Theorem: For any vectors $\mathbf{u}, \mathbf{v}$ in an inner product space: $|\langle \mathbf{u}, \mathbf{v} \rangle| \leq \|\mathbf{u}\| \|\mathbf{v}\|$

with equality if and only if $\mathbf{u}$ and $\mathbf{v}$ are linearly dependent.

Proof: We consider two cases.

Case 1: If $\mathbf{v} = \mathbf{0}$, both sides equal zero, so equality holds.

Case 2: Assume $\mathbf{v} \neq \mathbf{0}$. For any scalar $t$, consider the vector $\mathbf{u} - t\mathbf{v}$. By positivity of the norm: $\|\mathbf{u} - t\mathbf{v}\|^2 \geq 0$

Expanding using inner product properties: $\langle \mathbf{u} - t\mathbf{v}, \mathbf{u} - t\mathbf{v} \rangle \geq 0$ $\langle \mathbf{u}, \mathbf{u} \rangle - t\langle \mathbf{u}, \mathbf{v} \rangle - t\langle \mathbf{v}, \mathbf{u} \rangle + t^2\langle \mathbf{v}, \mathbf{v} \rangle \geq 0$

By symmetry, $\langle \mathbf{u}, \mathbf{v} \rangle = \langle \mathbf{v}, \mathbf{u} \rangle$: $\|\mathbf{u}\|^2 - 2t\langle \mathbf{u}, \mathbf{v} \rangle + t^2\|\mathbf{v}\|^2 \geq 0$

This is a quadratic in $t$ that is non-negative for all $t$. Choose the special value: $t = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{v}\|^2}$

This choice minimizes the quadratic. Substituting: $\|\mathbf{u}\|^2 - 2\frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{v}\|^2} \cdot \langle \mathbf{u}, \mathbf{v} \rangle + \frac{[\langle \mathbf{u}, \mathbf{v} \rangle]^2}{\|\mathbf{v}\|^4} \cdot \|\mathbf{v}\|^2 \geq 0$

Simplifying: $\|\mathbf{u}\|^2 - 2\frac{[\langle \mathbf{u}, \mathbf{v} \rangle]^2}{\|\mathbf{v}\|^2} + \frac{[\langle \mathbf{u}, \mathbf{v} \rangle]^2}{\|\mathbf{v}\|^2} \geq 0$ $\|\mathbf{u}\|^2 - \frac{[\langle \mathbf{u}, \mathbf{v} \rangle]^2}{\|\mathbf{v}\|^2} \geq 0$

Multiplying both sides by $\|\mathbf{v}\|^2 > 0$: $\|\mathbf{u}\|^2 \|\mathbf{v}\|^2 - [\langle \mathbf{u}, \mathbf{v} \rangle]^2 \geq 0$ $[\langle \mathbf{u}, \mathbf{v} \rangle]^2 \leq \|\mathbf{u}\|^2 \|\mathbf{v}\|^2$

Taking square roots: $|\langle \mathbf{u}, \mathbf{v} \rangle| \leq \|\mathbf{u}\| \|\mathbf{v}\|$

Equality condition: Equality holds if and only if the minimum of the quadratic is zero, which occurs when: $\|\mathbf{u} - t\mathbf{v}\|^2 = 0 \quad \text{for } t = \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\|\mathbf{v}\|^2}$

By definiteness, this means: $\mathbf{u} - t\mathbf{v} = \mathbf{0} \implies \mathbf{u} = t\mathbf{v}$

So $\mathbf{u}$ and $\mathbf{v}$ are linearly dependent.

Conversely, if $\mathbf{u} = c\mathbf{v}$ for some scalar $c$, then: $|\langle \mathbf{u}, \mathbf{v} \rangle| = |\langle c\mathbf{v}, \mathbf{v} \rangle| = |c|\|\mathbf{v}\|^2$ $\|\mathbf{u}\|\|\mathbf{v}\| = \|c\mathbf{v}\|\|\mathbf{v}\| = |c|\|\mathbf{v}\|^2$

So equality holds. ∎