## 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 .

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