06-240/Classnotes For Thursday November 9: Difference between revisions
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|'''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. |
|'''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. |
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|<math>A=\begin{pmatrix}0 |
|<math>A=\begin{pmatrix}0</math> |
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{| |
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|'''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>. |
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|'''Proof.''' Owed. |
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|'''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": |
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# Swap two rows/columns |
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# Multiply a row/column by a nonzero scalar. |
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# Add a multiple of one row/column to another row/column. |
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|'''Proof.''' Semi-owed. |
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'''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: |
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{| border="1px" cellspadding="5" cellspacing=0 style="font-size:90%;" |
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|+ |
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|align=center|'''Do''' |
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|align=center|'''Get''' |
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|align=center|'''Do''' |
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|align=center|'''Get''' |
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|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>. |
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|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> |
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|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. |
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|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> |
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|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. |
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|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> |
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|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). |
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|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> |
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|5. Turn the 2-2 entry to a <math>1</math> by multiplying the second row by <math>1/2</math>. |
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|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> |
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|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. |
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|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> |
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|7. Using three column operations clean the second row except the pivot. |
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|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> |
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|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. |
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|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> |
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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>. |
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==Class Notes== |
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[[Media:Lect015.pdf|Scan of Week 9 Lecture 2 notes]] |
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==Tutorial Notes== |
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[[Media:06-240-nov09tut-1.jpeg|Nov09 Lecture notes 1 of 3]] |
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[[Media:06-240-nov09tut-2.jpeg|Nov09 Lecture notes 2 of 3]] |
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[[Media:06-240-nov09tut-3.jpeg|Nov09 Lecture notes 3 of 3]] |
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[[Media:Tut009.pdf|Scan of Week 9 Tutorial notes]] |
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Revision as of 02:37, 28 May 2007
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Review of Last Class
| Problem. Find the rank (the dimension of the image) of a linear transformation whose matrix representation is the matrix A shown on the right. | Failed to parse (unknown function "\begin{pmatrix}"): {\displaystyle A=\begin{pmatrix}0} |
