Exercise 3.1.17. Suppose that and
are subspaces of
and are orthogonal complements. Is there a matrix
such that the row space of
is
and the nullspace of
is
? If so, show how to construct
using the basis vectors for
.
Answer: Suppose through
are the basis vectors for
. (Since
is a subspace of
we will have
.) Let
be an
by
matrix with row 1 equal to
, row 2 equal to
, and so on through row
equal to
.
The row space of is the set of all linear combinations of the rows, which is equivalent to the set of all linear combinations of
through
. But since
through
is a basis set for
they span
and the set of all linear combinations of
through
is
itself. The row space of
is therefore equal to
.
Now consider the null space of consisting of all vectors
such that
. If this is the case then the inner product of
with each row of
must be zero or (put another way)
must be orthogonal to each row of
. Since the rows of
are
through
this means that any vector
in the nullspace of
is orthogonal to all of
through
.
Since is orthogonal to all of
through
it is also orthogonal to all linear combinations of
through
, and since
through
form a basis set for
this means that
is orthogonal to any vector in
. All vectors in the nullspace are thus orthogonal to
. Moreover, any vector orthogonal to
(i.e., orthogonal to all vectors in
) must be orthogonal to all of
through
and is therefore in the nullspace of
.
The nullspace of thus contains all vectors orthogonal to
. But the set of all vectors orthogonal to
is
, the orthogonal complement to
. The nullspace of
is therefore equal to
.
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
.