## Linear Algebra and Its Applications, Exercise 3.2.2

Exercise 3.2.2. Use the formula $p = \bar{x}a = \left(a^Tb/a^Ta\right)a$ (where $p$ is the projection of $b$ on $a$) to confirm that $\|p\| = \|b\||\cos \theta|$ (where $\theta$ is the angle between $a$ and $b$).

Answer: Since $p = \left(a^Tb/a^Ta\right)a$ we can compute the square of the length of $p$ as

$\|p\|^2 = \left[\left(a^Tb/a^Ta\right)a\right]^T \left[\left(a^Tb/a^Ta\right)a\right]$

Since $\left(a^Tb/a^Ta\right)$ is a scalar quantity we can bring it out of the transpose expression and then bring it to the left of the expression:

$\|p\|^2 = \left[\left(a^Tb/a^Ta\right)a\right]^T \left[\left(a^Tb/a^Ta\right)a\right]$

$= \left(a^Tb/a^Ta\right) a^T \left(a^Tb/a^Ta\right) a$

$= \left(a^Tb/a^Ta\right) \left(a^Tb/a^Ta\right) a^Ta$

$= \left(a^Tb/a^Ta\right)^2 a^Ta$

This leaves us with the scalar quantity $a^Ta$ at the right, so we can further simplify this expression as follows:

$\|p\|^2 = \left(a^Tb/a^Ta\right)^2 a^Ta = \left(a^Tb\right)^2 a^Ta / \left(a^Ta\right)^2$

$= \left(a^Tb\right)^2 / a^Ta = \left(a^Tb\right)^2/\|a\|^2$

Taking the positive square root of both sides we then have

$\|p\| = |a^Tb/\|a\||$

At the same time from (2) on page 146 (theorem 3G) we have

$\cos \theta = a^Tb/\left(\|a\|\|b\|\right)$

so that

$\|b\|\cos \theta = a^Tb/\|a\|$

Substituting into the above expression for $p$ we then have

$\|p\| = |a^Tb/\|a\|| = |\|b\|\cos \theta| = \|b\||\cos \theta|$

(since $\|b\|$ is guaranteed to be positive but $\cos \theta$ is not).

UPDATE: I expanded the derivation above in response to a question posted on MathHub.

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, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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