Two Components, One Representation
Suppose a vector is written as a sum of two parts: one belonging to a subspace \(W\), and the other orthogonal to every vector in \(W\). The parts have different roles: the first lies within \(W\), while the second points in a direction perpendicular to \(W\). The key point here is that, once such a representation exists, there cannot be two different choices of those parts. This uniqueness follows from a basic fact about orthogonal complements, not from a formula for computing the components.
Let \(V\) be an inner product space over \(\mathbb{R}\) or \(\mathbb{C}\), and let \(W\) be a subspace of \(V\). The inner product is positive definite, so \(\langle \mathbf{x},\mathbf{x}\rangle=0\) holds exactly when \(\mathbf{x}=\mathbf{0}\). The orthogonal complement of \(W\) is the set of vectors orthogonal to every vector in \(W\):
Over \(\mathbb{C}\), the inner product convention is linear in its first slot and conjugate-linear in its second. Thus the displayed condition is equivalent to \(\langle \mathbf{w},\mathbf{z}\rangle=0\) for every \(\mathbf{w}\in W\). The set \(W^\perp\) is a subspace: it contains \(\mathbf{0}\), and if \(\mathbf{z}_1,\mathbf{z}_2\in W^\perp\) and \(a,b\) are scalars, then \(\langle a\mathbf{z}_1+b\mathbf{z}_2,\mathbf{w}\rangle=a\langle\mathbf{z}_1,\mathbf{w}\rangle+b\langle\mathbf{z}_2,\mathbf{w}\rangle=0\) for every \(\mathbf{w}\in W\).
In this statement, “unique” means that if the same \(\mathbf{v}\) also equals \(\mathbf{w}' + \mathbf{z}'\), with \(\mathbf{w}'\in W\) and \(\mathbf{z}'\in W^\perp\), then \(\mathbf{w}'=\mathbf{w}\) and \(\mathbf{z}'=\mathbf{z}\). This is a claim about the pair of components, not just about their sum.
The Intersection That Forces Uniqueness
Proof. The zero vector belongs to \(W\) and to \(W^\perp\), so it belongs to their intersection. Now let \(\mathbf{x}\in W\cap W^\perp\). Since \(\mathbf{x}\in W\) and \(\mathbf{x}\in W^\perp\), the definition of \(W^\perp\), applied with \(\mathbf{w}=\mathbf{x}\), gives \(\langle\mathbf{x},\mathbf{x}\rangle=0\). Positive definiteness of the inner product implies \(\mathbf{x}=\mathbf{0}\). Therefore the intersection contains no vector other than \(\mathbf{0}\), and \(W\cap W^\perp=\{\mathbf{0}\}\). \(\square\)
Proof. Subtract the two expressions for \(\mathbf{v}\) and rearrange: \(\mathbf{w}_1-\mathbf{w}_2=\mathbf{z}_2-\mathbf{z}_1\). Because \(W\) is a subspace, \(\mathbf{w}_1-\mathbf{w}_2\in W\). Because \(W^\perp\) is a subspace, \(\mathbf{z}_2-\mathbf{z}_1\in W^\perp\). The shared vector therefore belongs to \(W\cap W^\perp\), which the preceding proposition shows is \(\{\mathbf{0}\}\). Hence \(\mathbf{w}_1-\mathbf{w}_2=\mathbf{0}\), so \(\mathbf{w}_1=\mathbf{w}_2\); the equation \(\mathbf{w}_1-\mathbf{w}_2=\mathbf{z}_2-\mathbf{z}_1\) then gives \(\mathbf{z}_1=\mathbf{z}_2\). \(\square\)
The proof separates two questions that are easy to conflate. The existence theorem guarantees that a suitable pair can be found when \(V\) is finite-dimensional. The uniqueness argument says that any two suitable pairs must agree, and it works even without finite dimensionality. The argument does not provide a method for calculating the pair; it establishes why a valid pair, once obtained, is the only one.
Checking the Uniqueness in Examples
Worked Example: A Line in \(\mathbb{R}^2\)
Use the standard dot product on \(\mathbb{R}^2\). Let \(W=\operatorname{span}\{(1,1)^{\top}\}\), and decompose \(\mathbf{v}=(5,1)^{\top}\). A vector in \(W\) has the form \(a(1,1)^{\top}\). To find \(W^\perp\), a vector \((x,y)^{\top}\) must satisfy \((x,y)\cdot(1,1)=x+y=0\), so \(W^\perp=\operatorname{span}\{(1,-1)^{\top}\}\).
Write a candidate decomposition as \(\mathbf{v}=a(1,1)^{\top}+b(1,-1)^{\top}\). Comparing coordinates gives \(a+b=5\) and \(a-b=1\). Adding these equations gives \(2a=6\), so \(a=3\); then \(b=2\). Thus \(\mathbf{w}=(3,3)^{\top}\) and \(\mathbf{z}=(2,-2)^{\top}\).
Check: \(\mathbf{w}+\mathbf{z}=(3+2,3-2)^{\top}=(5,1)^{\top}=\mathbf{v}\), and \(\mathbf{w}\cdot(1,-1)=3-3=0\), so \(\mathbf{w}\in W\) and \(\mathbf{z}\in W^\perp\). Any other decomposition must use coefficients \(a',b'\) satisfying \(a'+b'=5\) and \(a'-b'=1\), which force \(a'=3\) and \(b'=2\) by the same additions and subtraction. The components are unique.
Worked Example: A Plane in \(\mathbb{R}^3\)
Let \(W=\{(x,y,z)^{\top}:x+y+z=0\}\) in \(\mathbb{R}^3\), with the standard dot product. First identify its orthogonal complement. Both \((1,-1,0)^{\top}\) and \((1,0,-1)^{\top}\) lie in \(W\). If \((r,s,t)^{\top}\in W^\perp\), orthogonality to these two vectors gives \(r-s=0\) and \(r-t=0\). Hence \(r=s=t\), so \((r,s,t)^{\top}=r(1,1,1)^{\top}\). Conversely, \((1,1,1)^{\top}\) is orthogonal to every \((x,y,z)^{\top}\in W\), since its dot product with that vector is \(x+y+z=0\). Therefore \(W^\perp=\operatorname{span}\{(1,1,1)^{\top}\}\).
For \(\mathbf{v}=(2,1,3)^{\top}\), take \(\mathbf{z}=(2,2,2)^{\top}\) and subtract to obtain \(\mathbf{w}=\mathbf{v}-\mathbf{z}=(0,-1,1)^{\top}\). The sum is \((0+2,-1+2,1+2)^{\top}=(2,1,3)^{\top}\). Also, \(0-1+1=0\), so \(\mathbf{w}\in W\), and \(\mathbf{z}=2(1,1,1)^{\top}\in W^\perp\).
To check uniqueness directly, suppose another orthogonal-complement component is \(\mathbf{z}'=c(1,1,1)^{\top}\). Its corresponding component in \(W\) must be \(\mathbf{v}-\mathbf{z}'=(2-c,1-c,3-c)^{\top}\). The coordinates of a vector in \(W\) sum to zero, so \(6-3c=0\), giving \(c=2\). Thus \(\mathbf{z}'=\mathbf{z}\) and \(\mathbf{v}-\mathbf{z}'=\mathbf{w}\).
Worked Example: Constants and Linear Polynomials
Consider \(V=P_1\), the real vector space of polynomials of degree at most one, with inner product \(\langle f,g\rangle=\int_0^1 f(t)g(t)\,dt\). This is an inner product: in particular, \(\int_0^1 f(t)^2\,dt>0\) for every nonzero polynomial \(f\in P_1\). Let \(W\) be the subspace of constant polynomials. A polynomial \(z(t)=at+b\) belongs to \(W^\perp\) exactly when it is orthogonal to every constant \(c\), or equivalently when \(\langle z,c\rangle=c\int_0^1(at+b)\,dt=c(a/2+b)=0\) for every real \(c\). This holds exactly when \(b=-a/2\).
For \(p(t)=4t\), choose the constant component \(w(t)=2\) and the orthogonal component \(z(t)=4t-2\). Their sum is \(2+(4t-2)=4t=p(t)\). Moreover, \(\int_0^1(4t-2)\,dt=[2t^2-2t]_0^1=0\), so \(z\) is orthogonal to every constant polynomial.
For a direct uniqueness check, suppose \(4t=c+z'(t)\), where \(c\) is constant and \(z'\in W^\perp\). Integrating both sides from \(0\) to \(1\) gives \(2=c+0\), because \(\int_0^1 4t\,dt=2\) and every element of \(W^\perp\) has integral zero. Hence \(c=2\), and then \(z'(t)=4t-2\). Both components are forced.
A Common Mistake: Matching the Sum Is Not Enough
A pair of vectors can add to \(\mathbf{v}\) and still fail to be an orthogonal decomposition. In the line example, \((4,4)^{\top}+(1,-3)^{\top}=(5,1)^{\top}\), so the sum is correct. But \((1,-3)\cdot(1,1)=1-3=-2\ne0\); the second vector is not in \(W^\perp\). This pair does not meet the conditions of the decomposition, so it cannot be used as an alternative to the valid pair \((3,3)^{\top}+(2,-2)^{\top}\).
Another useful check in a uniqueness proof is to keep the subtraction signs in order. From \(\mathbf{w}_1+\mathbf{z}_1=\mathbf{w}_2+\mathbf{z}_2\), the correct rearrangement is \(\mathbf{w}_1-\mathbf{w}_2=\mathbf{z}_2-\mathbf{z}_1\). Each side then belongs to one of the two subspaces. The conclusion follows because the shared vector lies in their intersection, not merely because the two sums are equal.
Why Uniqueness Matters
Uniqueness makes the decomposition meaningful as a way of assigning two definite components to a vector: one in \(W\), one in \(W^\perp\). Without uniqueness, the phrase “the \(W\)-component” would not identify a single vector. The proof also shows exactly which property secures this: the two subspaces intersect only at zero. The zero and full-space cases fit the same statement. If \(W=\{\mathbf{0}\}\), then \(W^\perp=V\), and the only possible decomposition is \(\mathbf{v}=\mathbf{0}+\mathbf{v}\). If \(W=V\), then \(W^\perp=\{\mathbf{0}\}\), and it is \(\mathbf{v}=\mathbf{v}+\mathbf{0}\). In each case, the pair is unique.
Check Your Understanding
Use the definitions and the intersection argument to justify each answer.
- Why does \(\mathbf{x}\in W\cap W^\perp\) imply \(\langle\mathbf{x},\mathbf{x}\rangle=0\)?
- If \(\mathbf{v}=\mathbf{w}_1+\mathbf{z}_1=\mathbf{w}_2+\mathbf{z}_2\), which rearranged equation places one vector in both \(W\) and \(W^\perp\)?
- For \(W=\operatorname{span}\{(1,0)^{\top}\}\) in \(\mathbb{R}^2\), identify \(W^\perp\) and decompose \((3,4)^{\top}\).
- Explain why a pair whose sum is \(\mathbf{v}\) is not necessarily an orthogonal decomposition of \(\mathbf{v}\).
- Does the uniqueness proof require \(V\) to be finite-dimensional? State which step determines your answer.
