Orthogonal Complements

Orthogonal complements are the backbone of orthogonal projection, which finds the closest point in a subspace to a given vector — this is exactly how Orthogonal Projections are constructed, by projecting along $U^\perp$ onto $U$. This idea powers least-squares regression in statistics: fitting a line or hyperplane to noisy data amounts to projecting a data vector onto the column space of a design matrix, with the residual living in the orthogonal complement. In numerical linear algebra, the QR decomposition and Gram–Schmidt-based algorithms exploit the same complement structure to solve linear systems stably. In signal processing, decomposing a signal into a "signal subspace" and a "noise subspace" (as in principal component analysis) is literally an orthogonal complement decomposition. The identity $(U^\perp)^\perp = U$ and the complement construction also underlie the theory of the adjoint, since the adjoint's kernel and the range of the original map are orthogonal complements of one another — a fact central to the study of self-adjoint and normal operators in quantum mechanics, where observables are represented by operators whose eigenspaces decompose the Hilbert space via mutually orthogonal complements. In coding theory, the dual code of a linear code over a finite field is defined via an orthogonal-complement-style construction with respect to a bilinear form, echoing the same algebraic pattern.

:::quiz[In the counterexample with $U = \{p \in \mathbb{R}[x] : p(0)=0\}$, what causes $(U^\perp)^\perp = U$ to fail?]

• The inner product used is not positive-definite
• $U$ is not a subspace
• $V$ is infinite-dimensional and $U$ is not closed ✓
• $U^\perp$ is not actually a subspace in this example

Ask the tutor: How does the direct sum $V = U \oplus U^\perp$ give a formula for the orthogonal projection onto $U$?

Picture $V = \mathbb{R}^3$ with the standard dot product, and let $U$ be the $xy$-plane, spanned by $e_1$ and $e_2$. A vector $v = (a,b,c)$ satisfies $\langle v, e_1\rangle = a$ and $\langle v, e_2 \rangle = b$; for $v$ to be orthogonal to every vector of $U$ (every linear combination of $e_1, e_2$), we need $a = b = 0$. So $U^\perp$ is exactly the $z$-axis. Notice that $U$ and $U^\perp$ together span all of $\mathbb{R}^3$, and they meet only at the origin — this is the direct sum decomposition in its simplest form. The key structural fact is that this splitting is not a coincidence of the plane-and-axis example: it happens for every subspace of every finite-dimensional inner product space. If $U$ is any subspace, no matter how tilted or oddly shaped its basis vectors look, there is always a complementary subspace $U^\perp$, orthogonal to it, such that the two together reconstruct $V$ uniquely. The proof strategy exploits the one tool we already have: an orthonormal basis. Extend an orthonormal basis of $U$ to one of $V$; the added vectors span exactly $U^\perp$. This is why Existence of Orthonormal Bases (via Gram–Schmidt) is the essential prerequisite — orthogonal complements are built by extending orthonormal bases, not by an independent argument. The double-complement fact $(U^\perp)^\perp = U$ says the relationship is symmetric: $U$ and $U^\perp$ are mutual complements, and there is no "bigger" orthogonal complement lurking outside $U$ waiting to be discovered by complementing twice. This symmetry is special to finite dimensions (or to closed subspaces of infinite-dimensional Hilbert spaces) — a fact the counterexample below makes precise.

Example 1: A tilted plane in $\mathbb{R}^3$. Let $V = \mathbb{R}^3$ with the standard dot product, and let

These two vectors are linearly independent (neither is a scalar multiple of the other), so $\dim U = 2$. Step 1: Find $U^\perp$ directly from the definition. A vector $v = (x,y,z)$ lies in $U^\perp$ iff it is orthogonal to $u_1$ and $u_2$ (orthogonality to all of $U$ reduces to orthogonality to a spanning set, by linearity):

From the first equation $y = -x$; from the second $z = -x$. So $v = (x, -x, -x) = x(1,-1,-1)$. Hence

a one-dimensional subspace (a line), as predicted since $\dim U^\perp = \dim V - \dim U = 3 - 2 = 1$. Step 2: Decompose an explicit vector. Take $v = (3, 2, 5)$. We want $v = u + w$ with $u \in U$, $w \in U^\perp$. Write $u = a\,u_1 + b\,u_2 = (a+b,\, a,\, b)$ and $w = c(1,-1,-1)$. Then

This gives the system

From the second equation $a = 2+c$; from the third $b = 5+c$. Substituting into the first:

Then $a = 2 - \tfrac43 = \tfrac23$ and $b = 5 - \tfrac43 = \tfrac{11}{3}$. So

Check: $u + w = \left(\tfrac{13}{3}-\tfrac43,\ \tfrac23+\tfrac43,\ \tfrac{11}{3}+\tfrac43\right) = (3, 2, 5) = v$. ✓ Also $\langle u_1, w\rangle = -\tfrac43+\tfrac43+0 = 0$ ✓ and $\langle u_2, w\rangle = -\tfrac43 + 0 + \tfrac43 = 0$ ✓, confirming $w \in U^\perp$. This concretely exhibits the direct sum decomposition $V = U \oplus U^\perp$ for a non-axis-aligned subspace, with all arithmetic tracked exactly.

Example 2: A subspace of $\mathbb{C}^3$ and the double-complement identity. Let $V = \mathbb{C}^3$ with the standard Hermitian inner product $\langle x, y\rangle = x_1\overline{y_1} + x_2 \overline{y_2} + x_3\overline{y_3}$, and let

So $\dim U = 1$. Step 1: Compute $U^\perp$. A vector $v = (x,y,z) \in U^\perp$ iff $\langle v, u\rangle = 0$, i.e.

So $v = (iy, y, z)$ for arbitrary $y, z \in \mathbb{C}$, giving

a $2$-dimensional subspace, matching $\dim U + \dim U^\perp = 1 + 2 = 3 = \dim V$. Step 2: Verify $(U^\perp)^\perp = U$ directly, without invoking the theorem, as a sanity check. Let $w_1 = (i,1,0)$, $w_2 = (0,0,1)$ span $U^\perp$. A vector $v=(x,y,z) \in (U^\perp)^\perp$ must satisfy $\langle v, w_1\rangle = 0$ and $\langle v, w_2 \rangle = 0$:

So $v = (x, ix, 0) = x(1, i, 0) = x\,u$. Hence $(U^\perp)^\perp = \operatorname{span}(u) = U$, confirming the theorem's conclusion (5) by direct computation, with the conjugate-linearity of the second argument correctly tracked throughout (note the sign flip from $\overline{i} = -i$).