## Linear Algebra and Its Applications, Exercise 3.4.19

Exercise 3.4.19. When doing Gram-Schmidt orthogonalization, an alternative approach to computing $c' = c - (q_1^Tc)q_1 - (q_2^Tc)q_2$ (equation 7 on page 173) is to instead compute $c'$ in two separate steps: $c'' = c - (q_1^Tc)q_1 \qquad c' = c'' - (q_2^Tc'')q_2$

Show that the second method is equivalent to the first.

Answer: We substitute the expression for $c''$ from the first equation into the second equation: $c' = c'' - (q_2^Tc'')q_2 = [c - (q_1^Tc)q_1] - \left[q_2^T\left[c - (q_1^Tc)q_1\right]\right]q_2$ $= c - (q_1^Tc)q_1 - \left[q_2^Tc - q_2^T(q_1^Tc)q_1\right]q_2$ $= c - (q_1^Tc)q_1 - \left[q_2^Tc - (q_1^Tc)q_2^Tq_1\right]q_2$ $= c - (q_1^Tc)q_1 - \left[q_2^Tc - (q_1^Tc) \cdot 0\right]q_2$ $= c - (q_1^Tc)q_1 - (q_2^Tc)q_2$

The resulting equation $c' = c - (q_1^Tc)q_1 - (q_2^Tc)q_2$ is the original calculation of $c'$ from equation 7 on page 173. So the second method using $c''$ is equivalent to the first.

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 .

This entry was posted in linear algebra and tagged . Bookmark the permalink.