06-240/Classnotes For Thursday November 9

06-240/Navigation Panel

#	Week of...	Notes and Links
1	Sep 11	About, Tue, HW1, Putnam, Thu
2	Sep 18	Tue, HW2, Thu
3	Sep 25	Tue, HW3, Photo, Thu
4	Oct 2	Tue, HW4, Thu
5	Oct 9	Tue, HW5, Thu
6	Oct 16	Why?, Iso, Tue, Thu
7	Oct 23	Term Test, Thu (double)
8	Oct 30	Tue, HW6, Thu
9	Nov 6	Tue, HW7, Thu
10	Nov 13	Tue, HW8, Thu
11	Nov 20	Tue, HW9, Thu
12	Nov 27	Tue, HW10, Thu
13	Dec 4	On the final, Tue, Thu
F	Dec 11	Final: Dec 13 2-5PM at BN3, Exam Forum
Register of Good Deeds
Add your name / see who's in!
edit the panel

Review of Last Class

Problem. Find the rank (the dimension of the image) of a linear transformation

T

whose matrix representation is the matrix A shown on the right.

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}}

.

Theorem 1. If $T:V\to W$ is a linear transformation and $P:V\to V$ and $Q:W\to W$ are invertible linear transformations, then the rank of $T$ is the same as the rank of $QTP$ .		Proof. Owed.
Theorem 2. The following row/column operations can be applied to a matrix $A$ 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 $A$ 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:


Do	Get	Do	Get
1. Bring a $1$ to the upper left corner by swapping the first two rows and multiplying the first row (after the swap) by $1/4$ .	${\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\8&2&0&10&2\\6&3&2&9&1\end{pmatrix}}$	2. Add $(-8)$ times the first row to the third row, in order to cancel the $8$ in position 3-1.	${\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\6&3&2&9&1\end{pmatrix}}$
3. Likewise add $(-6)$ times the first row to the fourth row, in order to cancel the $6$ in position 4-1.	${\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}}$	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).	${\begin{pmatrix}1&0&0&0&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}}$
5. Turn the 2-2 entry to a $1$ by multiplying the second row by $1/2$ .	${\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}}$	6. Using two row operations "clean" the second column; that is, cancel all entries in it other than the "pivot" $1$ at position 2-2.	${\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}}$
7. Using three column operations clean the second row except the pivot.	${\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}}$	8. Clean up the row and the column of the $4$ in position 3-3 by first multiplying the third row by $1/4$ and then performing the appropriate row and column transformations. Notice that by pure luck, the $4$ at position 4-5 of the matrix gets killed in action.	${\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&0&0\end{pmatrix}}$

But the matrix we now have represents a linear transformation $S$ satisfying $S(v_{1},\,v_{2},\,v_{3},\,v_{4}\,v_{5})=(w_{1},\,w_{2},\,w_{3},\,0,\,0)$ for some bases $(v_{i})_{i=1}^{5}$ of $V$ and $(w_{j})_{j=1}^{4}$ of $W$ . Thus the image (range) of $S$ is spanned by $\{w_{1},w_{2},w_{3}\}$ , and as these are independent, they form a basis of the image. Thus the rank of $S$ is $3$ . Going backward through the "matrix reduction" process above and repeatedly using theorems 1 and 2, we find that the rank of $T$ must also be $3$ .