## Linear Algebra and Its Applications, Exercise 3.4.26

Exercise 3.4.26. In the Gram-Schmidt orthogonalization process the third component $c'$ is computed as $c' = c - (q_1^Tc)q_1 - (q_2^Tc)q_2$. Verify that $c'$ is orthogonal to both $q_1$ and $q_2$.

Answer: Taking the dot product of $q_1$ and $c'$ we have $q_1^Tc' = q_1^T \left[ c - (q_1^Tc)q_1 - (q_2^Tc)q_2 \right] = q_1^Tc - q_1^T(q_1^Tc)q_1 - q_1^T(q_2^Tc)q_2$

Since $q_1^Tc$ and $q_2^Tc$ are scalars and $q_1$ and $q_2$ are orthonormal we then have $q_1^Tc' = q_1^Tc - q_1^T(q_1^Tc)q_1 - q_1^T(q_2^Tc)q_2 = q_1^Tc - (q_1^Tc)q_1^Tq_1 - (q_2^Tc)q_1^Tq_2$ $= q_1^Tc - (q_1^Tc) \cdot 1 - (q_2^Tc) \cdot 0 = q_1^Tc - q_1^Tc = 0$

So $c'$ is orthogonal to $q_1$.

Taking the dot product of $q_2$ and $c'$ we have $q_2^Tc' = q_2^T \left[ c - (q_1^Tc)q_1 - (q_2^Tc)q_2 \right] = q_2^Tc - q_2^T(q_1^Tc)q_1 - q_2^T(q_2^Tc)q_2$ $= q_1^Tc - (q_1^Tc)q_1^Tq_1 - (q_2^Tc)q_1^Tq_2 = q_2^Tc - q_2^Tc = 0$

So $c'$ is also orthogonal to $q_2$.

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.