06-240/Classnotes For Thursday November 9

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 {w1,w2,w3}, 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.