06-240/Classnotes For Thursday November 9

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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. .
Theorem 1. If is a linear transformation and and are invertible linear transformations, then the rank of is the same as the rank of .     Proof. Owed.
Theorem 2. The following row/column operations can be applied to a matrix by multiplying it on the left/right (respectively) by certain invertible "elementary matrices":
  1. Swap two rows/columns
  2. Multiply a row/column by a nonzero scalar.
  3. 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 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 to the upper left corner by swapping the first two rows and multiplying the first row (after the swap) by . 2. Add times the first row to the third row, in order to cancel the in position 3-1.
3. Likewise add times the first row to the fourth row, in order to cancel the in position 4-1. 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).
5. Turn the 2-2 entry to a by multiplying the second row by . 6. Using two row operations "clean" the second column; that is, cancel all entries in it other than the "pivot" at position 2-2.
7. Using three column operations clean the second row except the pivot. 8. Clean up the row and the column of the in position 3-3 by first multiplying the third row by and then performing the appropriate row and column transformations. Notice that by pure luck, the at position 4-5 of the matrix gets killed in action.

But the matrix we now have represents a linear transformation satisfying for some bases of and of . Thus the image (range) of is spanned by , and as these are independent, they form a basis of the image. Thus the rank of is . Going backward through the "matrix reduction" process above and repeatedly using theorems 1 and 2, we find that the rank of must also be .

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