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.