Pythagorean theorem

Dependencies:

1. Inner product space
2. Inner product is anti-linear in second argument

Let $V$ be an inner product space and $\mathbf{u}, \mathbf{v} \in V$. If $\langle \mathbf{u}, \mathbf{v} \rangle = 0$, then $\|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 = \|\mathbf{u}+\mathbf{v}\|^2$.

Proof

\[ \langle \mathbf{v}, \mathbf{u} \rangle = \overline{\langle \mathbf{u}, \mathbf{v} \rangle} \tag{by conjugate symmetry} = \overline{0} = 0 \]

\begin{align} & \|\mathbf{u}+\mathbf{v}\|^2 \\ &= \langle \mathbf{u}+\mathbf{v}, \mathbf{u}+\mathbf{v} \rangle \\ &= \langle \mathbf{u}, \mathbf{u} \rangle + \langle \mathbf{u}, \mathbf{v} \rangle + \langle \mathbf{v}, \mathbf{u} \rangle + \langle \mathbf{v}, \mathbf{v} \rangle \tag{by (anti-)linearity} \\ &= \langle \mathbf{u}, \mathbf{u} \rangle + \langle \mathbf{v}, \mathbf{v} \rangle \\ &= \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 \end{align}

Dependency for: None

Info:

• Depth: 6
• Number of transitive dependencies: 7

Transitive dependencies:

1. /complex-numbers/conjugation-is-homomorphic
2. Group
3. Ring
4. Field
5. Vector Space
6. Inner product space
7. Inner product is anti-linear in second argument