12-240/Classnotes for Tuesday November 15

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Problem. Find the rank the matrix


Solution. Using (invertible!) row/column operations we aim to bring A to look as close as possible to an identity matrix:

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}

Thus the rank of our matrix is 3.