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.

Linear Algebra and Its Applications, Exercise 3.4.26

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Linear Algebra and Its Applications, Exercise 3.4.26

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