## 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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