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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 25PM at BN3, Exam Forum

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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. Semiowed.

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 31.

${\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&6&8&6&2\\6&3&2&9&1\end{pmatrix}}$

3. Likewise add $(6)$ times the first row to the fourth row, in order to cancel the $6$ in position 41.

${\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&6&8&6&2\\0&3&4&3&1\end{pmatrix}}$

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 22 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 22.

${\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}}$

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 33 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 45 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 $3$.
Class Notes
Scan of Week 9 Lecture 2 notes
Tutorial Notes
Nov09 Lecture notes 1 of 3
Nov09 Lecture notes 2 of 3
Nov09 Lecture notes 3 of 3
Scan of Week 9 Tutorial notes