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": 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 $\{w_{1},w_{2},w_{3}\}$, 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

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Tutorial Notes

Nov09 Lecture notes 1 of 3

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