30 CHAPTER 1 • Linear Equations in Linear Algebra
Note: Exercises 41 and 42 help to prepare for later work on the column space of a matrix. (See Section 2.9 or
4.6.) The Study Guide points out that these exercises depend on the following idea, not explicitly mentioned
in the text: when a row operation is performed on a matrix A, the calculations for each new entry depend only
on the other entries in the same column. If a column of A is removed, forming a new matrix, the absence of
this column has no affect on any row-operation calculations for entries in the other columns of A. (The
absence of a column might affect the particular choice of row operations performed for some purpose, but that
is not being considered here.)
42. [M] Examine the calculations in Exercise 40. The third column of the original matrix, say A, is not a
pivot column. Let A
o
be the matrix formed by deleting column 3 of A, let B be the echelon form obtained
from A, and let B
o
be the matrix obtained by deleting column 3 of B. The sequence of row operations that
reduces A to B also reduces A
o
to B
o
. Since B
o
is in echelon form, it shows that A
o
has a pivot position in
each row. Therefore, the columns of A
o
span R
4
.
It is possible to delete column 2 of A instead of column 3. (See the remark for Exercise 41.) However,
only one column can be deleted. If two or more columns were deleted from A, the resulting matrix would
have fewer than four columns, so it would have fewer than four pivot positions. In such a case, not every
row could contain a pivot position, and the columns of the matrix would not span R
4
, by Theorem 4.
Notes: At the end of Section 1.4, the Study Guide gives students a method for learning and mastering linear
algebra concepts. Specific directions are given for constructing a review sheet that connects the basic
definition of “span” with related ideas: equivalent descriptions, theorems, geometric interpretations, special
cases, algorithms, and typical computations. I require my students to prepare such a sheet that reflects their
choices of material connected with “span”, and I make comments on their sheets to help them refine their
review. Later, the students use these sheets when studying for exams.
The MATLAB box for Section 1.4 introduces two useful commands
gauss and bgauss that allow a
student to speed up row reduction while still visualizing all the steps involved. The command
B = gauss(A,1) causes MATLAB to find the left-most nonzero entry in row 1 of matrix A, and use that
entry as a pivot to create zeros in the entries below, using row replacement operations. The result is a matrix
that a student might write next to A as the first stage of row reduction, since there is no need to write a new
matrix after each separate row replacement. I use the
gauss command frequently in lectures to obtain an
echelon form that provides data for solving various problems. For instance, if a matrix has 5 rows, and if row
swaps are not needed, the following commands produce an echelon form of A:
B = gauss(A,1), B = gauss(B,2), B = gauss(B,3), B = gauss(B,4)
If an interchange is required, I can insert a command such as B = swap(B,2,5) . The command bgauss
uses the left-most nonzero entry in a row to produce zeros above that entry. This command, together with
scale, can change an echelon form into reduced echelon form.
The use of
gauss and bgauss creates an environment in which students use their computer program
the same way they work a problem by hand on an exam. Unless you are able to conduct your exams in a
computer laboratory, it may be unwise to give students too early the power to obtain reduced echelon forms
with one command—they may have difficulty performing row reduction by hand during an exam. Instructors
whose students use a graphic calculator in class each day do not face this problem. In such a case, you may
wish to introduce
rref earlier in the course than Chapter 4 (or Section 2.8), which is where I finally allow
students to use that command.
1.5 SOLUTIONS
Notes: The geometry helps students understand Span{u, v}, in preparation for later discussions of subspaces.
The parametric vector form of a solution set will be used throughout the text. Figure 6 will appear again in
Sections 2.9 and 4.8.