Recipe

Find the Riesz vector for a linear functional

Use when: you have a linear functional \varphi on a finite-dimensional inner product space and need the single vector \mathbf w with \varphi(\mathbf v)=\langle\mathbf v,\mathbf w\rangle for every \mathbf v

Goal: the unique vector \mathbf w representing \varphi, confirmed to reproduce \varphi on every input

Taught in The Riesz representation theorem

1. 1

   Expand \varphi(\mathbf v) as c_1v_1+\cdots+c_nv_n, where v_i are \mathbf v's coordinates relative to an orthonormal basis \{\mathbf e_i\}.

   Example

   For \varphi(x,y)=3x-y on \mathbb R^2 with the standard dot product, the standard basis is already orthonormal, so c_1=3 and c_2=-1.

   • the basis already in the problem is orthonormal under the given inner product — read the coefficients c_i straight off \varphi's formula.
   • the given basis isn't orthonormal under this inner product — run Gram-Schmidt on it first, then re-expand \varphi against the orthonormal result.

   Why: the theorem's proof needs \varphi's values on the basis vectors themselves, and coordinates against an orthonormal basis are exactly the fastest way to read those off.

2. 2

   Set w_i=\overline{c_i} for each coordinate, then assemble \mathbf w=\sum_i w_i\mathbf e_i.

   Example

   On \mathbb C^2 with \varphi(z_1,z_2)=(1+i)z_1+2z_2, conjugating gives \mathbf w=(1-i,2); on \mathbb R^2's \varphi(x,y)=3x-y, conjugation is a no-op, giving \mathbf w=(3,-1).

   • the space is real — conjugation does nothing, so w_i=c_i.
   • the space is complex — conjugate each c_i before assigning it to w_i.

   Why: the inner product \langle\mathbf v,\mathbf w\rangle=\sum_iv_i\overline{w_i} is conjugate-linear in its second slot, so matching it to \varphi(\mathbf v)=\sum_ic_iv_i forces exactly this conjugation.

3. 3

   Pick a test vector (a basis vector is enough), compute \langle\mathbf v,\mathbf w\rangle directly, and confirm it matches \varphi(\mathbf v).

   Example

   With \mathbf w=(1-i,2) and \mathbf v=(1,0): \langle\mathbf v,\mathbf w\rangle=1\cdot\overline{(1-i)}=1+i, matching \varphi(1,0)=(1+i)(1)+2(0)=1+i.

   Why: this catches a conjugation slip or a coordinate mismatch before the vector is used anywhere else, such as in building an adjoint.

4. 4

   If asked whether a DIFFERENT vector could also represent \varphi, compute \langle\mathbf v,\mathbf w_{\text{candidate}}\rangle for one specific \mathbf v and compare it against \varphi(\mathbf v) on that same input.

   Example

   For candidate \mathbf w_2=(4,2) against \varphi(x,y)=3x-y: \langle(1,0),\mathbf w_2\rangle=4, but \varphi(1,0)=3, so \mathbf w_2 fails.

   Why: a single mismatched input is already enough to rule out a candidate, since the theorem guarantees only one vector reproduces \varphi on every input.