Exercise 2.4.12. Suppose that for a matrix the system
has at least one nonzero solution. Show that there exists at least one vector
for which the system
has no solution. Show an example of such a matrix
and vector
.
Answer: Suppose is an
by
matrix. If there exists a nonzero
for which
then the columns of
are linearly dependent. (If they were linearly independent then
would imply that
for all
.) We therefore have the rank
.
Since and the dimension of the row space
is equal to
the dimension of
is less than
. In other words, the rows of
do not span all of
.
But if that is the case then there exists at least one vector in
that is not in
and cannot be expressed as a linear combination of the rows of
. Therefore for the vector
there is no solution to the system
. (If there were such a solution then its coefficients
would define a linear combination of the rows of
equal to
.)
For example, suppose that we have the following 2 by 2 matrix
The system is equivalent to
This system has rank with
as a basic variable and
as a free variable.
From the first equation of the system we have or
. If we set the free variable
to 1 we then have
and
as a (nonzero) solution to
.
The vector is a basis for the row space
, which consists of all vectors of the form
. Geometically this is a line through the origin and the point
.
Pick a point in the x-y plane not on this line, for example , and consider the vector
and the system
The first equation of the system produces while the second equation produces
, a contradiction. There is thus no solution to
for the example values of
and
.
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
.