Linear Algebra and Its Applications, Review Exercise 2.11

Review exercise 2.11. a) Given the following matrix

$A = LU = \begin{bmatrix} 1&&& \\ 2&1&& \\ 2&1&1& \\ 3&2&4&1 \end{bmatrix} \begin{bmatrix} 1&2&0&1&2&1 \\ 0&0&2&2&0&0 \\ 0&0&0&0&0&1 \\ 0&0&0&0&0&0 \end{bmatrix}$

find its rank and a basis for the nullspace.

b) Are the first three rows of $U$ a basis for the row space of $A$ ? Are the first, third, and sixth columns of $U$ a basis for the column space of $A$ ? Are the four rows of $A$ a basis for the row space of $A$ ?

c) Find the largest set of linearly independent vectors $b$ that are a solution for $Ax = b$ .

d) In doing elimination on $A$ what value is used to multiply the third row before subtracting it from the fourth row?

Answer: a) The rank of $A$ is the same as the rank of $U$ . Since $U$ has pivots in three columns (1, 3, and 6) the rank of $U$ and thus of $A$ is $r = 3$ .

We now find the nullspace of $A$ . Since $U$ has pivots in columns 1, 3, and 6 the variables $x_1$ , $x_3$ , and $x_6$ are basic variables and $x_2$ , $x_4$ , and $x_5$ are free variables.

If we set $x_2 = 1$ and $x_4 = x_5 = 0$ then from the third row of $U$ we have $x_6 = 0$ . From the second row of $U$ we have $2x_3 + 2x_4 = 2x_3 + 0 = 0$ or $x_3 = 0$ . Finally, from the first row of $U$ we have $x_1 + 2x_2 + x_4 + 2x_5 + x_6 = x_1 + 2 + 0 + 0 + 0 = 0$ or $x_1 = -2$ . So one solution to $Ax = 0$ is $(-2, 1, 0, 0, 0, 0)$ .

Next we set $x_4 = 1$ and $x_2 = x_5 = 0$ . From the third row of $U$ we again have $x_6 = 0$ . From the second row of $U$ we have $2x_3 + 2x_4 = 2x_3 + 2 = 0$ or $x_3 = -1$ . Finally, from the first row of $U$ we have $x_1 + 2x_2 + x_4 + 2x_5 + x_6 = x_1 + 0 + 1 + 0 + 0 = 0$ or $x_1 = -1$ . So a second solution to $Ax = 0$ is $(-1, 0, -1, 1, 0, 0)$ .

Finally we set $x_5 = 1$ and $x_2 = x_4 = 0$ . From the third row of $U$ we again have $x_6 = 0$ . From the second row of $U$ we have $2x_3 + 2x_4 = 2x_3 + 0 = 0$ or $x_3 = 0$ . Finally, from the first row of $U$ we have $x_1 + 2x_2 + x_4 + 2x_5 + x_6 = x_1 + 0 + 0 + 2 + 0 = 0$ or $x_1 = -2$ . So a third solution to $Ax = 0$ is $(-2, 0, 0, 0, 1, 0)$ .

We thus have three solutions to $Ax = 0$ that together form a basis for the nullspace of $A$ :

$\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} -1 \\ 0 \\ -1 \\ 1 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} -2 \\ 0 \\ 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$

b) The three questions in (b) above are true or false as follows:

True: The rows of $U$ are created by multiplying rows of $A$ by scalars and subtracting them from other rows, and hence the rows of $U$ are linear combination of rows of $A$ . The space spanned by the rows of $U$ (the row space of $U$ ) is therefore the same as the space spanned by the rows of $A$ (the row space of $A$ ). The first three rows of $U$ are linearly independent and the dimension of the row space of $U$ is 3 (the rank of $U$ ). The first three rows of $U$ therefore serve as a basis for the row space of $U$ , and thus for the row space of $A$ as well.

False: Columns 1, 3, and 6 of $U$ are linearly independent and serve as a basis for the column space of $U$ ; however the column space of $U$ is not the same as the column space of $A$ and thus the basis for $U$ is not a basis for $A$ . (The corresponding columns 1, 3, and 6 of $A$ do form a basis for the column space of $A$ .)

False: The rank of $A$ is 3; this is the dimension of its column space and its row space. The four rows of $A$ are not linearly independent (otherwise the rank of $A$ would be 4, not 3) and thus cannot be a basis.

c) If $Ax = b$ for some $b$ then $b$ is in the column space of $A$ . From the form of $U$ we know that columns 1, 3, and 6 of $A$ are linearly independent and form a basis for the column space of $A$ .

We have

$A = LU = \begin{bmatrix} 1&&& \\ 2&1&& \\ 2&1&1& \\ 3&2&4&1 \end{bmatrix} \begin{bmatrix} 1&2&0&1&2&1 \\ 0&0&2&2&0&0 \\ 0&0&0&0&0&1 \\ 0&0&0&0&0&0 \end{bmatrix}$

$= \begin{bmatrix} 1&2&0&1&2&1 \\ 2&4&2&4&4&2 \\ 2&4&2&4&4&3 \\ 3&6&4&7&6&7 \end{bmatrix}$

So the following are linearly independent vectors $b$ such that there exist solutions to $Ax = b$ :

$c_1 \begin{bmatrix} 1 \\ 2 \\ 2 \\ 3 \end{bmatrix} \qquad c_2 \begin{bmatrix} 0 \\ 2 \\ 2 \\ 4 \end{bmatrix} \qquad c_3 \begin{bmatrix} 1 \\ 2 \\ 3 \\ 7 \end{bmatrix}$

where $c_1$ , $c_2$ , and $c_3$ are nonzero.

d) The multipliers used in elimination to reduce $A$ to the echelon form $U$ are in the matrix $L$ . In particular the value multiplying the third row prior to subtracting from the fourth row is $l_{43} = 4$ .

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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Linear Algebra and Its Applications, Review Exercise 2.11

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