The standard inner product on $\mathbb{R}^n$ is the dot product:

$\langle x, y \rangle = x \cdot y = \sum_{i=1}^n x_i y_i = x^T y.$

For $x = (1, 2, 3)$ and $y = (4, -1, 2)$: $\langle x, y \rangle = 4 - 2 + 6 = 8$.

This satisfies all three axioms: linearity is clear, symmetry gives $x \cdot y = y \cdot x$, and $\langle x, x \rangle = x_1^2 + \cdots + x_n^2 \geq 0$ with equality iff $x = 0$.

On $\mathbb{C}^n$:

$\langle x, y \rangle = \sum_{i=1}^n x_i \overline{y_i} = x^* y,$

where $x^* = \bar{x}^T$ is the conjugate transpose.

For $x = (1, i)$ and $y = (2, 3i)$: $\langle x, y \rangle = 1 \cdot \bar{2} + i \cdot \overline{3i} = 2 + i(-3i) = 2 + 3 = 5$.

Note: $\langle x, x \rangle = |x_1|^2 + \cdots + |x_n|^2 \geq 0$, so positive definiteness holds.

On $\mathbb{R}^n$ with positive weights $w_1, \ldots, w_n > 0$:

$\langle x, y \rangle_w = \sum_{i=1}^n w_i x_i y_i.$

For $w = (2, 3)$ on $\mathbb{R}^2$: $\langle (1, 1), (1, -1) \rangle_w = 2 \cdot 1 \cdot 1 + 3 \cdot 1 \cdot (-1) = -1$.

The vectors $(1, 1)$ and $(1, -1)$ are orthogonal in the standard inner product ($1 - 1 = 0$) but not in this weighted one.

On the space of continuous functions $C[a, b]$:

$\langle f, g \rangle = \int_a^b f(x) g(x) \, dx.$

For $f(x) = x$ and $g(x) = x^2$ on $[0, 1]$: $\langle f, g \rangle = \int_0^1 x^3 \, dx = \frac{1}{4}$.

For $f(x) = \sin(\pi x)$ and $g(x) = \sin(2\pi x)$ on $[0, 1]$: $\langle f, g \rangle = \int_0^1 \sin(\pi x) \sin(2\pi x) \, dx = 0$ (orthogonality of Fourier modes).

For $v = (3, 4, 0) \in \mathbb{R}^3$ with the standard inner product:

$\|v\| = \sqrt{9 + 16 + 0} = 5$.

$\|(1, 1, 1)\| = \sqrt{3}$, $\|(1, 0, 0)\| = 1$.

With $\langle f, g \rangle = \int_0^1 f(x) g(x) \, dx$:

$\|x\| = \sqrt{\int_0^1 x^2 \, dx} = \sqrt{1/3} = \frac{1}{\sqrt{3}}$.

$\|1\| = \sqrt{\int_0^1 1 \, dx} = 1$.

$\|\sin(\pi x)\| = \sqrt{\int_0^1 \sin^2(\pi x) \, dx} = \sqrt{1/2} = \frac{1}{\sqrt{2}}$.

$u = (3, 0)$, $v = (0, 4)$: $\|u + v\| = \|(3, 4)\| = 5 \leq 3 + 4 = \|u\| + \|v\| = 7$. The inequality is strict because $u$ and $v$ are not parallel.

$u = (1, 2)$, $v = (2, 4)$: $\|u + v\| = \|(3, 6)\| = 3\sqrt{5} = 3\sqrt{5}$ and $\|u\| + \|v\| = \sqrt{5} + 2\sqrt{5} = 3\sqrt{5}$. Equality holds because $v = 2u$ (they are parallel).

$u = (1, 0, 0)$, $v = (1, 1, 0)$: $\cos\theta = \frac{1}{\sqrt{2}}$, so $\theta = \pi/4 = 45°$.

$u = (1, 0, 0)$, $v = (0, 1, 0)$: $\cos\theta = 0$, so $\theta = \pi/2 = 90°$ (perpendicular).

$u = (1, 1, 1)$, $v = (-1, -1, -1)$: $\cos\theta = \frac{-3}{3} = -1$, so $\theta = \pi = 180°$ (anti-parallel).

With $\langle f, g \rangle = \int_0^1 f(x)g(x)\,dx$, the angle between $f(x) = 1$ and $g(x) = x$:

$\cos\theta = \frac{\langle 1, x \rangle}{\|1\| \cdot \|x\|} = \frac{1/2}{1 \cdot 1/\sqrt{3}} = \frac{\sqrt{3}}{2}$, so $\theta = \pi/6 = 30°$.

$M = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}$ is positive definite (eigenvalues $\frac{5 \pm \sqrt{5}}{2} > 0$).

$\langle x, y \rangle = x^T M y$. For $x = (1, 0)$, $y = (0, 1)$: $\langle x, y \rangle = (1, 0) \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = 1$.

$\|e_1\|^2 = \langle e_1, e_1 \rangle = 2$, $\|e_2\|^2 = 3$. The standard basis vectors are not unit vectors in this inner product.

$M = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$ has eigenvalues $3$ and $-1$. Since $-1 < 0$, $M$ is not positive definite.

$\langle (1, -1), (1, -1) \rangle = (1, -1) \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = (1, -1)\begin{pmatrix} -1 \\ 1 \end{pmatrix} = -2 < 0$.

This fails positive definiteness, so it is not an inner product.

$u = (1, 0)$, $v = (0, 1)$ in $\mathbb{R}^2$:

$\|u + v\|^2 + \|u - v\|^2 = \|(1,1)\|^2 + \|(1,-1)\|^2 = 2 + 2 = 4$. $2\|u\|^2 + 2\|v\|^2 = 2 + 2 = 4$ ✓.