Exercise 2.6.20. A nonlinear transformation from a vector space
to a vector space
is invertible a) if for any
in
there exists some
in
such that
and b) if
and
are in
then
implies that
. Describe which of the following transformations from
to
are invertible, and explain why or why not:
a)
b)
c)
d)
Answer: a) If is any real number then
exists and is a unique solution to
. The transformation
is therefore invertible.
b) We have for all
in
, so if
then there is no
for which
. The transformation
is therefore not invertible.
c) If is any real number then
exists and is a unique solution to
. The transformation
is therefore invertible.
d) We have for all
in
so if
or
then there is no
for which
. Also note even for
solutions to
are not unique, since (for example)
. The transformation
is therefore not invertible.
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
.