Exercises — Lesson 2: Dot Products, Norms, and Orthogonality
Exercise 2.1. [Hand] For \(x = (1,2,2)\) and \(y = (2,-2,1)\): compute \(\langle x, y \rangle\), \(\lVert x \rVert\), \(\lVert y \rVert\), and \(\cos\theta\). Are \(x\) and \(y\) orthogonal?
Solution.
\[ \begin{aligned} \langle x, y \rangle &= 2 - 4 + 2 = 0 \\[2pt] \lVert x \rVert &= \sqrt{9} = 3 = \lVert y \rVert \\[2pt] \cos\theta &= \frac{0}{9} = 0 \qquad \Longrightarrow \qquad \theta = \frac{\pi}{2} \end{aligned} \]
Exercise 2.2. [Hand] Project \(x = (4,1)\) onto \(y = (2,1)\); write \(x = cy + z\) and verify \(\langle z, y \rangle = 0\). Then compute \(\lVert cy \rVert^2 + \lVert z \rVert^2\) and confirm it equals \(\lVert x \rVert^2\).
Solution. With \(x = (4,1)\) and \(y = (2,1)\),
\[ \begin{aligned} c &= \frac{\langle x, y \rangle}{\langle y, y \rangle} = \frac{9}{5}, &\qquad cy &= \left( \frac{18}{5},\, \frac{9}{5} \right) \\[4pt] z &= x - cy = \left( \frac{20}{5},\, \frac{5}{5} \right) - \left( \frac{18}{5},\, \frac{9}{5} \right) = \left( \frac{2}{5},\, \frac{-4}{5} \right), &\qquad \langle z, y \rangle &= \frac{4}{5} - \frac{4}{5} = 0. \quad\checkmark \end{aligned} \]
The Pythagorean check:
\[ \begin{aligned} \lVert cy \rVert^2 &= \frac{324}{25} + \frac{81}{25} = \frac{405}{25} \\[2pt] \lVert z \rVert^2 &= \frac{4}{25} + \frac{16}{25} = \frac{20}{25} \\[2pt] \lVert x \rVert^2 &= 17 \\[2pt] \lVert cy \rVert^2 + \lVert z \rVert^2 &= \frac{425}{25} = 17. \quad\checkmark \end{aligned} \]
Exercise 2.3. [Proof, \(\star\)] (Parallelogram equality, LADR 6A) Prove that for all \(x, y \in \mathbb{R}^n\), \[ \lVert x+y \rVert^2 + \lVert x-y \rVert^2 = 2\lVert x \rVert^2 + 2\lVert y \rVert^2, \] using only the three inner-product properties. Interpret the identity for the parallelogram with sides \(x\) and \(y\).
Proof. Prove for all \(x, y \in \mathbb{R}^n\): \(\lVert x+y \rVert^2 + \lVert x-y \rVert^2 = 2\lVert x \rVert^2 + 2\lVert y \rVert^2\).
\[ \begin{aligned} \lVert x+y \rVert^2 &= \langle x+y,\, x+y \rangle = \langle x,\, x+y \rangle + \langle y,\, x+y \rangle = \langle x,x \rangle + \langle x,y \rangle + \langle y,x \rangle + \langle y,y \rangle \\[4pt] \lVert x-y \rVert^2 &= \langle x-y,\, x-y \rangle = \langle x,\, x-y \rangle + \langle -y,\, x-y \rangle = \langle x,x \rangle - \langle x,y \rangle - \langle y,x \rangle + \langle y,y \rangle \end{aligned} \]
Adding the two lines, the cross terms cancel:
\[ \lVert x+y \rVert^2 + \lVert x-y \rVert^2 = 2\langle x,x \rangle + 2\langle y,y \rangle = 2\lVert x \rVert^2 + 2\lVert y \rVert^2. \qquad\square \]
For the parallelogram with sides \(x\) and \(y\), the two diagonals are \(x+y\) and \(x-y\): the sum of the squared diagonals equals the sum of the squares of all four sides.
Exercise 2.4. [Proof] Show that if \(\langle x, v \rangle = 0\) for every \(v \in \mathbb{R}^n\), then \(x = 0\). (Hint: choose \(v\) wisely.)
Proof. Prove: if \(\langle x, v \rangle = 0\) for all \(v \in \mathbb{R}^n\), then \(x = 0\).
Suppose \(\langle x, v \rangle = 0\) for all \(v \in \mathbb{R}^n\). Let \(v \in \mathbb{R}^n\) and set \(v' = x - v\). Note \(v' \in \mathbb{R}^n\). Then
\[ \langle x, v' \rangle = \langle x,\, x - v \rangle = \langle x, x \rangle - \langle x, v \rangle = \langle x, x \rangle. \]
But \(v' \in \mathbb{R}^n \implies \langle x, v' \rangle = 0 \implies \langle x, x \rangle = 0 \implies x = 0\). \(\square\)