Linear Algebra and Its Applications, Review Exercise 2.33

Review exercise 2.33. Consider the following PA = LU factorization:

\begin{bmatrix} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&1 \\ 0&0&1&0 \end{bmatrix} \begin{bmatrix} 0&0&1&-3&2 \\ 2&-1&4&2&1 \\ 4&-2&9&1&4 \\ 2&-1&5&-1&5 \end{bmatrix}

= \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ 1&1&1&0 \\ 2&1&0&1 \end{bmatrix} \begin{bmatrix} 2&-1&4&2&1 \\ 0&0&1&-3&2 \\ 0&0&0&0&2 \\ 0&0&0&0&0 \end{bmatrix}

a) What is the rank of A?

b) Find a basis for the row space of A.

c) Are rows 1, 2, and 3 of A linearly independent: true or false?

d) Find a basis for the column space of A.

e) What is the dimension of the left nullspace of A?

f) Find the general solution to Ax = 0.

Answer: a) The echelon matrix U has three pivots (in columns 1, 3, and 5) and thus has rank r = 3. The rank of A is the same as the rank of U, namely r = 3.

b) The row space of A is the same as the row space of U. The first three rows of U are linearly independent and thus can serve as a basis for the row space of U and thus the row space of A:

\begin{bmatrix} 2 \\ -1 \\ 4 \\ 2 \\ 1 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 0 \\ 1 \\ -3 \\ 2 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 2 \end{bmatrix}

c) Rows 1 and 2 of A are clearly linearly independent. If row 3 is a linear combination of rows 1 and 2 then the coefficient for row 2 must be 2 since row 1 has a zero in the first two columns and cannot contribute to those entries. If we take 2 times row 2 and then add row 1 (i.e., using 1 as the coefficient for row 1) we have

\begin{bmatrix} 0 \\ 0 \\ 1 \\ -3 \\ 2 \end{bmatrix} + 2 \begin{bmatrix} 2 \\ -1 \\ 4 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 4 \\ -2 \\ 9\\ 1 \\ 4 \end{bmatrix}

The result is equal to row 3, so row 3 is a linear combination of rows 1 and 2. The three rows are not linearly independent.

d) As noted above, the echelon matrix U has pivots in columns 1, 3, and 5, so those three columns are linearly independent. The corresponding columns of A are then also linearly independent and form a basis for the column space of A:

\begin{bmatrix} 0 \\ 2 \\ 4 \\ 2 \end{bmatrix} \qquad \begin{bmatrix} 1 \\ 4 \\ 9 \\ 5 \end{bmatrix} \qquad \begin{bmatrix} 2 \\ 1 \\ 4 \\ 5 \end{bmatrix}

e) The rank of A is r = 3 so the dimension of the left nullspace of A is m - r = 4 - 3 = 1.

f) The general solution Ax = 0 is the same as the general solution to Ux = 0. From the echelon matrix U we see that x_1, x_3 and x_5 are basic variables and that x_2 and x_4 are free variables.

We first set x_2 = 1 and x_4 = 0. From row 3 of U we have x_5 = 0. From row 2 we have x_3 - 3x_4 + 2x_5 = x_3 - 3 \cdot 0 + 2 \cdot 0 = 0 or x_3 = 0. From row 1 we have

2x_1 - x_2 + 4x_3 - 2x_4 + x_5 = 2x_1 - 1 + 4 \cdot 0 + 2 \cdot 0 + 0 = 0

or x_1 = \frac{1}{2}. So one solution is (\frac{1}{2}, 1, 0, 0, 0).

We next set x_2 = 0 and x_4 = 1. From row 3 of U we have x_5 = 0. From row 2 we have x_3 - 3x_4 + 2x_5 = x_3 - 3 \cdot 1 + 2 \cdot 0 = 0 or x_3 = 3. From row 1 we have

2x_1 - x_2 + 4x_3 + 2x_4 + x_5 = 2x_1 - 0 + 4 \cdot 3 + 2 \cdot 1 + 0 = 0

or x_1 = -7. So a second solution is (-7, 0, 3, 1, 0).

We have two solutions to Ux = 0 and thus to Ax = 0. The general solution is then

x = x_2 \begin{bmatrix} \frac{1}{2} \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -7 \\ 0 \\ 3 \\ 1 \\ 0 \end{bmatrix}

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.

 Buy me a snack to sponsor more posts like this!

This entry was posted in linear algebra and tagged , , , , , , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s