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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