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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