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

Answer: If we put 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 are thus linearly independent and form a basis for the column space of :

The rank of is therefore and the dimension of the nullspace of is therefore .

In the system

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

Setting , from the fourth equation we have or , from the second equation we have or , and from the first equation we have or . The vector

is thus a solution to the homogeneous system and a basis for the nullspace of .

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

The dimension of the left nullspace of is therefore .

To find the left nullspace we must find a solution to the system or (alternately) . To do this we do Gaussian elimination on the matrix

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

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

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

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

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

Setting , from the fourth equation we have or , from the second equation we have or , and from the first equation we have . The vector

is thus a solution to the homogeneous system and a basis for the left nullspace of .

We now turn to

This matrix has rank and the first column

is a basis for the column space of .

Since the rank of is the dimension of the nullspace is . In the homogeneous system or

Gaussian elimination produces the following

so that is a free variable and is basic.

Setting we have or . The vector

is thus a solution to and a basis for the nullspace of .

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

The dimension of the left nullspace of is therefore .

To find the left nullspace we must find a solution to the system or (alternately) . To do this we do Gaussian elimination on the matrix

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

The resulting echelon matrix has a pivot in column 1, with and being free variables. Setting and , from the first equation we have or . Setting and , from the first equation we have or . The vectors

are thus solutions to and form a basis for the left nullspace of .

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.