Exercise 3.4.10. Given the two orthonormal vectors and
and an arbitrary vector
, what linear combination of
and
is the least distance from
? Show that the difference between
and that combination (i.e., the error vector) is orthogonal to both
and
.
Answer: This exercise is similar to the previous one. Any linear combination of and
is in the plane formed by
and
, and the combination closest to
is the projection
of
onto that plane. Because
and
are orthonormal that projection is equal to the sum of the separate projections of
onto
and
respectively:
So is the closest combination to
.
The error vector is then
Taking the dot product of the error vector with we have
Since and
are orthonormal we have
and
. So we have
Similarly we have
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, Fifth Edition
and the accompanying free online course, and Dr Strang’s other books
.