>f(x) \in B</math>. Show that <math>T</math> is both one-to-one and onto and conclude that <math>dim(P_{n - 1}) = dim(W)</math>.

'''Approach 2: Use the Rank-Nullity Theorem'''

Let <math>K = \{1, x, x^2, ..., x^{n - 1}, x^n\}</math> be the standard ordered basis of <math>P_n</math> and <math>f(x) \in P_{n}(R)</math>. Then <math>f(x) = \displaystyle\sum_{i=1}^{n} c_ig_i(x)</math> where <math>c_i \in R</math> and <math>g_i(x) \in K</math>. Define <math>T: P_{n}(R) \to R</math> by <math>T(\displaystyle\sum_{i=1}^{n} c_ig_i(x))= \displaystyle\sum_{i=1}^{n} c_ig_i(a)</math>. Then it is easy to show that <math>T</math> is both well-defined and linear. Afterwards, show that <math>rank(T) = 1</math> and use the rank-nullity theorem to conclude that <math>dim(W) = n</math>.

'''Approach 3: Find a Basis with the Decomposed Polynomial'''

This approach is straightforward. Show that <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> is a basis of <math>W</math>.

'''Approach 4: Find a Basis without the Decomposed Polynomial'''

This approach requires a little more cleverness when constructing the basis: <math>S = \{x - a, (x^2 - a^2), (x^3 - a^3), ..., (x^n - a^n)\}</math>.

====Cite Carefully====

'''Boris's Section Only'''

If you use in your proof '''Corollary 1 of the Fundamental Theorem of Algebra''', then please cite it as "Corollary 1 of the Fundamental Theorem of Algebra". Do not cite it as the "Fundamental Theorem of Algebra" since that means you are citing the fundamental theorem instead of its corollary.

==Nikita==

14-240/Tutorial-November4

2014-11-30T22:47:51Z

Bug: /* Cite Carefully */

{{14-240/Navigation}}

==Boris==

====Question 26 on Page 57 in Homework 5====

Let <math>a \in R</math> and <math>W = \{f \in P_n(R): f(a) = 0\}</math> be a subspace of <math>P_n(R)</math>. Find <math>dim(W)</math>.

First, let <math>f(x) \in W</math>. Then we can decompose <math>f(x)</math> since there is a <math>g(x) \in P_{n - 1}(R)</math> such that <math>f(x) = (x - a)g(x)</math>. From here, there are several approaches:

'''Approach 1: Use Isomorphisms'''

We show that <math>W</math> is isomorphic to <math>P_{n - 1}(R)</math>. Let <math>B = \{1, x, x^2, ..., x^{n - 1}\}</math> be the standard ordered basis of <math>P_{n - 1}(R)</math> and <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> be a subset of <math>W</math>. Then there is a unique linear transformation <math>T:P_{n - 1} \to W</math> such that <math>T(f(x)) = (x - a)f(x)</math> where <math>f(x) \in B</math>. Show that <math>T</math> is both one-to-one and onto and conclude that <math>dim(P_{n - 1}) = dim(W)</math>.

'''Approach 2: Use the Rank-Nullity Theorem'''

Let <math>K = \{1, x, x^2, ..., x^{n - 1}, x^n\}</math> be the standard ordered basis of <math>P_n</math> and <math>f(x) \in P_{n}(R)</math>. Then <math>f(x) = \displaystyle\sum_{i=1}^{n} c_ig_i(x)</math> where <math>c_i \in R</math> and <math>g_i(x) \in K</math>. Define <math>T: P_{n}(R) \to R</math> by <math>T(\displaystyle\sum_{i=1}^{n} c_ig_i(x))= \displaystyle\sum_{i=1}^{n} c_ig_i(a)</math>. Then it is easy to show that <math>T</math> is both well-defined and linear. Afterwards, show that <math>rank(T) = 1</math> and use the rank-nullity theorem to conclude that <math>dim(W) = n</math>.

'''Approach 3: Find a Basis with the Decomposed Polynomial'''

This approach is straightforward. Show that <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> is a basis of <math>W</math>.

'''Approach 4: Find a Basis without the Decomposed Polynomial'''

This approach requires a little more cleverness when constructing the basis: <math>S = \{x - a, (x^2 - a^2), (x^3 - a^3), ..., (x^n - a^n)\}</math>.

====Cite Carefully====

'''Boris's Section Only'''

If you use in your proof '''Corollary 1 of the Fundamental Theorem of Algebra''', then please cite as "Corollary 1 of the Fundamental Theorem of Algebra". Do not cite it as the "Fundamental Theorem of Algebra" since that means you are citing the fundamental theorem instead of its corollary.

==Nikita==

14-240/Tutorial-November4

2014-11-30T22:47:08Z

Bug:

{{14-240/Navigation}}

==Boris==

====Question 26 on Page 57 in Homework 5====

Let <math>a \in R</math> and <math>W = \{f \in P_n(R): f(a) = 0\}</math> be a subspace of <math>P_n(R)</math>. Find <math>dim(W)</math>.

First, let <math>f(x) \in W</math>. Then we can decompose <math>f(x)</math> since there is a <math>g(x) \in P_{n - 1}(R)</math> such that <math>f(x) = (x - a)g(x)</math>. From here, there are several approaches:

'''Approach 1: Use Isomorphisms'''

We show that <math>W</math> is isomorphic to <math>P_{n - 1}(R)</math>. Let <math>B = \{1, x, x^2, ..., x^{n - 1}\}</math> be the standard ordered basis of <math>P_{n - 1}(R)</math> and <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> be a subset of <math>W</math>. Then there is a unique linear transformation <math>T:P_{n - 1} \to W</math> such that <math>T(f(x)) = (x - a)f(x)</math> where <math>f(x) \in B</math>. Show that <math>T</math> is both one-to-one and onto and conclude that <math>dim(P_{n - 1}) = dim(W)</math>.

'''Approach 2: Use the Rank-Nullity Theorem'''

Let <math>K = \{1, x, x^2, ..., x^{n - 1}, x^n\}</math> be the standard ordered basis of <math>P_n</math> and <math>f(x) \in P_{n}(R)</math>. Then <math>f(x) = \displaystyle\sum_{i=1}^{n} c_ig_i(x)</math> where <math>c_i \in R</math> and <math>g_i(x) \in K</math>. Define <math>T: P_{n}(R) \to R</math> by <math>T(\displaystyle\sum_{i=1}^{n} c_ig_i(x))= \displaystyle\sum_{i=1}^{n} c_ig_i(a)</math>. Then it is easy to show that <math>T</math> is both well-defined and linear. Afterwards, show that <math>rank(T) = 1</math> and use the rank-nullity theorem to conclude that <math>dim(W) = n</math>.

'''Approach 3: Find a Basis with the Decomposed Polynomial'''

This approach is straightforward. Show that <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> is a basis of <math>W</math>.

'''Approach 4: Find a Basis without the Decomposed Polynomial'''

This approach requires a little more cleverness when constructing the basis: <math>S = \{x - a, (x^2 - a^2), (x^3 - a^3), ..., (x^n - a^n)\}</math>.

====Cite Carefully====

'''Boris's Section Only'''

If you use in your proof '''Corollary 1 of the Fundamental Theorem of Algebra''', then please cite as "Corollary 1 of the Fundamental Theorem of Algebra". Do not cite it as the "Fundamental Theorem of Algebra" since the fundamental theorem and its corollaries are not the same.

==Nikita==

14-240/Tutorial-November11

2014-11-30T22:35:03Z

Bug:

{{14-240/Navigation}}

==Boris==

====Useful Definitions====

Let <math>V</math> be a finite dimensional vector space over a field <math>F</math>, <math>B = \{v_1, v_2, v_3, ..., v_n\}</math> be an ordered basis of <math>V</math> and <math>v \in V</math>. Then <math>v = \displaystyle\sum_{i=1}^{n} c_iv_i</math> where <math>c_i \in F</math>. Then the '''coordinate vector''' of <math>v</math> relative to <math>B</math> is the column vector <math> \begin{pmatrix}c_1\\c_2\\c_3\\...\\c_n\end{pmatrix}</math>.

Let <math>W</math> be a finite dimensional vector space over the same field <math>F</math> and <math>K = \{v_1, v_2, v_3, ..., v_m\}</math> be an ordered basis of <math>W</math>. Define a linear transformation <math>T:V \to W</math>. Then <math>T(v_j) = \displaystyle\sum_{i=1}^{m} c_{ij}T(v_j)</math> where <math>c_{ij} \in F</math>. Then the '''matrix representation''' of <math>T</math> in the ordered bases <math>B, K</math> is the matrix <math>\begin{pmatrix}c_{11} & c_{12} & c_{13} & ... & c_{1n}\\c_{21} & c_{22} & c_{23} & ... & c_{2n}\\c_{31} & c_{32} & c_{33} & ... & c_{3n}\\c_{21} & c_{22} & c_{23} & ... & c_{2n}\\... & ... & ... & ...& ... \\c_{m1} & c_{m2} & c_{m3} & ... & c_{mn}\end{pmatrix}</math>.

====Boris's Problems====

Let <math>B</math> be the standard ordered basis of <math>P_n(F)</math> and <math>K</math> be the standard ordered basis of <math>F</math>.

'''Q1'''. What is the '''coordinate vector''' of <math>x^2 + x^5</math> relative to <math>B</math>?

'''Q2'''. Let <math>T:P_n \to F</math> be a linear transformation that is defined by <math>T(f(x)) = f(0)</math>. What is the '''matrix representation''' of <math>T</math> in <math>B, K</math>?

==Nikita==

14-240/Tutorial-November4

2014-11-30T22:26:40Z

Bug:

14-240/Tutorial-November4

2014-11-30T22:26:09Z

Bug: /* Boris */

{{14-240/Navigation}}

{{14-240/Navigation}}

==Boris==

====Question 26 on Page 57 in Homework 5====

Let <math>a \in R</math> and <math>W = \{f \in P_n(R): f(a) = 0\}</math> be a subspace of <math>P_n(R)</math>. Find <math>dim(W)</math>.

First, let <math>f(x) \in W</math>. Then we can decompose <math>f(x)</math> since there is a <math>g(x) \in P_{n - 1}(R)</math> such that <math>f(x) = (x - a)g(x)</math>. From here, there are several approaches:

'''Approach 1: Use Isomorphisms'''

We show that <math>W</math> is isomorphic to <math>P_{n - 1}(R)</math>. Let <math>B = \{1, x, x^2, ..., x^{n - 1}\}</math> be the standard ordered basis of <math>P_{n - 1}(R)</math> and <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> be a subset of <math>W</math>. Then there is a unique linear transformation <math>T:P_{n - 1} \to W</math> such that <math>T(f(x)) = (x - a)f(x)</math> where <math>f(x) \in B</math>. Show that <math>T</math> is both one-to-one and onto and conclude that <math>dim(P_{n - 1}) = dim(W)</math>.

'''Approach 2: Use the Rank-Nullity Theorem'''

Let <math>K = \{1, x, x^2, ..., x^{n - 1}, x^n\}</math> be the standard ordered basis of <math>P_n</math> and <math>f(x) \in P_{n}(R)</math>. Then <math>f(x) = \displaystyle\sum_{i=1}^{n} c_ig_i(x)</math> where <math>c_i \in R</math> and <math>g_i(x) \in K</math>. Define <math>T: P_{n}(R) \to R</math> by <math>T(\displaystyle\sum_{i=1}^{n} c_ig_i(x))= \displaystyle\sum_{i=1}^{n} c_ig_i(a)</math>. Then it is easy to show that <math>T</math> is both well-defined and linear. Afterwards, show that <math>rank(T) = 1</math> and use the rank-nullity theorem to conclude that <math>dim(W) = n</math>.

'''Approach 3: Find a Basis with the Decomposed Polynomial'''

This approach is straightforward. Show that <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> is a basis of <math>W</math>.

'''Approach 4: Find a Basis without the Decomposed Polynomial'''

This approach requires a little more cleverness when constructing the basis: <math>S = \{x - a, (x^2 - a^2), (x^3 - a^3), ..., (x^n - a^n)\}</math>.

==Nikita==

14-240/Tutorial-November11

2014-11-30T22:25:36Z

Bug:

{{14-240/Navigation}}

==Boris==

====Background====

Let <math>V</math> be a finite dimensional vector space over a field <math>F</math>, <math>B = \{v_1, v_2, v_3, ..., v_n\}</math> be an ordered basis of <math>V</math> and <math>v \in V</math>. Then <math>v = \displaystyle\sum_{i=1}^{n} c_iv_i</math> where <math>c_i \in F</math>. Then the '''coordinate vector''' of <math>v</math> relative to <math>B</math> is the column vector <math> \begin{pmatrix}c_1\\c_2\\c_3\\...\\c_n\end{pmatrix}</math>.

Let <math>W</math> be a finite dimensional vector space over the same field <math>F</math> and <math>K = \{v_1, v_2, v_3, ..., v_m\}</math> be an ordered basis of <math>W</math>. Define a linear transformation <math>T:V \to W</math>. Then <math>T(v_j) = \displaystyle\sum_{i=1}^{m} c_{ij}T(v_j)</math> where <math>c_{ij} \in F</math>. Then the '''matrix representation''' of <math>T</math> in the ordered bases <math>B, K</math> is the matrix <math>\begin{pmatrix}c_{11} & c_{12} & c_{13} & ... & c_{1n}\\c_{21} & c_{22} & c_{23} & ... & c_{2n}\\c_{31} & c_{32} & c_{33} & ... & c_{3n}\\c_{21} & c_{22} & c_{23} & ... & c_{2n}\\... & ... & ... & ...& ... \\c_{m1} & c_{m2} & c_{m3} & ... & c_{mn}\end{pmatrix}</math>.

====Boris's Problems====

Let <math>B</math> be the standard ordered basis of <math>P_n(F)</math> and <math>K</math> be the standard ordered basis of <math>F</math>.

'''Q1'''. What is the '''coordinate vector''' of <math>x^2 + x^5</math> relative to <math>B</math>?

'''Q2'''. Let <math>T:P_n \to F</math> be a linear transformation that is defined by <math>T(f(x)) = f(0)</math>. What is the '''matrix representation''' of <math>T</math> in <math>B, K</math>?

==Nikita==

14-240/Tutorial-November4

2014-11-30T22:25:20Z

Bug: /* Boris */

14-240/Tutorial-November11

2014-11-30T00:08:52Z

Bug: /* Coordinate and Matrix Representation Problems */

14-240/Tutorial-November11

2014-11-30T00:02:23Z

Bug: /* Coordinate and Matrix Representation Problems */

{{14-240/Navigation}}

==Boris==

====Coordinate and Matrix Representation Problems====

Let <math>V</math> be a finite dimensional vector space over a field <math>F</math>, <math>B = \{v_1, v_2, v_3, ..., v_n\}</math> be an ordered basis of <math>V</math> and <math>v \in V</math>. Then <math>v = \displaystyle\sum_{i=1}^{n} c_iv_i</math> where <math>c_i \in F</math>. Then the '''coordinate vector''' of <math>v</math> relative to <math>B</math>is the column vector <math> \begin{pmatrix}c_1\\c_2\\c_3\\...\\c_n\end{pmatrix}</math>.

Let <math>W</math> be a finite dimensional vector space over the same field <math>F</math> and <math>K = \{v_1, v_2, v_3, ..., v_m\}</math> be an ordered basis of <math>W</math>. Define a linear transformation <math>T:V \to W</math>. Then <math>T(v_j) = \displaystyle\sum_{i=1}^{m} c_{ij}T(v_j)</math> where <math>c_{ij} \in F</math>. Then the '''matrix representation''' of <math>T</math> in the ordered bases <math>B, K</math> is the matrix <math>\begin{pmatrix}c_{11} & c_{12} & c_{13} & ... & c_{1n}\\c_{21} & c_{22} & c_{23} & ... & c_{2n}\\c_{31} & c_{32} & c_{33} & ... & c_{3n}\\c_{21} & c_{22} & c_{23} & ... & c_{2n}\\... & ... & ... & ...& ... \\c_{m1} & c_{m2} & c_{m3} & ... & c_{mn}\end{pmatrix}</math>.

Boris's Problems:

Let <math>S</math> be the standard ordered basis of <math>P_n(F)</math>.

'''Q1'''. What is the '''coordinate vector''' of <math>x^2 + x^5</math> relative to <math>S</math>?

'''Q2'''. Let <math>T:P_n \to F</math> be a linear transformation that is defined by <math>T(f) = f(0)</math>. What is the '''matrix representation''' of <math>T</math> in <math>S</math>?

14-240/Tutorial-November11

2014-11-29T23:52:14Z

Bug: /* Coordinate and Matrix Representation Problems */

{{14-240/Navigation}}

==Boris==

====Coordinate and Matrix Representation Problems====

Let <math>V</math> be a finite dimensional vector space over a field <math>F</math>, <math>B = \{v_1, v_2, v_3, ..., v_n\}</math> be an ordered basis of <math>V</math> and <math>v \in V</math>. Then <math>v = \displaystyle\sum_{i=1}^{n} c_iv_i</math> where <math>c_i \in F</math>. Then the '''coordinate representation''' of <math>v</math> is the column vector <math> \begin{pmatrix}c_1\\c_2\\c_3\\...\\c_n\end{pmatrix}</math>.

Let <math>W</math> be a finite dimensional vector space over the same field <math>F</math> and <math>K = \{v_1, v_2, v_3, ..., v_m\}</math> be an ordered basis of <math>W</math>. Define a linear transformation <math>T:V \to W</math>. Then <math>T(v_j) = \displaystyle\sum_{i=1}^{m} c_{ij}T(v_j)</math> where <math>c_{ij} \in F</math>. Then the '''matrix representation''' of <math>T</math> in the ordered bases <math>B, K</math> is the matrix <math>\begin{pmatrix}c_{11} & c_{12} & c_{13} & ... & c_{1n}\\c_{21} & c_{22} & c_{23} & ... & c_{2n}\\c_{31} & c_{32} & c_{33} & ... & c_{3n}\\c_{21} & c_{22} & c_{23} & ... & c_{2n}\\... & ... & ... & ...& ... \\c_{m1} & c_{m2} & c_{m3} & ... & c_{mn}\end{pmatrix}</math>.

14-240/Tutorial-November11

2014-11-29T23:51:30Z

Bug: /* Boris */

{{14-240/Navigation}}

==Boris==

====Coordinate and Matrix Representation Problems====

Recall:

Let <math>V</math> be a finite dimensional vector space over a field <math>F</math>, <math>B = \{v_1, v_2, v_3, ..., v_n\}</math> be an ordered basis of <math>V</math> and <math>v \in V</math>. Then <math>v = \displaystyle\sum_{i=1}^{n} c_iv_i</math> where <math>c_i \in F</math>. Then the '''coordinate representation''' of <math>v</math> is the column vector <math> \begin{pmatrix}c_1\\c_2\\c_3\\...\\c_n\end{pmatrix}</math>.

Let <math>W</math> be a finite dimensional vector space over the same field <math>F</math> and <math>K = \{v_1, v_2, v_3, ..., v_m\}</math> be an ordered basis of <math>W</math>. Define a linear transformation <math>T:V \to W</math>. Then <math>T(v_j) = \displaystyle\sum_{i=1}^{m} c_{ij}T(v_j)</math> where <math>c_{ij} \in F</math>. Then the '''matrix representation''' of <math>T</math> in the ordered bases <math>B, K</math> is the matrix <math>\begin{pmatrix}c_{11} & c_{12} & c_{13} & ... & c_{1n}\\c_{21} & c_{22} & c_{23} & ... & c_{2n}\\c_{31} & c_{32} & c_{33} & ... & c_{3n}\\c_{21} & c_{22} & c_{23} & ... & c_{2n}\\... & ... & ... & ...& ... \\c_{m1} & c_{m2} & c_{m3} & ... & c_{mn}\end{pmatrix}</math>.

14-240/Tutorial-November11

2014-11-29T23:42:13Z

Bug:

14-240/Tutorial-November11

2014-11-29T23:26:00Z

Bug:

14-240/Tutorial-November11

2014-11-29T23:24:03Z

Bug: /* Coordinate and Matrix Representation Problems */

14-240/Tutorial-November11

2014-11-29T23:23:47Z

Bug: /* Coordinate and Matrix Representation Problems */

14-240/Tutorial-November11

2014-11-29T23:23:10Z

Bug: /* Coordinate and Matrix Representation Problems */

14-240/Tutorial-November11

2014-11-29T23:21:44Z

Bug: /* Boris */

14-240/Tutorial-November4

2014-11-29T22:19:01Z

Bug: /* Question 26 on Page 57 in Homework 5 */

14-240/Tutorial-November4

2014-11-29T22:17:49Z

Bug: /* Question 26 on Page 57 in Homework 5 */

{{14-240/Navigation}}

==Boris==

====Question 26 on Page 57 in Homework 5====

Let <math>a \in R</math> and <math>W = \{f \in P_n(R): f(a) = 0\}</math> be a subspace of <math>P_n(R)</math>. Find <math>dim(W)</math>.

First, let <math>f(x) \in W</math>. Then we can decompose <math>f(x)</math> since there is a <math>g(x) \in P_{n - 1}(R)</math> such that <math>f(x) = (x - a)g(x)</math>. From here, there are several approaches:

'''Approach 1: Use Isomorphisms'''

We show that <math>W</math> is isomorphic to <math>P_{n - 1}(R)</math>. Let <math>B = \{1, x, x^2, ..., x^{n - 1}\}</math> be the standard ordered basis of <math>P_{n - 1}(R)</math> and <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> be a subset of <math>W</math>. Then there is a unique linear transformation <math>T:P_{n - 1} \to W</math> such that <math>T(f(x)) = (x - a)f(x)</math> where <math>f(x) \in B</math>. Show that <math>T</math> is one-to-one and onto and conclude that <math>dim(P_{n - 1}) = dim(W)</math>.

'''Approach 2: Use the Rank-Nullity Theorem'''

Let <math>K = \{1, x, x^2, ..., x^{n - 1}, x^n\}</math> be the standard ordered basis of <math>P_n</math> and <math>f(x) \in P_{n}(R)</math>. Then <math>f(x) = \displaystyle\sum_{i=1}^{n} c_ig_i(x)</math> where <math>c_i \in R</math> and <math>g_i(x) \in K</math>. Define <math>T: P_{n}(R) \to R</math> by <math>T(\displaystyle\sum_{i=1}^{n} c_ig_i(x))= \displaystyle\sum_{i=1}^{n} c_ig_i(a)</math>. Then it is easy to show that <math>T</math> is both well-defined and linear. Afterwards, show that <math>rank(T) = 1</math> and use the rank-nullity theorem to conclude that <math>dim(W) = n</math>.

'''Approach 3: Find a Basis with the Decomposed Polynomial'''

This approach is straightforward. Show that <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> is a basis of <math>W</math>.

'''Approach 4: Find a Basis without the Decomposed Polynomial'''

This approach requires a little more cleverness when constructing the basis: <math>S = \{x - a, (x^2 - a^2), (x^3 - a^3), ..., (x^n - a^n)\}</math>.

14-240/Tutorial-November4

2014-11-29T22:11:20Z

Bug: /* Question 26 on Page 57 in Homework 5 */

{{14-240/Navigation}}

==Boris==

====Question 26 on Page 57 in Homework 5====

Let <math>a \in R</math> and <math>W = \{f \in P_n(R): f(a) = 0\}</math> be a subspace of <math>P_n(R)</math>. Find <math>dim(W)</math>.

First, let <math>f(x) \in W</math>. Then we can decompose <math>f(x)</math> since there is a <math>g(x) \in P_{n - 1}(R)</math> such that <math>f(x) = (x - a)g(x)</math>. From here, there are several approaches:

'''Approach 1: Use Isomorphisms'''

We show that <math>W</math> is isomorphic to <math>P_{n - 1}(R)</math>. Let <math>B = \{1, x, x^2, ..., x^{n - 1}\}</math> be the standard ordered basis of <math>P_{n - 1}(R)</math> and <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> be a subset of <math>W</math>. Then there is a unique linear transformation <math>T:P_{n - 1} \to W</math> such that <math>T(f(x)) = (x - a)f(x)</math> where <math>f(x) \in B</math>. Show that <math>T</math> is one-to-one and onto and conclude that <math>dim(P_{n - 1}) = dim(W)</math>.

'''Approach 2: Use the Rank-Nullity Theorem'''

Let <math>K = \{1, x, x^2, ..., x^{n - 1}, x^n\}</math> be the standard ordered basis of <math>P_n</math> and <math>f(x) \in P_{n}(R)</math>. Then <math>f(x) = \displaystyle\sum_{i=1}^{n} c_ig_i(x)</math> where <math>c_i \in R</math> and <math>g_i(x) \in K</math>. Define <math>T: P_{n}(R) \to R</math> by <math>T(\displaystyle\sum_{i=1}^{n} c_ig_i(x))= \displaystyle\sum_{i=1}^{n} c_ig_i(a)</math>. Then it is easy to show that <math>T</math> is both well-defined and linear. Afterwards, show that <math>rank(T) = 1</math> and use the rank-nullity theorem to conclude that <math>dim(W) = n</math>.

'''Approach 3: Find a Basis with the Decomposed Polynomial'''

'''Approach 4: Find a Basis without the Decomposed Polynomial'''

14-240/Tutorial-November4

2014-11-29T22:10:36Z

Bug: /* Question 26 on Page 57 in Homework 5 */

{{14-240/Navigation}}

==Boris==

====Question 26 on Page 57 in Homework 5====

Let <math>a \in R</math> and <math>W = \{f \in P_n(R): f(a) = 0\}</math> be a subspace of <math>P_n(R)</math>. Find <math>dim(W)</math>.

First, let <math>f(x) \in W</math>. Then we can decompose <math>f(x)</math> since there is a <math>g(x) \in P_{n - 1}(R)</math> such that <math>f(x) = (x - a)g(x)</math>. From here, there are several approaches:

'''Approach 1: Use Isomorphisms'''

We show that <math>W</math> is isomorphic to <math>P_{n - 1}(R)</math>. Let <math>B = \{1, x, x^2, ..., x^{n - 1}\}</math> be the standard ordered basis of <math>P_{n - 1}(R)</math> and <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> be a subset of <math>W</math>.

Then there is a unique linear transformation <math>T:P_{n - 1} \to W</math> such that <math>T(f(x)) = (x - a)f(x)</math> where <math>f(x) \in B</math>. Show that <math>T</math> is one-to-one and onto and conclude that <math>dim(P_{n - 1}) = dim(W)</math>.

'''Approach 2: Use the Rank-Nullity Theorem'''

Let <math>K = \{1, x, x^2, ..., x^{n - 1}, x^n\}</math> be the standard ordered basis of <math>P_n</math> and <math>f(x) \in P_{n}(R)</math>. Then <math>f(x) = \displaystyle\sum_{i=1}^{n} c_ig_i(x)</math> where <math>c_i \in R</math> and <math>g_i(x) \in K</math>. Define <math>T: P_{n}(R) \to R</math> by <math>T(\displaystyle\sum_{i=1}^{n} c_ig_i(x))= \displaystyle\sum_{i=1}^{n} c_ig_i(a)</math>. Then it is easy to show that <math>T</math> is both well-defined and linear. Afterwards, show that <math>rank(T) = 1</math> and use the rank-nullity theorem to conclude that <math>dim(W) = n</math>.

'''Approach 3: Find a Basis with the Decomposed Polynomial'''

'''Approach 4: Find a Basis without the Decomposed Polynomial'''

14-240/Tutorial-November4

2014-11-29T22:10:25Z

Bug: /* Question 26 on Page 57 in Homework 5 */

{{14-240/Navigation}}

==Boris==

====Question 26 on Page 57 in Homework 5====

Let <math>a \in R</math> and <math>W = \{f \in P_n(R): f(a) = 0\}</math> be a subspace of <math>P_n(R)</math>. Find <math>dim(W)</math>.

First, let <math>f(x) \in W</math>. Then we can decompose <math>f(x)</math> since there is a <math>g(x) \in P_{n - 1}(R)</math> such that <math>f(x) = (x - a)g(x)</math>. From here, there are several approaches:

'''Approach 1: Use Isomorphisms'''

We show that <math>W</math> is isomorphic to <math>P_{n - 1}(R)</math>. Let <math>B = \{1, x, x^2, ..., x^{n - 1}\}</math> be the standard ordered basis of <math>P_{n - 1}(R)</math> and <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> be a subset of <math>W</math>.

Then there is a unique linear transformation <math>T:P_{n - 1} \to W</math> such that <math>T(f(x)) = (x - a)f(x)</math> where <math>f(x) \in B</math>. Show that <math>T</math> is one-to-one and onto and conclude that <math>dim(P_{n - 1}) = dim(W)</math>.

'''Approach 2: Use the Rank-Nullity Theorem'''

Let <math>K = \{1, x, x^2, ..., x^{n - 1}, x^n\}</math> be the standard ordered basis of <math>P_n</math> and <math>f(x) \in P_{n}(R)</math>. Then <math>f(x) = \displaystyle\sum_{i=1}^{n} c_ig_i(x)</math> where <math>c_i \in R</math> and <math>g_i(x) \in K</math>. Define <math>T: P_{n}(R) \to R</math> by <math>T(\displaystyle\sum_{i=1}^{n} c_ig_i(x))= \displaystyle\sum_{i=1}^{n} c_ig_i(a)</math>. Then it is easy to show that <math>T</math> is both well-defined and linear. Afterwards, show that <math>rank(T) = 1</math> and use the rank-nullity theorem to conclude that <math>dim(W) = n</math>.

'''Approach 3: Find a Basis with the Decomposed Polynomial'''

'''Approach 4: Find a Basis without the Decomposed Polynomial'''

14-240/Tutorial-November4

2014-11-29T22:10:00Z

Bug: /* Question 26 on Page 57 in Homework 5 */

{{14-240/Navigation}}

==Boris==

====Question 26 on Page 57 in Homework 5====

Let <math>a \in R</math> and <math>W = \{f \in P_n(R): f(a) = 0\}</math> be a subspace of <math>P_n(R)</math>. Find <math>dim(W)</math>.

First, let <math>f(x) \in W</math>. Then we can decompose <math>f(x)</math> since there is a <math>g(x) \in P_{n - 1}(R)</math> such that <math>f(x) = (x - a)g(x)</math>. From here, there are several approaches:

'''Approach 1: Use Isomorphisms'''

We show that <math>W</math> is isomorphic to <math>P_{n - 1}(R)</math>. Let <math>B = \{1, x, x^2, ..., x^{n - 1}\}</math> be the standard ordered basis of <math>P_{n - 1}(R)</math> and <math>S = \{x - a, (x - a)x, (x - a)x^2, ..., (x - a)x^{n - 1}\}</math> be a subset of <math>W</math>. Then there is a unique linear transformation <math>T:P_{n - 1} \to W</math> such that <math>T(f(x)) = (x - a)f(x)</math> where <math>f(x) \in B</math>. Show that <math>T</math> is one-to-one and onto and conclude that <math>dim(P_{n - 1}) = dim(W)</math>.

'''Approach 2: Use the Rank-Nullity Theorem'''

Let <math>K = \{1, x, x^2, ..., x^{n - 1}, x^n\}</math> be the standard ordered basis of <math>P_n</math> and <math>f(x) \in P_{n}(R)</math>. Then <math>f(x) = \displaystyle\sum_{i=1}^{n} c_ig_i(x)</math> where <math>c_i \in R</math> and <math>g_i(x) \in K</math>. Define <math>T: P_{n}(R) \to R</math> by <math>T(\displaystyle\sum_{i=1}^{n} c_ig_i(x))= \displaystyle\sum_{i=1}^{n} c_ig_i(a)</math>. Then it is easy to show that <math>T</math> is both well-defined and linear. Afterwards, show that <math>rank(T) = 1</math> and use the rank-nullity theorem to conclude that <math>dim(W) = n</math>.

'''Approach 3: Find a Basis with the Decomposed Polynomial'''

'''Approach 4: Find a Basis without the Decomposed Polynomial'''

14-240/Tutorial-November4

2014-11-29T21:59:00Z

Bug: /* Boris */

14-240/Tutorial-November4

2014-11-29T21:58:10Z

Bug: /* Question 26 on Page 57 in Homework 5 */

14-240/Tutorial-November4

2014-11-29T21:47:55Z

Bug: /* Question 26 on Page 57 in Homework 5 */

14-240/Tutorial-November4

2014-11-29T21:42:02Z

Bug: /* Question 26 on Page 57 in Homework 5 */

14-240/Tutorial-November4

2014-11-29T21:41:48Z

Bug: /* Question 26 on Page 57 in Homework 5 */

14-240/Tutorial-November4

2014-11-29T21:40:20Z

Bug: /* Question 26 on Page 57 in Homework 5 */

14-240/Tutorial-November4

2014-11-29T21:40:02Z

Bug: /* Question 26 on Page 57 in Homework 5 */

14-240/Tutorial-November4

2014-11-29T21:29:05Z

Bug: /* Question 26 on Page 57 in Homework 5 */

14-240/Tutorial-November4

2014-11-29T21:28:50Z

Bug: /* Question 26 on Page 57 in Homework 5 */

14-240/Tutorial-November4

2014-11-29T21:27:44Z

Bug: /* Question 26 on Page 57 in Homework 5 */