# 06-240/Classnotes For Thursday November 9

## Review of Last Class

 Problem. Find the rank (the dimension of the image) of a linear transformation ${\displaystyle T}$ whose matrix representation is the matrix A shown on the right. ${\displaystyle A={\begin{pmatrix}0&2&4&2&2\\4&4&4&8&0\\8&2&0&10&2\\6&3&2&9&1\end{pmatrix}}}$.
 Theorem 1. If ${\displaystyle T:V\to W}$ is a linear transformation and ${\displaystyle P:V\to V}$ and ${\displaystyle Q:W\to W}$ are invertible linear transformations, then the rank of ${\displaystyle T}$ is the same as the rank of ${\displaystyle QTP}$. Proof. Owed. Theorem 2. The following row/column operations can be applied to a matrix ${\displaystyle A}$ 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 ${\displaystyle A}$ 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 ${\displaystyle 1}$ to the upper left corner by swapping the first two rows and multiplying the first row (after the swap) by ${\displaystyle 1/4}$. ${\displaystyle {\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\8&2&0&10&2\\6&3&2&9&1\end{pmatrix}}}$ 2. Add ${\displaystyle (-8)}$ times the first row to the third row, in order to cancel the ${\displaystyle 8}$ in position 3-1. ${\displaystyle {\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\6&3&2&9&1\end{pmatrix}}}$ 3. Likewise add ${\displaystyle (-6)}$ times the first row to the fourth row, in order to cancel the ${\displaystyle 6}$ in position 4-1. ${\displaystyle {\begin{pmatrix}1&1&1&2&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}}}$ 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). ${\displaystyle {\begin{pmatrix}1&0&0&0&0\\0&2&4&2&2\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}}}$ 5. Turn the 2-2 entry to a ${\displaystyle 1}$ by multiplying the second row by ${\displaystyle 1/2}$. ${\displaystyle {\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&-6&-8&-6&2\\0&-3&-4&-3&1\end{pmatrix}}}$ 6. Using two row operations "clean" the second column; that is, cancel all entries in it other than the "pivot" ${\displaystyle 1}$ at position 2-2. ${\displaystyle {\begin{pmatrix}1&0&0&0&0\\0&1&2&1&1\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}}}$ 7. Using three column operations clean the second row except the pivot. ${\displaystyle {\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&4&0&8\\0&0&2&0&4\end{pmatrix}}}$ 8. Clean up the row and the column of the ${\displaystyle 4}$ in position 3-3 by first multiplying the third row by ${\displaystyle 1/4}$ and then performing the appropriate row and column transformations. Notice that by pure luck, the ${\displaystyle 4}$ at position 4-5 of the matrix gets killed in action. ${\displaystyle {\begin{pmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&0&0\end{pmatrix}}}$

But the matrix we now have represents a linear transformation ${\displaystyle S}$ satisfying ${\displaystyle S(v_{1},\,v_{2},\,v_{3},\,v_{4}\,v_{5})=(w_{1},\,w_{2},\,w_{3},\,0,\,0)}$ for some bases ${\displaystyle (v_{i})_{i=1}^{5}}$ of ${\displaystyle V}$ and ${\displaystyle (w_{j})_{j=1}^{4}}$ of ${\displaystyle W}$. Thus the image (range) of ${\displaystyle S}$ is spanned by ${\displaystyle \{w_{1},w_{2},w_{3}\}}$, and as these are independent, they form a basis of the image. Thus the rank of ${\displaystyle S}$ is ${\displaystyle 3}$. Going backward through the "matrix reduction" process above and repeatedly using theorems 1 and 2, we find that the rank of ${\displaystyle T}$ must also be ${\displaystyle 3}$.