## 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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