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 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":
  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 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.

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