## 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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