Linear Algebra and Its Applications, Exercise 3.4.10

Exercise 3.4.10. Given the two orthonormal vectors q_1 and q_2 and an arbitrary vector b, what linear combination of q_1 and q_2 is the least distance from b? Show that the difference between b and that combination (i.e., the error vector) is orthogonal to both q_1 and q_2.

Answer: This exercise is similar to the previous one. Any linear combination of q_1 and q_2 is in the plane formed by q_1 and q_2, and the combination closest to b is the projection p of b onto that plane. Because q_1 and q_2 are orthonormal that projection is equal to the sum of the separate projections of b onto q_1 and q_2 respectively:

p = (q_1^Tb)q_1 + (q_2^Tb)q_2

So (q_1^Tb)q_1 + (q_2^Tb)q_2 is the closest combination to b.

The error vector is then

e = b - p = b - (q_1^Tbq_1 + q_2^Tbq_2)

Taking the dot product of the error vector with q_1 we have

q_1^Te = q_1^T [b - (q_1^Tbq_1 + q_2^Tbq_2)] = q_1^Tb - q_1^T(q_1^Tb)q_1 - q_1^T(q_2^Tb)q_2

= q_1^Tb - (q_1^Tb)q_1^Tq_1 - (q_2^Tb)q_1^Tq_2

Since q_1 and q_2 are orthonormal we have q_1^Tq_1 = 1 and q_1^Tq_2 = 0. So we have

q_1^Te = q_1^Tb - (q_1^Tb) \cdot 1 - (q_2^Tb) \cdot 0 = q_1^Tb - q_1^Tb = 0

Similarly we have

q_2^Te = q_2^T [b - (q_1^Tbq_1 + q_2^Tbq_2)] = q_2^Tb - q_2^T(q_1^Tb)q_1 - q_2^T(q_2^Tb)q_2

= q_2^Tb - (q_1^Tb)q_2^Tq_1 - (q_2^Tb)q_2^Tq_2 = q_2^Tb - q_2^Tb = 0

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.

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