Pythagorean Identity for Orthogonal Sums

From Right Angles to Orthogonal Sums

For two perpendicular vectors in the plane, the Pythagorean theorem relates the length of their sum to their individual lengths. Inner product spaces extend this relationship beyond the plane: vectors may be polynomials or complex coordinate vectors, and a finite list may contain more than two vectors. The key condition is orthogonality. When every distinct pair in a list is orthogonal, the cross terms in the squared norm of the sum all vanish.

Let \(V\) be an inner product space over \(\mathbb{F}\), where \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\). The inner product is written \(\langle\mathbf{u},\mathbf{v}\rangle\), with the convention that it is linear in the first slot and conjugate-linear in the second. The induced norm is \(\|\mathbf{v}\|=\sqrt{\langle\mathbf{v},\mathbf{v}\rangle}\), so \(\|\mathbf{v}\|^2=\langle\mathbf{v},\mathbf{v}\rangle\). In a real inner product space, orthogonality means \(\langle\mathbf{u},\mathbf{v}\rangle=0\); in a complex inner product space, the same definition applies.

The phrase “pairwise orthogonal” matters: each vector must be orthogonal to every other distinct vector in the list. It is not enough for the vectors to be orthogonal to one chosen member, nor does the definition require a vector to be orthogonal to itself. In particular, a list with just one vector is pairwise orthogonal because there are no distinct pairs to check.

The Pythagorean Identity

Proof. Write \(\mathbf{s}=\mathbf{v}_1+\cdots+\mathbf{v}_p\). By the definition of the induced norm and linearity in the first slot, followed by conjugate-linearity in the second slot,

Separate the terms with \(i=j\) from those with \(i\ne j\). The diagonal terms are \(\langle\mathbf{v}_i,\mathbf{v}_i\rangle=\|\mathbf{v}_i\|^2\). Every off-diagonal term is zero because the list is pairwise orthogonal. Thus the double sum is exactly \(\sum_{i=1}^{p}\|\mathbf{v}_i\|^2\), as required. If \(p=0\), the sum is the zero vector and the sum of squared norms is the empty sum \(0\), so the identity also holds under the usual empty-sum convention. \(\square\)

This proof works over both fields. In the complex case, the conjugation in the second slot does not create an extra term: every cross inner product is already zero. Also, if \(\langle\mathbf{v}_i,\mathbf{v}_j\rangle=0\), conjugate symmetry gives \(\langle\mathbf{v}_j,\mathbf{v}_i\rangle=\overline{\langle\mathbf{v}_i,\mathbf{v}_j\rangle}=0\). Therefore both orders of each cross term vanish.

For two orthogonal vectors, the theorem says \(\|\mathbf{u}+\mathbf{v}\|^2=\|\mathbf{u}\|^2+\|\mathbf{v}\|^2\). The finite-list version is not a different assumption applied repeatedly: its proof directly accounts for every pair of terms in the expansion. No finite-dimensionality assumption is needed; only a finite list and an inner product are required.

Using the Identity in a Polynomial Space

The same calculation applies when vectors are not coordinate columns. For real polynomials on \([-1,1]\), use the inner product \(\langle f,g\rangle=\int_{-1}^{1}f(x)g(x)\,dx\). The induced squared norm is \(\|f\|^2=\int_{-1}^{1}f(x)^2\,dx\). To apply the identity, verify orthogonality with this inner product, rather than by comparing polynomial coefficients.

Complex Inner Products and Cross Terms

In \(\mathbb{C}^n\), the standard inner product under the course convention is \(\langle\mathbf{u},\mathbf{v}\rangle=\sum_k u_k\overline{v_k}\). Because one slot is conjugated, a complex orthogonality check should use the inner product itself; real dot-product intuition alone does not determine the calculation. Once the inner product is zero, however, the Pythagorean identity applies just as it does over \(\mathbb{R}\).

When the Identity Does Not Apply

Pairwise orthogonality is a hypothesis, not a conclusion that follows from having a sum. If two vectors are not orthogonal, their inner products remain in the expansion. For two vectors in a real inner product space, the full expansion is \(\|\mathbf{u}+\mathbf{v}\|^2=\|\mathbf{u}\|^2+2\langle\mathbf{u},\mathbf{v}\rangle+\|\mathbf{v}\|^2\). In a complex space, the two cross terms combine as \(2\operatorname{Re}\langle\mathbf{u},\mathbf{v}\rangle\), where \(\operatorname{Re}\) denotes real part. Thus they vanish for an orthogonal pair, but need not vanish otherwise.

Common mistake: applying the identity without checking orthogonality. Let \(\mathbf{a}=(1,0)^{\top}\) and \(\mathbf{b}=(1,1)^{\top}\) in \(\mathbb{R}^2\). A tempting but incorrect calculation is to add their squared norms and claim \(\|\mathbf{a}+\mathbf{b}\|^2=1+2=3\). The condition fails because \(\mathbf{a}\cdot\mathbf{b}=1\), not \(0\). The actual sum is \((2,1)^{\top}\), whose squared norm is \(2^2+1^2=5\). The missing cross contribution is \(2(\mathbf{a}\cdot\mathbf{b})=2\), so the correct expansion gives \(1+2+2=5\).

The identity is useful because it converts one norm calculation into several smaller ones whenever orthogonality is known. It applies to real and complex inner product spaces and does not require coordinates or finite dimension. For a finite list, the proof is always the same: expand the inner product of the sum with itself, retain the diagonal terms, and use pairwise orthogonality to eliminate the rest.