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The Pythagorean Theorem

The classical Pythagorean theorem says that for a right triangle with legs $a, b$ and hypotenuse $c$ , we have $a^2 + b^2 = c^2$ . The vector version says exactly the same thing: if $\mathbf{u} \perp \mathbf{v}$ , then $\|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2$ .

The proof is a direct computation using the dot product. Expand $\|\mathbf{u} + \mathbf{v}\|^2 = (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = \mathbf{u}\cdot\mathbf{u} + 2\mathbf{u}\cdot\mathbf{v} + \mathbf{v}\cdot\mathbf{v}$ . The cross-term $2\mathbf{u} \cdot \mathbf{v}$ vanishes exactly when $\mathbf{u} \perp \mathbf{v}$ .

This result holds in all dimensions — not just 2D. Whenever two vectors are orthogonal, the squared length of their sum equals the sum of their squared lengths. This is the key behind many decompositions and error bounds.

Formal View

Theorem 6.8 — Pythagorean Theorem

Vectors

\mathbf{u}, \mathbf{v} \in \mathbb{R}^m

are orthogonal if and only if

\|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2.

Proof: $\|\mathbf{u}+\mathbf{v}\|^2 = (\mathbf{u}+\mathbf{v})\cdot(\mathbf{u}+\mathbf{v}) = \|\mathbf{u}\|^2 + 2\,\mathbf{u}\cdot\mathbf{v} + \|\mathbf{v}\|^2$ . The $2\mathbf{u}\cdot\mathbf{v}$ term vanishes iff $\mathbf{u} \perp \mathbf{v}$ .

Why This Matters

The Pythagorean theorem is used to bound errors and measure approximation quality throughout applied mathematics.

Parseval's theorem in signal processing is the Pythagorean theorem for Fourier components — total power equals sum of powers at each frequency
In the best approximation theorem, the Pythagorean theorem proves that the projection is strictly closer than any other vector in the subspace
Variance decomposition in statistics ( $R^2$ ) follows from the Pythagorean theorem applied to residuals and fitted values

Learning Resources

Proof of the Pythagorean theorem via dot products

Khan Academy

The algebraic and geometric proof of Pythagoras using dot products.

15 min

Orthogonal complements and projections

MIT OpenCourseWare

Strang on projections and the Pythagorean structure of decompositions.

45 min

Quiz

Question 1

For any two vectors $\mathbf{u}$ and $\mathbf{v}$ : $\|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2$ .

Question 2

If $\mathbf{u} \perp \mathbf{v}$ , $\|\mathbf{u}\| = 3$ , and $\|\mathbf{v}\| = 4$ , what is $\|\mathbf{u} + \mathbf{v}\|$ ?

Common Mistakes

Assuming $\|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2$ always — this only holds when $\mathbf{u} \perp \mathbf{v}$ .
Forgetting the cross term when expanding $\|\mathbf{u} + \mathbf{v}\|^2$ — always expand using the dot product, not treating it like ordinary algebra.