## 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.

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