## Linear Algebra and Its Applications, Review Exercise 2.27

Review exercise 2.27. Find bases for each of the following matrices:

$A_1 = \begin{bmatrix} 1&2&0&3 \\ 0&2&2&2 \\ 0&0&0&0 \\ 0&0&0&4 \end{bmatrix} \qquad A_2 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \begin{bmatrix} 1&4 \end{bmatrix}$

Answer: If we put $A_1$ in echelon form (by exchanging rows 3 and 4) the resulting matrix would have pivots in columns 1, 2, and 4. Columns 1, 2, and 4 of $A_1$ are thus linearly independent and form a basis for the column space of $A_1$:

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

The rank of $A_1$ is therefore $r = 3$ and the dimension of the nullspace of $A_1$ is therefore $n-r = 4-3 = 1$.

In the system

$A_1x = \begin{bmatrix} 1&2&0&3 \\ 0&2&2&2 \\ 0&0&0&0 \\ 0&0&0&4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = 0$

it is apparent from inspection that $x_3$ is a free variable and the rest are basic.

Setting $x_3 = 1$, from the fourth equation we have $4x_4 = 0$ or $x_4 = 0$, from the second equation we have $x_2 + x_3 + x_4 = x_2 + 1 + 0 = 0$ or $x_2 = -1$, and from the first equation we have $x_1 + 2x_2 + 3x_4 = x_1 - 2 + 0 = 0$ or $x_1 = 2$. The vector

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

is thus a solution to the homogeneous system $A_1x = 0$ and a basis for the nullspace of $A_1$.

Since the rank of $A_1$ is $r = 3$ the dimension of the row space of $A_1$ is also 3; from inspection it is apparent that rows 1, 2, and 4 of $A_1$ are linearly independent and are a basis for the row space of $A_1$:

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

The dimension of the left nullspace of $A_1$ is therefore $m-r = 4-3 = 1$.

To find the left nullspace we must find a solution to the system $y^TA_1 = 0$ or (alternately) $A_1^Ty = 0$. To do this we do Gaussian elimination on the matrix

$A_1^T = \begin{bmatrix} 1&0&0&0 \\ 2&2&0&0 \\ 0&2&0&0 \\ 3&2&0&4 \end{bmatrix}$

We begin by multiplying row 1 by 2 and subtracting the result from row 2:

$\begin{bmatrix} 1&0&0&0 \\ 2&2&0&0 \\ 0&2&0&0 \\ 3&2&0&4 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&0&0&0 \\ 0&2&0&0 \\ 0&2&0&0 \\ 3&2&0&4 \end{bmatrix}$

We then multiply row 1 by 3 and subtract the result from row 4:

$\begin{bmatrix} 1&0&0&0 \\ 0&2&0&0 \\ 0&2&0&0 \\ 3&2&0&4 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&0&0&0 \\ 0&2&0&0 \\ 0&2&0&0 \\ 0&2&0&4 \end{bmatrix}$

We then multiply row 2 by 1 and subtract the result from row 3:

$\begin{bmatrix} 1&0&0&0 \\ 0&2&0&0 \\ 0&2&0&0 \\ 0&2&0&4 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&0&0&0 \\ 0&2&0&0 \\ 0&0&0&0 \\ 0&2&0&4 \end{bmatrix}$

and then multiply row 2 by 1 and subtract the result from row 3:

$\begin{bmatrix} 1&0&0&0 \\ 0&2&0&0 \\ 0&0&0&0 \\ 0&2&0&4 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&0&0&0 \\ 0&2&0&0 \\ 0&0&0&0 \\ 0&0&0&4 \end{bmatrix}$

Although the resulting matrix is not in echelon form, it is apparent from inspection that $y_3$ is a free variable and the rest are basic.

Setting $y_3 = 1$, from the fourth equation we have $4y_4 = 0$ or $y_4 = 0$, from the second equation we have $2y_2 = 0$ or $y_2 = 0$, and from the first equation we have $x_1 = 0$. The vector

$\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$

is thus a solution to the homogeneous system $A_1^Ty = 0$ and a basis for the left nullspace of $A_1$.

We now turn to

$A_2 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \begin{bmatrix} 1&4 \end{bmatrix} = \begin{bmatrix} 1&4 \\ 1&4 \\ 1&4 \end{bmatrix}$

This matrix has rank $r = 1$ and the first column

$\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$

is a basis for the column space of $A_2$.

Since the rank of $A_2$ is $r = 1$ the dimension of the nullspace is $n-r = 2-1 = 1$. In the homogeneous system $A_2x = 0$ or

$\begin{bmatrix} 1&4 \\ 1&4 \\ 1&4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = 0$

Gaussian elimination produces the following

$\begin{bmatrix} 1&4 \\ 1&4 \\ 1&4 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&4 \\ 0&0 \\ 0&0 \end{bmatrix}$

so that $x_2$ is a free variable and $x_1$ is basic.

Setting $x_2 = 1$ we have $x_1 + 4x_2 = x_1 + 4 = 0$ or $x_1 = -4$. The vector

$\begin{bmatrix} -4 \\ 1 \end{bmatrix}$

is thus a solution to $A_2x = 0$ and a basis for the nullspace of $A_2$.

Since the rank of $A_2$ is $r = 1$ the dimension of the row space of $A_2$ is also 1; from inspection it is apparent that row 1 of $A_2$ is a basis for the row space of $A_2$:

$\begin{bmatrix} 1 \\ 4 \end{bmatrix}$

The dimension of the left nullspace of $A_2$ is therefore $m-r = 3-1 = 2$.

To find the left nullspace we must find a solution to the system $y^TA_2 = 0$ or (alternately) $A_2^Ty = 0$. To do this we do Gaussian elimination on the matrix

$A_2^T = \begin{bmatrix} 1&1&1 \\ 4&4&4 \end{bmatrix}$

We multiply row 1 by 4 and subtract the result from row 2:

$\begin{bmatrix} 1&1&1 \\ 4&4&4 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&1 \\ 0&0&0 \end{bmatrix}$

The resulting echelon matrix has a pivot in column 1, with $y_2$ and $y_3$ being free variables. Setting $y_2 = 1$ and $y_3 = 0$, from the first equation we have $y_1 + y_2 + y_3 = y_1 + 1 + 0 = 0$ or $y_1 = -1$. Setting $y_2 = 0$ and $y_3 = 0$, from the first equation we have $y_1 + y_2 + y_3 = y_1 + 0 + 1 = 0$ or $y_1 = -1$. The vectors

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

are thus solutions to $A_2^Ty =0$ and form a basis for the left nullspace of $A_2$.

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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