4.30 Inner Product

Let \(k=\R\) or \(\mathbb{C}\) and \(V=\R^{2}(k=\R)\)

\(v=(x_0,y_0)=x_0e_1+y_0e_2\) and \(d^2=x_0^2+y_0^2\) where \(d=\sqrt{x_0^2+y_0^2}\)

Definition of Inner Product

An inner product k-vector space \(V\) is a k-vector space \(V\) together with a map \(\lang,\rang:V\times V\to k\) such that:

\(\forall v,w,z\in V,\forall \lambda\in k\)
- \(\lang v+w,z\rang=\lang v,z\rang\underbrace{+}_{k}\lang w,z\rang\)
- \(\lang \underbrace{\lambda v}_{\text{action of k in V}},z\rang=\lambda \underbrace{\cdot}_{\text{product in k}} \lang v,z\rang\)
Remark: We are saying that for any fixed \(z\in V\), \(\lang \quad,z\rang:V\to k\) is k-linear and \(\lang\quad,z\rang(v)=\lang v,z\rang\) is k-linear
\(\forall v,w\in V ,\lang v,w\rang = \overline{\lang w,v\rang}\) (Remark: If \(k=\R\), \(\langle v,w\rangle=\langle w,v\rangle\))

Consequence of 1 and 2:
- \(\forall v,w,z\in V,\forall \lambda\in k\), we have \(\lang v,w+z\rang \overset{(2)}{=}\overline{\lang w+z,v\rang}\overset{(1)}{=}\overline{\lang w,v\rang+\lang z,v\rang}=\overline{\lang w,v\rang}+\overline{\lang z,v\rang}=\overline{\overline{\lang v,w\rang}}+\overline{\overline{\lang v,z\rang}}=\lang v,w\rang+\lang v,z\rang\)
- \(\lang v,\lambda w\rang\overset{(2)}{=}\overline{\lang \lambda w,v\rang}\overset{(1)}{=}\overline{\lambda\lang w,v\rang}=\bar \lambda\cdot \overline{\lang w,v\rang}=\bar \lambda\cdot \overline{\overline{\lang v,w\rang}}=\bar \lambda\cdot \lang v,w\rang\)
We can see that inner product is linear on first position, like half linear
\(\forall v\in V,\lang v,v\rang \geq 0\Rightarrow \begin{cases} \lang v,v\rang \in\R \\ \lang v,v\rang \geq 0 \end{cases}\) and \(\lang v,v\rang =0\iff v=0_V\)

Example

\(k=\R,V=\R^2\), define \(\langle(x_{1},..,x_{n})=v,(y_{1},...,y_{n})=w\rangle=\sum_{i=1}^{n}x_{i}y_{i}\)

Verification: \(z=(z_1,...,z_n)\), then
1. \(\langle v+z, w \rangle = \langle (x_1+z_1, x_2+z_2, \dots, x_n+z_n), (y_1, \dots, y_n) \rangle = \sum_{i=1}^n (x_i+z_i) y_i = \sum_{i=1}^n x_i y_i + \sum_{i=1}^n z_i y_i = \langle v, w \rangle + \langle z, w \rangle\)
  
  Given \(\lambda \in \mathbb{R}\), \(\langle \lambda v, w \rangle = \langle (\lambda x_1, \lambda x_2, \dots, \lambda x_n), (y_1, \dots, y_n) \rangle = \sum_{i=1}^n (\lambda x_i) y_i = \lambda \sum_{i=1}^n x_i y_i = \lambda \langle v, w \rangle\) 2. \(\langle v, w \rangle = \sum_{i=1}^n x_i y_i = \sum_{i=1}^n y_i x_i = \langle w, v \rangle\) 3. \(\langle v, v \rangle = \sum_{i=1}^n x_i^2 \ge 0\) and \(\langle v, v \rangle = 0 \iff \sum_{i=1}^n x_i^2 = 0 \iff x_i^2 = 0\) for all \(i\) \(\iff x_i = 0\) for all \(i\) \(\iff v = (x_1, \dots, x_n) = (0, \dots, 0)\)
\(k = \mathbb{C}\), \(V = \mathbb{C}^n\), \(\langle \cdot, \cdot \rangle: V \times V \to \mathbb{C}\) and define \(\langle (x_1, \dots, x_n), (y_1, \dots, y_n) \rangle = \sum_{i=1}^n x_i \overline{y_i}\)
Verify that it is an inner product.

Let \(v=(v_1,...,v_n)\), \(w=(w_1,...,w_n)\) and \(z=(z_1,...,z_n)\)
1. \(\langle v+w,z\rangle=\langle\left(v_{1}+w_{1},\ldots,v_{n}+w_{n}\right),\left(z_{1} ,\ldots,z_{n}\right)\rangle=\sum_{i=1}^{n}\left(v_{i}+w_{i}\right)\bar{z_{i}}=\sum _{i=1}^{n}v_{i}\bar{z_{i}}+\sum_{i=1}^{n}w_{i}\bar{z_{i}}=\langle v,z\rangle+\langle w,z\rangle\)
  
  Given \(\lambda \in \mathbb{R}\), \(\langle\lambda v,w\rangle=\langle(\lambda v_{1},\lambda v_{2},\dots,\lambda v_{n} ),(w_{1},\dots,w_{n})\rangle=\sum_{i=1}^{n}(\lambda v_{i})\bar{w_{i}}=\lambda\sum _{i=1}^{n}v_{i}\bar{w_{i}}=\lambda\langle v,w\rangle\) 2. \(\forall v,w\in V,\langle v,w\rangle=\sum_{i=1}^{n}v_{i}\bar{w_{i}}=\overline{\sum_{i=1}^{n} \bar{v_{i}}w_{i}}=\overline{\langle w,v\rangle}\) 3. \(\langle v,v\rangle=\sum_{i=1}^{n}v_{i}^{2}\ge0\) and \(\langle v,v\rangle=0\;\char"27FA \;\sum_{i=1}^{n}v_{i}^{2}=0\;\char"27FA \;v_{i} ^{2}=0\) for all \(i\) \(\;\char"27FA \;v_{i}=0\) for all \(i\) \(\;\char"27FA \;v=(v_{1},\dots,v_{n})=(0,\dots,0)\)
These are called the STANDARD INNER PRODUCT in \(\mathbb{R}^n\) and \(\mathbb{C}^n\) respectively.
\(V = k^{n \times n}\), \(\langle \cdot, \cdot \rangle: V \times V \to k\), let \(A, B \in k^{n \times n}\), then \(\langle A, B \rangle \underset{DEF}{=} \operatorname{tr}(A B^*)\) where \(B^{*}\underset{DEF}{=}\overline{B^{T}}\)
\(A = (a_{ij}), B = (b_{ij}) \implies (B^*)_{ij} = \overline{b_{ji}}\)
Since \(A \cdot B^* \in k^{n \times n}\), then \((A \cdot B^*)_{ij} = \sum_{l=1}^n a_{il} (B^*)_{lj} = \sum_{l=1}^n a_{il} \overline{b_{jl}}\)
Then \(\operatorname{tr}(A \cdot B^*) = \sum_{i=1}^n (A \cdot B^*)_{ii} = \sum_{i=1}^n \left( \sum_{l=1}^n a_{il} \overline{b_{il}} \right)\)
Exercise: prove that it is an inner product.
\(V = k^{m \times 1}\), \(Q \in k^{m \times m}\), then \(\langle \cdot, \cdot \rangle: V \times V \to k\), then \(\langle X, Y \rangle \underset{DEF}{=} Y^* Q^* Q X\) where \(X \in k^{m \times 1}, Y \in k^{m \times 1}, Y^* \in k^{1 \times m}, Q^* \in k^{m \times m}, Q \in k^{m \times m}, X \in k^{m \times 1}\)
Exercise: prove that it is an inner product.
\(k = \mathbb{R}\), \(V = \mathbb{R}^2\)
\(\langle v, w \rangle \underset{DEF}{=} x_1 y_1 - x_2 y_1 - x_1 y_2 + 4 x_2 y_2\) where \(v=(x_{1},x_{2}),w=(y_{1},y_{2})\)
1. \(\langle v+z, w \rangle = \langle v, w \rangle + \langle z, w \rangle\) EASY.
2. \(\langle w, v \rangle = x_1 y_1 - y_2 x_1 - y_1 x_2 + 4 y_2 x_2 = \langle v, w \rangle\)
3. \(\langle v, v \rangle = x_1^2 - x_1 x_2 - x_1 x_2 + 4 x_2^2 = x_1^2 - 2x_1 x_2 + 4x_2^2\) \(= (x_1 - x_2)^2 + 3x_2^2 \ge 0\)
  
  Supp. that \(\langle v, v \rangle = 0\). Then \(\underbrace{(x_1 - x_2)^2}_{\ge 0} + \underbrace{3x_2^2}_{\ge 0} = 0\)\(\iff x_1 - x_2 = 0\) and \(x_2 = 0\).
  Then \(\begin{cases} x_1 - x_2 = 0 \\ x_2 = 0 \end{cases} \implies x_1 = 0 \text{ and } x_2 = 0\)\(\iff v = (x_1, x_2) = (0, 0) = 0\)
\(k=\mathbb{R}\), \(V=\{f: [0,1] \to \mathbb{R} \text{ continuous}\}\) and \(\langle \cdot, \cdot \rangle: V \times V \to \mathbb{R}\) where\(\langle f, g \rangle \underset{DEF}{=} \int_0^1 f(t) g(t) dt\)

Matrix of Inner Product

Let \(V\) be a \(k\)-v.s., \(\dim_{k} V = n\), let \(\langle \cdot, \cdot \rangle\) be an inner product. Let \(B = \{v_{1}, \dots, v_{n}\}\) basis of \(V\).

Define \(G\in M_{n\times n}(k)\) as follows: \(G_{jk} := \langle v_k, v_j \rangle\) for \(1 \le j, k \le n\). \(G\) is the matrix of \(\langle \cdot, \cdot \rangle\) in the basis \(B\).
Notation: \(G = [\langle \cdot, \cdot \rangle]_B\).

Now let \(v, w \in V\).
Let \(x \in k^{n \times 1}\) be the coordinates of \(v\) in the basis \(B\). \(v = \sum_{k=1}^{n}x_{k}v_{k}\), \(x = \begin{pmatrix} x_{1} \\ \vdots \\ x_{n} \end{pmatrix}\)
Let \(y \in k^{n \times 1}\) be the coordinates of \(w\) in the basis \(B\). \(w = \sum_{l=1}^{n}y_{l}v_{l}\), \(y = \begin{pmatrix} y_{1} \\ \vdots \\ y_{n} \end{pmatrix}\)

\(\langle v,w\rangle=\langle\sum_{k=1}^{n}x_{k}v_{k},\sum_{l=1}^{n}y_{l}v_{l}\rangle\) \(=\sum_{k=1}^{n}x_{k}\langle v_{k},\sum_{l=1}^{n}y_{l}v_{l}\rangle\) \(=\sum_{k=1}^{n}\sum_{l=1}^{n}x_{k}\overline{y_{l}}\langle v_{k},v_{l}\rangle\).

Using the definition \(G_{jk} = \langle v_k, v_j \rangle\), we have \(\langle v, w \rangle = \sum_{k=1}^{n}\sum_{l=1}^{n}x_{k}\overline{y_l}G_{lk}\)

Remark

\(G^* = \overline{G^t}\). \(G = (g_{ij})\). \((G^*)_{ij} = \overline{g_{ji}} = \overline{\langle v_j, v_i \rangle} = \langle v_i, v_j \rangle = (G)_{ij}\). So \(G^* = G\).

Definition

Given a matrix \(A \in M_m(k)\), we say that \(A\) is HERMITIAN iff \(A^* = A\).

Remark

\(k = \mathbb{R}\) and \(A \in M_m(\mathbb{R})\). Then \(A^* = \overline{A^t} = A^t\) since we are in \(\R\).
\(A\) is Hermitian iff \(A^t = A\) (\(A\) is symmetric)

Definition

An inner product \(k\)-vector space is a \(k\)-v. space \(V\) with an inner product.
Notation: \((V, \langle \cdot, \cdot \rangle)\) is a \(k\)-IPVS.

Names: \((V, \langle \cdot, \cdot \rangle)\) \(k\)-IPVS.
If \(k = \mathbb{R}\), \((V, \langle \cdot, \cdot \rangle)\) is an EUCLIDEAN SPACE.
If \(k = \mathbb{C}\), \((V, \langle \cdot, \cdot \rangle)\) is a UNITARY SPACE.

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