## Linear Algebra and Its Applications, Exercise 3.2.5

Exercise 3.2.5. If $a = (1, 1, \dots, 1)$ is a vector in $\mathbb{R}^n$ then what is the angle $\theta$ between $a$ and the coordinate axes? What is the matrix $P$ that projects vectors in $\mathbb{R}^n$ onto $a$?

Answer: Consider the $i^{th}$ coordinate axis and the unit vector $e_i$ lying along that axis, whose $i^{th}$ entry is 1 and whose other entries are zero. If $\theta$ is the angle between $a$ and $e_i$ then we have $\cos \theta = a^Te_i / \|a\|\|e_i\|$. We have $\|e_i\| = 1$ and $\|a\|^2 = \sum_{j=1}^n 1^2 = n$ so that $\|a\| = \sqrt{n}$. We also have $a^Te_i = 1$ since the $i^{th}$ entries of $a$ and $e_i$ are both 1 and all other entries of $e_i$ are zero.

We thus have $\cos \theta = a^Te_i / \|a\|\|e_i\| = 1/\sqrt{n}$ so that $\theta = \arccos 1/\sqrt{n}$.

The matrix $P$ that projects vectors in $\mathbb{R}^n$ onto $a$ is $P = aa^T/a^Ta = \frac{1}{a^Ta}(aa^T) = \frac{1}{n} \begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix} \begin{bmatrix} 1&\cdots&1 \end{bmatrix}$ $= \frac{1}{n} \begin{bmatrix} 1&\cdots&1 \\ \vdots&\ddots&\vdots \\ 1&\cdots&1 \end{bmatrix} = \begin{bmatrix} \frac{1}{n}&\cdots&\frac{1}{n} \\ \vdots&\ddots&\vdots \\ \frac{1}{n}&\cdots&\frac{1}{n} \end{bmatrix}$

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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