Masteries · Jason GlowneyArithmoí Book 1

Masteries

1.1.23 Geometric Visualization of the Inner Product

To vi­su­al­ize what an inner prod­uct of two 3×33 \times 3 ma­tri­ces ∈U(3)\in U(3) might look like, we use the ex­am­ples for G(X1),G(X2)G\left( X_{1} \right),G\left( X_{2} \right) ro­tat­ing the iden­ti­ty matrix I3I_{3}, with en­tries (A,B,C)(A,B,C) along its di­ag­o­nal, rep­re­sent­ing the ver­tices of the equi­lat­er­al tri­an­gle. Set­ting the ro­ta­tion angles ∠X1=(iπ3), ∠X2=(−iπ4)\angle X_{1} = \left( \frac{i\pi}{3} \right),\ \angle X_{2} = \left( - \frac{i\pi}{4} \right), this yields the fol­low­ing

G(X1(iπ3))ϵ(A,B,C)=(A(i32+12)000B(12)C(i32)0B(i32)C(12))=Xˆ1,\begin{array}{r} G\left( X_{1}\left( \frac{i\pi}{3} \right) \right)\epsilon(A,B,C) = \begin{pmatrix} A\left( \frac{i\sqrt{3}}{2} + \frac{1}{2} \right) & 0 & 0 \\ 0 & B\left( \frac{1}{2} \right) & C\left( \frac{i\sqrt{3}}{2} \right) \\ 0 & B\left( \frac{i\sqrt{3}}{2} \right) & C\left( \frac{1}{2} \right) \end{pmatrix} = {\widehat{X}}_{1}, \end{array}
(1.1.168)

and

G(X2(−iπ4))ϵ(A,B,C)=(A(12)0−C(i2)0B(12−i2)0−A(i2)0C(12))=Xˆ2.\begin{array}{r} G\left( X_{2}\left( - \frac{i\pi}{4} \right) \right)\epsilon(A,B,C) = \begin{pmatrix} A\left( \frac{1}{\sqrt{2}} \right) & 0 & - C\left( \frac{i}{\sqrt{2}} \right) \\ 0 & B\left( \frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}} \right) & 0 \\ - A\left( \frac{i}{\sqrt{2}} \right) & 0 & C\left( \frac{1}{\sqrt{2}} \right) \end{pmatrix} = {\widehat{X}}_{2}. \end{array}
(1.1.169)

Taking the inner prod­uct ⟨Xˆ1,Xˆ2⟩\left\langle {\widehat{X}}_{1},{\widehat{X}}_{2} \right\rangle

⟨Xˆ1,Xˆ2⟩=tr(Xˆ1∗Xˆ2)=A2(12−i32)2+B2(12−i2)2+C222,\begin{array}{r} \left\langle {\widehat{X}}_{1},{\widehat{X}}_{2} \right\rangle = tr\left( {\widehat{X}}_{1}^{*}{\widehat{X}}_{2} \right) = \frac{A^{2}\left( \frac{1}{2} - \frac{i\sqrt{3}}{2} \right)}{\sqrt{2}} + \frac{B^{2}\left( \frac{1}{2} - \frac{i}{2} \right)}{\sqrt{2}} + \frac{C^{2}}{2\sqrt{2}}, \end{array}
(1.1.170)

we find com­plex-valued A,BA,B vertex com­po­nents and a real-valued com­po­nent for the CC vertex. This is con­sis­tent with the effect of ro­tat­ing in the X1/X2X_{1}/X_{2}-plane, for the A,BA,B ver­tices, having been on the com­plex axes of ro­ta­tion for X1,X2X_{1},X_{2} re­spec­tive­ly, while the X3X_{3}-plane did not act as a com­plex axis of ro­ta­tion on the CC vertex. In Figure 1.1.15, we have plot­ted both ro­ta­tions of the equi­lat­er­al tri­an­gle iden­ti­ty matrix in the X1/X2X_{1}/X_{2}-plane.

Figure
Figure 1.1.15 – The inner prod­uct ⟨Xˆ1,Xˆ2⟩,\left\langle {\widehat{X}}_{1},{\widehat{X}}_{2} \right\rangle, (Xˆ1=(A1,B1,C1),Xˆ2=(A2,B2,C2))\left( {\widehat{X}}_{1} = \left( A_{1},B_{1},C_{1} \right),{\widehat{X}}_{2} = \left( A_{2},B_{2},C_{2} \right) \right).

The equi­lat­er­al tri­an­gle Xˆ2=(A2,B2,C2){\widehat{X}}_{2} = \left( A_{2},B_{2},C_{2} \right) looks larger than Xˆ1=(A1,B1,C1){\widehat{X}}_{1} = \left( A_{1},B_{1},C_{1} \right), but from the stand­point of the Hilbert-Schmidt norms, they are the same. This can be seen by first con­sid­er­ing ‖Xˆ1‖\left\| {\widehat{X}}_{1} \right\|

‖Xˆ1‖=(∑j=1m∑k=1n|xj,k|2)12=(|A|2+|B2|2+|B32|2+|C32|2+|C2|2)12=A2+B2+C2,\begin{array}{r} \left\| {\widehat{X}}_{1} \right\| = \left( \sum_{j = 1}^{m}{\sum_{k = 1}^{n}\left| x_{j,k} \right|^{2}} \right)^{\frac{1}{2}} = \left( |A|^{2} + \left| \frac{B}{2} \right|^{2} + \left| \frac{B\sqrt{3}}{2} \right|^{2} + \left| \frac{C\sqrt{3}}{2} \right|^{2} + \left| \frac{C}{2} \right|^{2} \right)^{\frac{1}{2}} = \sqrt{A^{2} + B^{2} + C^{2},} \end{array}
(1.1.171)

then for ‖Xˆ2‖\left\| {\widehat{X}}_{2} \right\|,

‖Xˆ2‖=(∑j=1m∑k=1n|xj,k|2)12=(|A2|2+|−Ai2|2+|B(1−i)2|2+|−Ci2|2+|C2|2)12=A2+B2+C2,\begin{array}{r} \left\| {\widehat{X}}_{2} \right\| = \left( \sum_{j = 1}^{m}{\sum_{k = 1}^{n}\left| x_{j,k} \right|^{2}} \right)^{\frac{1}{2}} = \left( \left| \frac{A}{\sqrt{2}} \right|^{2} + {\left| \frac{- Ai}{\sqrt{2}} \right|^{2} + \left| \frac{B(1 - i)}{\sqrt{2}} \right|}^{2} + \left| \frac{- Ci}{\sqrt{2}} \right|^{2} + \left| \frac{C}{\sqrt{2}} \right|^{2} \right)^{\frac{1}{2}} = \sqrt{A^{2} + B^{2} + C^{2}}, \end{array}
(1.1.172)

re­call­ing that A,B,C∈ℝ+A,B,C \in \mathbb{R}^{+}, and that A=B=CA = B = C rep­re­sent scalar mag­ni­tudes. Closer in­spec­tion also re­veals that the inner prod­uct ⟨Xˆ1,Xˆ2⟩=tr(Xˆ1∗Xˆ2)\left\langle {\widehat{X}}_{1},{\widehat{X}}_{2} \right\rangle = tr\left( {\widehat{X}}_{1}^{*}{\widehat{X}}_{2} \right), is es­sen­tial­ly the sum of the dot prod­ucts of the matched column vec­tors from Xˆ1{\widehat{X}}_{1} and Xˆ2{\widehat{X}}_{2}, as

∑j=13Xˆ11,jXˆ2j,1+∑j=13Xˆ12,jXˆ2j,2+∑j=13Xˆ13,jXˆ2j,3=⟨Xˆ1,Xˆ2⟩=tr(Xˆ1∗Xˆ2).\begin{array}{r} \sum_{j = 1}^{3}{\overline{{{\widehat{X}}_{1}}{}_{1,j}}{{\widehat{X}}_{2}}{}_{j,1} +}\sum_{j = 1}^{3}{\overline{{{\widehat{X}}_{1}}{}_{2,j}}{{\widehat{X}}_{2}}{}_{j,2}} + \sum_{j = 1}^{3}{\overline{{{\widehat{X}}_{1}}{}_{3,j}}{{\widehat{X}}_{2}}{}_{j,3} =}\left\langle {\widehat{X}}_{1},{\widehat{X}}_{2} \right\rangle = tr\left( {\widehat{X}}_{1}^{*}{\widehat{X}}_{2} \right). \end{array}
(1.1.173)

In this sec­tion, we have in­tro­duced the U(3)U(3) sym­me­try group as the in­ter­con­nec­tion of three com­plex unit cir­cles which form a com­plex unit sphere. Ro­ta­tions in U(3)U(3) are the result of group gen­er­a­tors acting on the equi­lat­er­al tri­an­gle as uni­tary linear trans­for­ma­tions. Hilbert spaces and their inner-prod­uct map­pings in­tro­duce com­plete vector spaces, a de­par­ture from the con­fines of Eu­clid­ean space.