An inner product of two vectors of the same size, usually denoted by \( \left\langle {\bf x} , {\bf y} \right\rangle ,\) is a generalization of the dot product if it satisfies the following properties:
\( \left\langle {\bf v} , {\bf v} \right\rangle \ge 0 , \) and equal if and only if
\( {\bf v} = {\bf 0} . \)
The fourth condition in the list above is known as the positive-definite condition. A vector space together with the inner product is called an inner product space. Every inner product space is a metric space. The metric or norm is given by
The nonzero vectors u and v of the same size are orthogonal (or perpendicular) when their inner product is zero:
\( \left\langle {\bf u} , {\bf v} \right\rangle = 0 . \) We abbreviate it as \( {\bf u} \perp {\bf v} . \)
If A is an n × npositive definite matrix and u and v are n-vectors, then we can define the weighted Euclidean inner product
In particular, if w1, w2, ... , wn are positive real numbers,
which are called weights, and if u = ( u1, u2, ... , un) and
v = ( v1, v2, ... , vn) are vectors in ℝn, then the formula
defines an inner product on \( \mathbb{R}^n , \) that is called the weighted Euclidean inner product with weights
w1, w2, ... , wn.
Example 4:
The Euclidean inner product and the weighted Euclidean inner product (when \( \left\langle {\bf u} , {\bf v} \right\rangle = \sum_{k=1}^n a_k u_k v_k , \)
for some positive numbers \( a_k , \ (k=1,2,\ldots , n \) ) are special cases of a general class
of inner products on \( \mathbb{R}^n \) called matrix inner product. Let A be an
invertible n-by-n matrix. Then the formula
defines an inner product, which is called the evaluation inner product at \( x_0 , x_1 , \ldots , x_n . \) ■
The invention of Cartesian coordinates in 1649 by René Descartes (Latinized name: Cartesius) revolutionized mathematics by providing the first systematic link between Euclidean geometry and algebra.
An inner product of two vectors of the same size, usually denoted by \( \left\langle {\bf x} , {\bf y} \right\rangle ,\) is a generalization of the dot product if it satisfies the following properties:
\( \left\langle {\bf v} , {\bf v} \right\rangle \ge 0 , \) and equal if and only if
\( {\bf v} = {\bf 0} . \)
The fourth condition in the list above is known as the positive-definite condition. A vector space together with the inner product is called an inner product space. Every inner product space is a metric space. The metric or norm is given by
The nonzero vectors u and v of the same size are orthogonal (or perpendicular) when their inner product is zero:
\( \left\langle {\bf u} , {\bf v} \right\rangle = 0 . \) We abbreviate it as \( {\bf u} \perp {\bf v} . \)
If A is an n × npositive definite matrix and u and v are n-vectors, then we can define the weighted Euclidean inner product
In particular, if w1, w2, ... , wn are positive real numbers,
which are called weights, and if u = ( u1, u2, ... , un) and
v = ( v1, v2, ... , vn) are vectors in ℝn, then the formula
defines an inner product on \( \mathbb{R}^n , \) that is called the weighted Euclidean inner product with weights
w1, w2, ... , wn.
Riesz representation theorem:
Let V be a finite dimensional vector space over the field 𝔽
(𝔽 = ℝ, ℂ) on which 〈·, ·〉 is an inner product. Let φ : V ⇾ 𝔽 be a linear functional on V. Then there exists a unique vector u ∈ V such that φ(v) = 〈u, v〉 for all v ∈ V.
Using the Gram–Schmidt orthogonalization process we can find
an orthonormal basis of V, say, {v1, v2, … , vn}. Now for an arbitrary vector v ∈ V, we have
\[
{\bf v} = \langle {\bf v}_1 , {\bf v} \rangle {\bf v}_1 + \langle {\bf v}_2 , {\bf v} \rangle {\bf v}_2 + \cdots + \langle {\bf v}_n , {\bf v} \rangle {\bf v}_n .
\]
Then it follows that
\[
\varphi ({\bf v}) = \langle {\bf v}_1 , {\bf v} \rangle \varphi ({\bf v}_1 ) + \langle {\bf v}_2 , {\bf v} \rangle \varphi ({\bf v}_2 ) + \cdots + \langle {\bf v}_n , {\bf v} \rangle \varphi ({\bf v}_n ) = \left\langle {\bf v}_1 \varphi^{\ast} ({\bf v}_1 ) + \cdots + {\bf v}_n \varphi^{\ast} ({\bf v}_n ) , {\bf v} \right\rangle
\]
Denoting by
\[
{\bf u} = {\bf v}_1 \varphi^{\ast} ({\bf v}_1 ) + \cdots + {\bf v}_n \varphi^{\ast} ({\bf v}_n ) ,
\]
the result follows.
Suppose there exist two vectors u1 and u2 in the Riesz representation theorem. Then for w = u1 − u2, we have 〈w, v〉 = 0 for all v ∈ V. But then for w ∈ V, it follows that 〈w, w〉 = ∥ w ∥² = 0, which implies that u1 = u2.