Finding Orthogonal Decomposition Components

Computing the Two Components

Suppose a vector $\mathbf{x}$ is to be written as a sum of one vector in a subspace $W$ and another vector orthogonal to every vector in $W$. The orthogonal decomposition theorem guarantees that such a sum exists uniquely in a finite-dimensional inner product space. If an orthonormal basis of $W$ is known, the two parts can be calculated directly: take inner products with the basis vectors to find the coefficients of the part in $W$, then subtract it from $\mathbf{x}$.

We work over $\mathbb{R}$. Let $V$ be a finite-dimensional real inner product space, with inner product $\langle\cdot,\cdot\rangle$. A list $(\mathbf{q}_1,\dots,\mathbf{q}_k)$ is an orthonormal basis of $W$ if it is a basis of $W$, each $\mathbf{q}_i$ has norm $1$, and distinct vectors in the list are orthogonal. Here $\|\mathbf{q}\|=\sqrt{\langle\mathbf{q},\mathbf{q}\rangle}$. The orthogonal complement of $W$ in $V$, denoted $W^\perp$, is the set of vectors in $V$ orthogonal to every vector in $W$.

Component formula: If $W$ is a subspace of a finite-dimensional real inner product space $V$, and $(\mathbf{q}_1,\dots,\mathbf{q}_k)$ is an orthonormal basis of $W$, then every $\mathbf{x}\in V$ has the decomposition $$ \mathbf{x}=\mathbf{x}_W+\mathbf{x}_{W^\perp}, \qquad \mathbf{x}_W=\sum_{i=1}^{k}\langle\mathbf{x},\mathbf{q}_i\rangle\mathbf{q}_i, \qquad \mathbf{x}_{W^\perp}=\mathbf{x}-\mathbf{x}_W. $$ The first component belongs to $W$, and the second belongs to $W^\perp$. An empty basis, when $W=\{\mathbf{0}\}$, gives an empty sum equal to $\mathbf{0}$.

The coefficient of $\mathbf{q}_i$ is the inner product $\langle\mathbf{x},\mathbf{q}_i\rangle$, not an arbitrary scaling factor. Orthonormality is what makes the coefficients this simple. For a general basis, taking these inner products does not in general give the coefficients of the component in $W$.

Why the Formula Gives the Components

The formula both constructs a decomposition and identifies its two parts uniquely. The proof uses only the basis property, orthonormality, and the definition of the orthogonal complement.

Theorem (Orthonormal-basis formula for orthogonal components): Let $V$ be a finite-dimensional real inner product space, let $W$ be a subspace of $V$, and let $(\mathbf{q}_1,\dots,\mathbf{q}_k)$ be an orthonormal basis of $W$. For every $\mathbf{x}\in V$, the formula above gives the unique expression $\mathbf{x}=\mathbf{w}+\mathbf{z}$ with $\mathbf{w}\in W$ and $\mathbf{z}\in W^\perp$.

Proof. Define $\mathbf{w}=\sum_{i=1}^{k}\langle\mathbf{x},\mathbf{q}_i\rangle\mathbf{q}_i$ and $\mathbf{z}=\mathbf{x}-\mathbf{w}$. Because each $\mathbf{q}_i\in W$ and $W$ is a subspace, $\mathbf{w}\in W$. By construction, $\mathbf{x}=\mathbf{w}+\mathbf{z}$. For each $j$ with $1\le j\le k$, orthonormality gives $\langle\mathbf{q}_i,\mathbf{q}_j\rangle=1$ when $i=j$ and $0$ otherwise. Therefore $\langle\mathbf{z},\mathbf{q}_j\rangle=\langle\mathbf{x},\mathbf{q}_j\rangle-\sum_{i=1}^{k}\langle\mathbf{x},\mathbf{q}_i\rangle\langle\mathbf{q}_i,\mathbf{q}_j\rangle=0$. Every vector in $W$ is a linear combination of the basis vectors, so $\mathbf{z}$ is orthogonal to every vector in $W$, and $\mathbf{z}\in W^\perp$. For uniqueness, suppose also $\mathbf{x}=\mathbf{w}'+\mathbf{z}'$ with $\mathbf{w}'\in W$ and $\mathbf{z}'\in W^\perp$. Then $\mathbf{w}-\mathbf{w}'=\mathbf{z}'-\mathbf{z}$. The left side belongs to $W$; the right side belongs to $W^\perp$. A vector in both $W$ and $W^\perp$ must be zero, since it is orthogonal to itself and hence has squared norm zero. Thus $\mathbf{w}=\mathbf{w}'$ and $\mathbf{z}=\mathbf{z}'$. $\square$

This proof also explains a practical shortcut: once $\mathbf{x}_W$ has been calculated, it is enough to subtract it from $\mathbf{x}$ to obtain the other component. The subtraction is justified because the theorem shows that the result is orthogonal to $W$, not merely because it is the leftover vector.

Worked Components in Three Settings

Worked Example: A Subspace of $\mathbb{R}^3$

Let $W\subseteq\mathbb{R}^3$ have orthonormal basis $\mathbf{q}_1=\frac{1}{\sqrt{2}}(1,1,0)^\top$ and $\mathbf{q}_2=\frac{1}{\sqrt{6}}(1,-1,2)^\top$. Find the two components of $\mathbf{x}=(2,0,1)^\top$.

The basis vectors have norm $1$, since their squared norms are $2/2=1$ and $6/6=1$. Their inner product is $(1-1+0)/\sqrt{12}=0$. Thus the stated list is orthonormal. The coefficients are

\langle\mathbf{x},\mathbf{q}_1\rangle =\frac{2+0+0}{\sqrt{2}}=\sqrt{2}, \qquad \langle\mathbf{x},\mathbf{q}_2\rangle =\frac{2+0+2}{\sqrt{6}}=\frac{4}{\sqrt{6}}.

Using the component formula gives

\mathbf{x}_W =\sqrt{2}\,\mathbf{q}_1+\frac{4}{\sqrt{6}}\,\mathbf{q}_2 =(1,1,0)^\top+\frac{2}{3}(1,-1,2)^\top =\left(\frac{5}{3},\frac{1}{3},\frac{4}{3}\right)^\top.

Subtract to find the other component: $\mathbf{x}_{W^\perp}=\mathbf{x}-\mathbf{x}_W=(\frac{1}{3},-\frac{1}{3},-\frac{1}{3})^\top$.

Check: The inner products with the basis of $W$ are $\langle\mathbf{x}_{W^\perp},\mathbf{q}_1\rangle=(\frac13-\frac13)/\sqrt2=0$ and $\langle\mathbf{x}_{W^\perp},\mathbf{q}_2\rangle=(\frac13+\frac13-\frac23)/\sqrt6=0$. Also $\mathbf{x}_W+\mathbf{x}_{W^\perp}=(2,0,1)^\top=\mathbf{x}$.

Worked Example: A Polynomial Subspace

In $P_2$, the real polynomials of degree at most $2$, use the inner product $\langle p,q\rangle=\int_{-1}^{1}p(t)q(t)\,dt$. Let $W=\operatorname{span}\{1,t\}$ and $\mathbf{x}(t)=t^2+t+1$. Find its components in $W$ and $W^\perp$.

An orthonormal basis of $W$ is $q_1(t)=1/\sqrt{2}$ and $q_2(t)=\sqrt{3/2}\,t$. Indeed, $\langle q_1,q_1\rangle=2/2=1$, $\langle q_2,q_2\rangle=(3/2)(2/3)=1$, and $\langle q_1,q_2\rangle=0$ because the integral of the odd function $t$ over $[-1,1]$ is zero. The coefficients are

\begin{aligned} \langle\mathbf{x},q_1\rangle &=\frac{1}{\sqrt{2}}\int_{-1}^{1}(t^2+t+1)\,dt =\frac{1}{\sqrt{2}}\left(\frac{2}{3}+0+2\right) =\frac{4\sqrt{2}}{3},\\ \langle\mathbf{x},q_2\rangle &=\sqrt{\frac{3}{2}}\int_{-1}^{1}(t^3+t^2+t)\,dt =\sqrt{\frac{3}{2}}\left(0+\frac{2}{3}+0\right) =\sqrt{\frac{2}{3}}. \end{aligned}

Therefore $\mathbf{x}_W=(4\sqrt2/3)q_1+\sqrt{2/3}\,q_2=4/3+t$. Subtracting gives $\mathbf{x}_{W^\perp}=t^2+t+1-(4/3+t)=t^2-1/3$.

Check: The remainder is orthogonal to both basis polynomials: $\langle t^2-1/3,1\rangle=2/3-(1/3)2=0$, and $\langle t^2-1/3,t\rangle=\int_{-1}^{1}(t^3-t/3)\,dt=0$, since the integrand is odd. Finally, $(4/3+t)+(t^2-1/3)=t^2+t+1$.

Worked Example: A Matrix Subspace

In the space of real $2\times2$ matrices, use the Frobenius inner product $\langle\mathbf{A},\mathbf{B}\rangle=\sum_{i=1}^{2}\sum_{j=1}^{2}a_{ij}b_{ij}$. Let $W$ be spanned by $\mathbf{Q}_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}$ and $\mathbf{Q}_2=\frac{1}{\sqrt2}\begin{bmatrix}0&1\\1&0\end{bmatrix}$. Find the components of $\mathbf{X}=\begin{bmatrix}2&3\\1&4\end{bmatrix}$.

The two spanning matrices are orthonormal: $\langle\mathbf{Q}_1,\mathbf{Q}_1\rangle=1$, $\langle\mathbf{Q}_2,\mathbf{Q}_2\rangle=(1+1)/2=1$, and $\langle\mathbf{Q}_1,\mathbf{Q}_2\rangle=0$. The coefficients are $\langle\mathbf{X},\mathbf{Q}_1\rangle=2$ and $\langle\mathbf{X},\mathbf{Q}_2\rangle=(3+1)/\sqrt2=2\sqrt2$. Hence

\mathbf{X}_W =2\mathbf{Q}_1+2\sqrt2\,\mathbf{Q}_2 =\begin{bmatrix}2&2\\2&0\end{bmatrix}, \qquad \mathbf{X}_{W^\perp} =\mathbf{X}-\mathbf{X}_W =\begin{bmatrix}0&1\\-1&4\end{bmatrix}.

Check: The inner products of the remainder with $\mathbf{Q}_1$ and $\mathbf{Q}_2$ are $0$ and $(1-1)/\sqrt2=0$, respectively. Adding the displayed matrices gives $\mathbf{X}$.

A Consequence: The Components Meet at Right Angles

Since $\mathbf{x}_W\in W$ and $\mathbf{x}_{W^\perp}\in W^\perp$, the two components are orthogonal. The norm of the original vector therefore splits into the sum of the squared norms of its components.

Theorem (Pythagorean identity for the components): Under the hypotheses of the orthonormal-basis formula, $\|\mathbf{x}\|^2=\|\mathbf{x}_W\|^2+\|\mathbf{x}_{W^\perp}\|^2$.

Proof. Write $\mathbf{x}=\mathbf{x}_W+\mathbf{x}_{W^\perp}$. Expanding the squared norm using the inner product gives $\|\mathbf{x}\|^2=\langle\mathbf{x}_W,\mathbf{x}_W\rangle+\langle\mathbf{x}_W,\mathbf{x}_{W^\perp}\rangle+\langle\mathbf{x}_{W^\perp},\mathbf{x}_W\rangle+\langle\mathbf{x}_{W^\perp},\mathbf{x}_{W^\perp}\rangle$. The middle terms are zero because the components are orthogonal. The first and last terms are their squared norms, proving the identity. $\square$

For an orthonormal basis of $W$, the squared norm of the component in $W$ can also be read from its coefficients: $\|\mathbf{x}_W\|^2=\sum_{i=1}^{k}\langle\mathbf{x},\mathbf{q}_i\rangle^2$. To verify this, expand $\langle\mathbf{x}_W,\mathbf{x}_W\rangle$; every cross term with distinct basis vectors vanishes, and each diagonal term is the square of its coefficient. This gives a useful numerical check alongside verifying that the remainder is orthogonal.

Common Pitfall: Forgetting to Normalize

For $\mathbf{x}=(2,0)^\top$ and $W=\operatorname{span}\{(1,1)^\top\}$, a tempting but incorrect calculation is to use $(1,1)^\top$ directly as though it were a unit basis vector. That gives the claimed component $2(1,1)^\top=(2,2)^\top$, using the inner product $(2,0)\cdot(1,1)=2$. But the remainder would be $(0,-2)^\top$, and $(0,-2)\cdot(1,1)=-2\ne0$. So this is not an orthogonal decomposition.

The unit vector in the direction of $(1,1)^\top$ is $\mathbf{q}=(1,1)^\top/\sqrt2$. Its coefficient is $\mathbf{x}\cdot\mathbf{q}=\sqrt2$, so the correct component is $\sqrt2\,\mathbf{q}=(1,1)^\top$, and the orthogonal-complement component is $(1,-1)^\top$. The check $(1,-1)\cdot(1,1)=0$ confirms the correction. The formula requires an orthonormal basis, not merely a basis of the subspace.

Check Your Understanding

For each proposed decomposition, identify the component in the subspace and verify that the other component is orthogonal to it.

In $\mathbb{R}^2$, let $W$ have orthonormal basis $\frac{1}{\sqrt2}(1,1)^\top$, and take $\mathbf{x}=(3,1)^\top$. Find both components.
In $P_2$ with $\langle p,q\rangle=\int_{-1}^{1}p(t)q(t)\,dt$, decompose $t^2$ relative to $W=\operatorname{span}\{1,t\}$.
For the matrix inner product in the worked example, find the two components of $\begin{bmatrix}1&2\\2&0\end{bmatrix}$ relative to the same $W$.
Explain why subtracting the component in $W$ produces a vector in $W^\perp$, using the orthonormal-basis formula.
State which part of the formula fails if the vectors spanning $W$ are not normalized or are not mutually orthogonal.

Finding Orthogonal Decomposition Components

What You'll Learn

Computing the Two Components

Why the Formula Gives the Components

Worked Components in Three Settings