## 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.

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