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06-240/Classnotes For Thursday November 16

2006-11-18T20:30:59Z

216.58.41.136:

{{06-240/Navigation}}

[[Media:06-240-nov16lec-1.jpeg|Nov16 Lecture notes: 1 of 2]]

[[Media:06-240-nov16lec-2.jpeg|Nov16 Lecture notes: 2 of 2]]

[[Media:06-240-nov16tut-1.jpeg|Nov16 Tutorial notes: 1 of 4]]

[[Media:06-240-nov16tut-2.jpeg|Nov16 Tutorial notes: 2 of 4]]

[[Media:06-240-nov16tut-3.jpeg|Nov16 Tutorial notes: 3 of 4]]

[[Media:06-240-nov16tut-4.jpeg|Nov16 Tutorial notes: 4 of 4]]

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2006-11-15T20:53:33Z

216.58.41.136:

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|align=center|1
|Sep 11
|[[06-240/About This Class|About]], [[06-240/Classnotes For Tuesday, September 12|Tue]], [[06-240/Homework Assignment 1|HW1]], [[06-240/Putnam Competition|Putnam]], [[06-240/Classnotes for Thursday, September 14|Thu]]
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|Sep 18
|[[06-240/Classnotes For Tuesday, September 19|Tue]], [[06-240/Homework Assignment 2|HW2]], [[06-240/Classnotes For Thursday, September 21|Thu]]
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|Sep 25
|[[06-240/Classnotes For Tuesday September 26|Tue]], [[06-240/Homework Assignment 3|HW3]], [[06-240/Class Photo|Photo]], [[06-240/Classnotes For Thursday, September 28|Thu]]
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|Oct 2
|[[06-240/Classnotes For Tuesday October 3|Tue]], [[06-240/Homework Assignment 4|HW4]], [[06-240/Classnotes For Thursday October 5|Thu]]
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|Oct 9
|[[06-240/Classnotes For Tuesday October 10|Tue]], [[06-240/Homework Assignment 5|HW5]], [[06-240/Classnotes For Thursday October 12|Thu]]
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|Oct 16
|[[06-240/Linear Algebra - Why We Care|Why?]], [http://en.wikipedia.org/wiki/Isomorphism Iso], [[06-240/Classnotes For Tuesday October 17|Tue]], [[06-240/Classnotes For Thursday October 19|Thu]]
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|align=center|7
|Oct 23
|[[06-240/Term Test|Term Test]], [[06-240/Classnotes For Thursday October 26|Thu (double)]]
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|Oct 30
|[[06-240/Classnotes For Tuesday October 31|Tue]], [[06-240/Homework Assignment 6|HW6]], [[06-240/Classnotes For Thursday November 2|Thu]]
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|Nov 6
|[[06-240/Homework Assignment 7|HW7]], [[06-240/Classnotes For Tuesday November 7|Tue]], [[06-240/Classnotes For Thursday November 9|Thu]]
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|Nov 13
|[[06-240/Classnotes For Tuesday November 14|Tue]], HW8
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|Nov 20
|HW9
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|Nov 27
|HW10
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|Dec 4
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|Dec 11
|Final: Dec 13 2-5PM at BN3
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06-240/Classnotes For Thursday November 9

2006-11-15T02:51:44Z

216.58.41.136:

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==Tutorial Notes==

[[Media:06-240-nov09tut-1.jpeg|Nov09 Lecture notes 1 of 3]]

[[Media:06-240-nov09tut-2.jpeg|Nov09 Lecture notes 2 of 3]]

[[Media:06-240-nov09tut-3.jpeg|Nov09 Lecture notes 3 of 3]]

==Review of Last Class==
{|
|-
|'''Problem.''' Find the rank (the dimension of the image) of a linear transformation <math>T</math> whose matrix representation is the matrix A shown on the right.
|<math>A=\begin{pmatrix}0&2&4&2&2\\4&4&4&8&0\\8&2&0&10&2\\6&3&2&9&1\end{pmatrix}</math>.
|}

{|
|- valign=bottom
|'''Theorem 1.''' If <math>T:V\to W</math> is a linear transformation and <math>P:V\to V</math> and <math>Q:W\to W</math> are ''invertible'' linear transformations, then the rank of <math>T</math> is the same as the rank of <math>QTP</math>.
|   
|'''Proof.''' Owed.
|- valign=bottom
|'''Theorem 2.''' The following row/column operations can be applied to a matrix <math>A</math> by multiplying it on the left/right (respectively) by certain ''invertible'' "elementary matrices":
# Swap two rows/columns
# Multiply a row/column by a nonzero scalar.
# Add a multiple of one row/column to another row/column.
|   
|'''Proof.''' Semi-owed.
|}

'''Solution of the problem.''' using these (invertible!) row/column operations we aim to bring <math>A</math> to look as close as possible to an identity matrix, hoping it will be easy to determine the rank of the matrix we get at the end:

{| border="1px" cellspadding="5" cellspacing=0 style="font-size:90%;"
|+
|align=center|'''Do'''
|align=center|'''Get'''
|align=center|'''Do'''
|align=center|'''Get'''
|- valign=top
|1. Bring a <math>1</math> to the upper left corner by swapping the first two rows and multiplying the first row (after the swap) by <math>1/4</math>.
|align=center|<math>\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\8&2&0&10&2\\6&3&2&9&1\end{pmatrix}</math>
|2. Add <math>(-8)</math> times the first row to the third row, in order to cancel the <math>8</math> in position 3-1.
|align=center|<math>\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\6&3&2&9&1\end{pmatrix}</math>
|- valign=top
|3. Likewise add <math>(-6)</math> times the first row to the fourth row, in order to cancel the <math>6</math> in position 4-1.
|align=center|<math>\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}</math>
|4. With similar column operations (you need three of those) cancel all the entries in the first row (except, of course, the first, which is used in the canceling).
|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}</math>
|- valign=top
|5. Turn the 2-2 entry to a <math>1</math> by multiplying the second row by <math>1/2</math>.
|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}</math>
|6. Using two row operations "clean" the second column; that is, cancel all entries in it other than the "pivot" <math>1</math> at position 2-2.
|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}</math>
|- valign=top
|7. Using three column operations clean the second row except the pivot.
|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}</math>
|8. Clean up the row and the column of the <math>4</math> in position 3-3 by first multiplying the third row by <math>1/4</math> and then performing the appropriate row and column transformations. Notice that by pure luck, the <math>4</math> at position 4-5 of the matrix gets killed in action.
|align=center|<math>\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&0&0\end{pmatrix}</math>
|}

But the matrix we now have represents a linear transformation <math>S</math> satisfying <math>S(v_1,\,v_2,\,v_3,\,v_4\,v_5)=(w_1,\,w_2,\,w_3,\,0,\,0)</math> for some bases <math>(v_i)_{i=1}^5</math> of <math>V</math> and <math>(w_j)_{j=1}^4</math> of <math>W</math>. Thus the image (range) of <math>S</math> is spanned by <math>\{w_1,w_2,w_3\}</math>, and as these are independent, they form a basis of the image. Thus the rank of <math>S</math> is <math>3</math>. Going backward through the "matrix reduction" process above and repeatedly using theorems 1 and 2, we find that the rank of <math>T</math> must also be <math>3</math>.