From Coordinates Back to a Vector
A coordinate list records how much of each basis vector is used to build a vector. If the basis is orthonormal, the list gives a direct reconstruction rule: multiply each basis vector by its corresponding coordinate and add. This tutorial focuses on using that rule in reverse: starting from a list of coordinates, construct the vector and check that the construction is correct by taking inner products with the basis vectors.
Let \(V\) be a finite-dimensional inner product space over \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). Over \(\mathbb{C}\), use the convention that the inner product is linear in its first slot and conjugate-linear in its second. An ordered basis matters: the first coordinate corresponds to the first basis vector, the second to the second, and so on.
In particular, an orthonormal basis is a basis, so it spans \(V\) and its vectors are linearly independent. Every vector therefore has a unique expression \(\mathbf{v}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\). The reconstruction problem is to start with the entries \(c_1,\dots,c_n\) and form that combination. The inner product provides a useful independent check: the entry \(c_j\) must be recovered by taking the inner product of the reconstructed vector with \(\mathbf{b}_j\).
Proof. Fix an index \(j\). Linearity in the first slot and orthonormality give \(\langle\mathbf{v},\mathbf{b}_j\rangle=\sum_{i=1}^{n}c_i\langle\mathbf{b}_i,\mathbf{b}_j\rangle=c_j\), since the inner product is \(1\) for \(i=j\) and \(0\) for every other \(i\). This proves that the constructed vector has the required inner products. For uniqueness, suppose \(\mathbf{w}\) is another vector with \(\langle\mathbf{w},\mathbf{b}_j\rangle=c_j\) for every \(j\). Then \(\langle\mathbf{v}-\mathbf{w},\mathbf{b}_j\rangle=0\) for every \(j\). Because the basis spans \(V\), \(\mathbf{v}-\mathbf{w}\) is a linear combination of the \(\mathbf{b}_j\); linearity in the second slot over \(\mathbb{R}\), and conjugate-linearity in that slot over \(\mathbb{C}\), imply that \(\mathbf{v}-\mathbf{w}\) is orthogonal to itself. Positive definiteness of the inner product then gives \(\langle\mathbf{v}-\mathbf{w},\mathbf{v}-\mathbf{w}\rangle=0\), so \(\mathbf{v}-\mathbf{w}=\mathbf{0}\) and \(\mathbf{v}=\mathbf{w}\). \(\square\)
The calculation uses the same rule over both fields because the coefficients multiply basis vectors in the first slot, where the inner product is linear. For complex spaces, conjugation belongs to the second slot; it does not mean that the coefficients in the reconstruction formula should be conjugated.
Worked Reconstructions
Worked Example: A Nonstandard Orthonormal Basis in R²
In \(\mathbb{R}^2\) with the usual dot product, use \(\mathbf{b}_1=\frac{1}{\sqrt{2}}(1,1)\) and \(\mathbf{b}_2=\frac{1}{\sqrt{2}}(1,-1)\). These vectors have squared lengths \(\frac{1+1}{2}=1\) and \(\frac{1+1}{2}=1\), and their dot product is \(\frac{1-1}{2}=0\). They form a basis: for any \((r,s)\), the coefficients \((r+s)/\sqrt{2}\) and \((r-s)/\sqrt{2}\) reconstruct \((r,s)\). Reconstruct the vector whose coordinate list is \((3,-1)\).
Multiply each basis vector by its matching coordinate and add:
Check the first coordinate by taking the dot product with \(\mathbf{b}_1\): \(\langle\mathbf{v},\mathbf{b}_1\rangle=(\sqrt{2}+2\sqrt{2})/\sqrt{2}=3\). For the second coordinate, \(\langle\mathbf{v},\mathbf{b}_2\rangle=(\sqrt{2}-2\sqrt{2})/\sqrt{2}=-1\). Both checks recover the corresponding entries of the list.
Worked Example: Complex Coordinates
In \(\mathbb{C}^2\), use \(\langle\mathbf{x},\mathbf{y}\rangle=x_1\overline{y_1}+x_2\overline{y_2}\), which is linear in the first slot. Let \(\mathbf{b}_1=(1,i)/\sqrt{2}\) and \(\mathbf{b}_2=(1,-i)/\sqrt{2}\). Each has squared norm \(1\), and \(\langle\mathbf{b}_1,\mathbf{b}_2\rangle=(1+i\cdot i)/2=0\). They are linearly independent and hence form a basis of \(\mathbb{C}^2\). Reconstruct the vector with coordinates \((1+i,2-i)\).
Use the coordinates as written, without conjugating them:
For the first check, the conjugate of the second component of \(\mathbf{b}_1\) is \(-i/\sqrt{2}\). Thus \(\langle\mathbf{v},\mathbf{b}_1\rangle=3/2+(-2-i)(-i)/2=3/2+(-1+2i)/2=1+i\). For the second check, the conjugate of the second component of \(\mathbf{b}_2\) is \(i/\sqrt{2}\). Hence \(\langle\mathbf{v},\mathbf{b}_2\rangle=3/2+(-2-i)i/2=3/2+(1-2i)/2=2-i\). The inner products recover both complex coordinates.
Worked Example: Coordinates in a Polynomial Space
Consider \(P_1\), the real vector space of polynomials of degree at most \(1\), with inner product \(\langle f,g\rangle=\int_{-1}^{1}f(t)g(t)\,dt\). Use \(\mathbf{b}_1(t)=1/\sqrt{2}\) and \(\mathbf{b}_2(t)=\sqrt{3/2}\,t\). Their squared norms are \(\int_{-1}^{1}1/2\,dt=1\) and \(\int_{-1}^{1}(3/2)t^2\,dt=(3/2)(2/3)=1\). Their inner product is \(\sqrt{3}/2\int_{-1}^{1}t\,dt=0\). Since they are two linearly independent polynomials in \(P_1\), they form a basis. Reconstruct the vector with coordinates \((2,-1)\).
Multiply and combine the basis polynomials:
To check the first coordinate, multiply \(p(t)\) by \(\mathbf{b}_1(t)\) and integrate: \(\langle p,\mathbf{b}_1\rangle=\int_{-1}^{1}(1-\frac{\sqrt{3}}{2}t)\,dt=2\). The term involving \(t\) integrates to zero because its values cancel symmetrically on \([-1,1]\). For the second coordinate, \(\langle p,\mathbf{b}_2\rangle=\int_{-1}^{1}(\sqrt{3}\,t-\frac{3}{2}t^2)\,dt=0-\frac{3}{2}\cdot\frac{2}{3}=-1\). Thus the polynomial has the requested coordinates.
What the Verification Establishes
Taking inner products with the basis vectors does more than provide a numerical check. It establishes that the reconstructed vector is the one represented by the list. If two candidate vectors give the same inner product with every basis vector, their difference is orthogonal to every vector in a spanning basis. The uniqueness argument in the theorem then forces that difference to be zero.
The order of the list is essential. If \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2)\), then \((a,b)\) means \(a\mathbf{b}_1+b\mathbf{b}_2\), not \(b\mathbf{b}_1+a\mathbf{b}_2\). The entries are scalars, while the basis vectors are the objects being scaled and added. This distinction applies whether the vectors are columns, functions, or other elements of an inner product space.
The theorem also gives a useful special case: a vector whose inner products with every basis vector are zero must be the zero vector. Indeed, the reconstruction formula with every \(c_j=0\) gives the zero vector, and uniqueness rules out any other vector with those inner products. This is a statement about a complete basis; checking inner products against only part of a basis does not in general determine the vector.
Common Mistake: Conjugating Complex Coordinates
In the complex example, a tempting but incorrect reconstruction is to conjugate each coordinate and use \((1-i)\mathbf{b}_1+(2+i)\mathbf{b}_2\). That gives \(\mathbf{v}_{\mathrm{wrong}}=(3,2-i)/\sqrt{2}\). Checking its first coordinate shows the error: \(\langle\mathbf{v}_{\mathrm{wrong}},\mathbf{b}_1\rangle=3/2+(2-i)(-i)/2=3/2+(-1-2i)/2=1-i\), not the required \(1+i\). The correction is to use the given list unchanged in the first slot: \(\mathbf{v}=(1+i)\mathbf{b}_1+(2-i)\mathbf{b}_2\). The conjugation in a complex inner product is applied to the second argument when evaluating the check; it is not applied to the supplied coefficients during reconstruction.
Check Your Understanding
For each reconstruction, keep the basis order fixed and verify the resulting vector by taking inner products with the basis vectors.
- In \(\mathbb{R}^2\), let \(\mathbf{b}_1=(1,1)/\sqrt{2}\) and \(\mathbf{b}_2=(1,-1)/\sqrt{2}\). Reconstruct the vector with coordinates \((1,2)\).
- In \(\mathbb{C}^2\) with the inner product linear in the first slot, explain why the coefficient of \(\mathbf{b}_j\) is recovered by \(\langle\mathbf{v},\mathbf{b}_j\rangle\), not by conjugating that inner product.
- For the polynomial basis in the worked example, reconstruct the vector with coordinates \((0,2)\) and check both inner products.
- Suppose two vectors have the same inner products with every vector in an orthonormal basis. Prove that the two vectors are equal.
- Why can inner products with only one vector of a two-vector basis fail to determine a vector uniquely?
