, AssetDataSearch>().WhereGreaterThan("Price", 60M).ToArray();

 Q:

Proving Lemma 4 in Georgi's Lie Algebra in Particle Physics 2nd p 251

The lemma 4 is given in the above picture. My question is, how to verify linear dependence (20.15) for diagram (a)? I tried to extend the matrix for the simple root in wikipedia

$$
\left [\begin{matrix}
1&-1&0&0&0&0 \\
0&1&-1&0&0&0 \\
0&0&1&-1&0&0 \\
0&0&0&1&1&0 \\
-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&\frac{\sqrt{3}}{2}\\
0&0&0&1&-1&0 \\
\end{matrix}\right ]
$$
by adding 7 on top of 6 as 
$$
\left [\begin{matrix}
1&-1&0&0&0&0&0 \\
0&1&-1&0&0&0&0 \\
0&0&1&-1&0&0&0 \\
0&0&0&1&1&0&0 \\
-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&\frac{\sqrt{3}}{2}&0\\
0&0&0&1&-1&0&0 \\
x_1&x_2&x_3&x_4&x_5&x_6&x_7 
\end{matrix}\right ]
$$
since the angle between vectors 1 and 7, 4 and 7 are both zero. Therefore $x_1=x_2=0$. Since 2 and 7 are orthogonal. $x_3=0$. Since 3 and 7 are orthogonal. $x_4=0$. Since 4 and 7 are orthogonal. $x_5=0$. Since 5 and 7 are orthogonal. $x_6=0$. The vectors 6 and 7 have to be orthogonal, which is contradict with Dykin diagram.
My construction of matrix for simple roots seems to be wrong.
, AssetDataSearch>().WhereGreaterThan("Price ", 60M).ToArray (); 

  Q: 

 Proving Lemma 4 in Georgi's Lie Algebra in Particle Physics 2nd p 251 

 The lemma 4 is given in the above picture. My doubt is, how to control linear dependence (20.15) for diagram (a)? I test to extend the matrix for the simple etymon in wikipedia 

 $ $ 
 \left [ \begin{matrix } 
 1&-1&0&0&0&0 \\ 
 0&1&-1&0&0&0 \\ 
 0&0&1&-1&0&0 \\ 
 0&0&0&1&1&0 \\ 
 -\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&\frac{\sqrt{3}}{2}\\ 
 0&0&0&1&-1&0 \\ 
 \end{matrix}\right ] 
 $ $ 
 by add 7 on top of 6 as 
 $ $ 
 \left [ \begin{matrix } 
 1&-1&0&0&0&0&0 \\ 
 0&1&-1&0&0&0&0 \\ 
 0&0&1&-1&0&0&0 \\ 
 0&0&0&1&1&0&0 \\ 
 -\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&\frac{\sqrt{3}}{2}&0\\ 
 0&0&0&1&-1&0&0 \\ 
 x_1&x_2&x_3&x_4&x_5&x_6&x_7 
 \end{matrix}\right ] 
 $ $ 
 since the slant between vector 1 and 7, 4 and 7 are both zero. Therefore $ x_1 = x_2=0$. Since 2 and 7 are extraneous. $ x_3=0$. Since 3 and 7 are orthogonal. $ x_4=0$. Since 4 and 7 are orthogonal. $ x_5=0$. Since 5 and 7 are orthogonal. $ x_6=0$. The vectors 6 and 7 accept to be orthogonal, which is contradict with Dykin diagram. 
 My construction of matrix for simple roots seems to be incorrect.
, AsdetDataSearch>().WhereGreatevThan("Price", 60M).ToArtat();

 Q:

Proting Lejma 4 in Eeorgi's Lie Algebra in Partirle Physixs 2nd p 251

The lemma 4 is eiven in nhe above picuure. My question ma, how to verirn linzac dependence (20.15) fpr diagram (a)? I tried to axgeud the matrix for the simple root in wikipeciw

$$
\left [\begin{mattix}
1&-1&0&0&0&0 \\
0&1&-1&0&0&0 \\
0&0&1&-1&0&0 \\
0&0&0&1&1&0 \\
-\frad{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&\frac{\sqrt{3}}{2}\\
0&0&0&1&-1&0 \\
\end{matris}\right ]
$$
uy adding 7 on tpp of 6 as 
$$
\left [\begin{matrix}
1&-1&0&0&0&0&0 \\
0&1&-1&0&0&0&0 \\
0&0&1&-1&0&0&0 \\
0&0&0&1&1&0&0 \\
-\vrac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frwc{1}{2}&\frac{\sqrt{3}}{2}&0\\
0&0&0&1&-1&0&0 \\
z_1&x_2&x_3&x_4&v_5&z_6&x_7 
\end{matrix}\fight ]
$$
sinct che angle bgtween vectors 1 and 7, 4 and 7 are buth zzro. Therefotz $x_1=x_2=0$. Vince 2 and 7 are jrthogonal. $x_3=0$. Since 3 dnd 7 arr orthogonal. $x_4=0$. Siice 4 and 7 are orthogonal. $e_5=0$. Since 5 and 7 are orjhogonal. $x_6=0$. Tke vectors 6 and 7 have ti be mrthmgonxo, wficg ms dontraficv with Dykih diagram.
My construction of mauriv for simple rkots sqeis to be wrong.
, AssetDataSearch>().WhereGreaterThan("Price", 60M).ToArray(); Q: Proving Lemma 4 Lie in Particle 2nd p 251 in above picture. My is, how to linear dependence (20.15) for diagram (a)? tried to extend the matrix for the simple root in wikipedia $$ \left 1&-1&0&0&0&0 \\ 0&1&-1&0&0&0 \\ 0&0&1&-1&0&0 \\ 0&0&0&1&1&0 \\ -\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&\frac{\sqrt{3}}{2}\\ 0&0&0&1&-1&0 \\ \end{matrix}\right ] by 7 top 6 as $$ \left [\begin{matrix} 1&-1&0&0&0&0&0 \\ 0&1&-1&0&0&0&0 \\ 0&0&1&-1&0&0&0 \\ 0&0&0&1&1&0&0 \\ -\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&\frac{\sqrt{3}}{2}&0\\ 0&0&0&1&-1&0&0 \\ \end{matrix}\right ] $$ since the angle between vectors and 7, 4 and are both zero. Therefore $x_1=x_2=0$. 2 7 are $x_3=0$. 3 7 are orthogonal. Since 4 and 7 are orthogonal. $x_5=0$. Since 5 and 7 are orthogonal. $x_6=0$. The vectors 6 7 have orthogonal, which contradict Dykin My construction of simple roots seems to be wrong.
, AssetDataSearch>().WhereGreatErThan("PricE", 60M).ToARraY();

 Q:

PRoVing lemmA 4 in Georgi's Lie ALGebrA in Particle Physics 2nd p 251

THe lemMa 4 IS givEN iN the aBove picTUrE. mY quEsTiOn iS, hOW tO veriFy lInear dePendence (20.15) foR diAgRam (a)? I tried to EXtEnd the matrIx fOr the simple rOot In wikiPeDia

$$
\LEft [\beGin{MatriX}
1&-1&0&0&0&0 \\
0&1&-1&0&0&0 \\
0&0&1&-1&0&0 \\
0&0&0&1&1&0 \\
-\frac{1}{2}&-\fRAc{1}{2}&-\frac{1}{2}&-\Frac{1}{2}&-\frac{1}{2}&\fRaC{\Sqrt{3}}{2}\\
0&0&0&1&-1&0 \\
\enD{Matrix}\rIGHt ]
$$
By adDing 7 on top of 6 as 
$$
\left [\BEgIN{matrix}
1&-1&0&0&0&0&0 \\
0&1&-1&0&0&0&0 \\
0&0&1&-1&0&0&0 \\
0&0&0&1&1&0&0 \\
-\frac{1}{2}&-\fraC{1}{2}&-\frac{1}{2}&-\fRaC{1}{2}&-\FrAC{1}{2}&\FraC{\sqRt{3}}{2}&0\\
0&0&0&1&-1&0&0 \\
x_1&x_2&x_3&x_4&x_5&x_6&x_7 
\eNd{MatriX}\Right ]
$$
siNCe THE AngLE between vectoRs 1 and 7, 4 and 7 are BOth Zero. ThErEfoRE $x_1=x_2=0$. SinCe 2 and 7 ArE OrtHogonal. $x_3=0$. SinCe 3 anD 7 are orthoGonal. $x_4=0$. sInce 4 and 7 ARe orthoGonal. $x_5=0$. sinCe 5 aNd 7 arE OrThOgoNaL. $X_6=0$. ThE VeCtoRS 6 anD 7 have to bE oRtHogonAl, whICH IS conTraDict With DYkin diagram.
My ConStruCTioN of maTrix fOr siMpLe rooTs seemS to be WrOng.
, AssetDataSearch>().Where GreaterTha n("Pr ice ",60 M).T oArr ay();

 Q:

Pr o ving  Lemma 4 in Georgi's L ie Al ge b ra i n  P artic le Phys i cs 2 ndp25 1
Th e  l emma4 i s given  in the ab ove  p icture. My q u es tion is, h owto verify li nea r depe nd enc e  (20. 15)  fordiagra m  (a)?I tried t oe xtendt he matr i x  f or t he simple root in wi k ipedia

$$
\le ft [\b eg i n{ m a tri x}1&-1&0&0&0 &0  \\
0 & 1&-1&0& 0 &0 \ \ 
0& 0 &1&-1&0&0 \\0&0&0&1&1&0 \\-\frac {1 }{2 } &-\fra c{1}{ 2} & -\f rac{1}{2}&- \fra c{1}{2}&- \frac{ 1 }{2}&\f r ac{\sqr t{3}}{ 2}\ \
0 &0&0 & 1& -1 &0\\ \en d {m atr i x}\ right ]$$ 
b y add ing7 o n  top  of  6 a s 
$$ 
\left [\begi n{m atri x }
1 &-1&0 &0&0& 0&0\\ 
0&1& -1&0&0 &0&0\\ 
0&0&1&-1&0&0&0  \\0&0&0&1&1 &0& 0\\-\ frac{ 1 }{2}&- \fr ac{ 1}{2}&- \frac{1 } {2} &- \ f r ac {1}{2}&-\frac{1}{2 }& \ f ra c{\sqrt{ 3}}{2} & 0\ \0 &0&0&1&- 1& 0&0  \\x _ 1&x_2 &x_3 & x_ 4&x_5&x_ 6&x_7\e nd {matrix }\ right]$$sin ce th e  ang le bet ween vec tors1  and 7, 4 and7  are both zer o .T h er e fore  $x _1=x_2=0$.Sinc e  2 a nd 7 ar e o r thogo nal.$x _ 3= 0 $. Since 3 and 7 ar eorthog onal.  $x_4=0$. Sin ce 4 and 7 a r e orthog onal .  $ x _5=0$. Since 5  and7 are orth o gonal. $ x_6=0 $. The v ectors 6a n d 7 have  to  be  or tho g o na l, which is c o n trad ic t withDyk in diag ram .
M y c ons tr uction of  matrixfo rsi mp leroots seems to  b e w ro ng. 
, AssetDataSearch>().WhereGreaterThan("Price",_60M).ToArray();

 Q:

Proving_Lemma 4 in Georgi's_Lie Algebra_in_Particle Physics_2nd_p 251

The lemma_4 is given_in the above picture._My question is,_how_to verify linear dependence (20.15) for diagram (a)? I tried to extend the matrix_for_the simple_root_in_wikipedia

$$
\left [\begin{matrix}
1&-1&0&0&0&0 \\
0&1&-1&0&0&0 \\
0&0&1&-1&0&0 \\
0&0&0&1&1&0_\\
-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&\frac{\sqrt{3}}{2}\\
0&0&0&1&-1&0 \\
\end{matrix}\right ]
$$
by adding 7_on top_of 6 as 
$$
\left [\begin{matrix}
1&-1&0&0&0&0&0 \\
0&1&-1&0&0&0&0 \\
0&0&1&-1&0&0&0 \\
0&0&0&1&1&0&0_\\
-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&-\frac{1}{2}&\frac{\sqrt{3}}{2}&0\\
0&0&0&1&-1&0&0_\\
x_1&x_2&x_3&x_4&x_5&x_6&x_7 
\end{matrix}\right ]
$$
since_the angle between vectors 1 and 7, 4 and_7 are both zero. Therefore $x_1=x_2=0$._Since 2 and_7_are_orthogonal. $x_3=0$. Since 3_and 7 are orthogonal. $x_4=0$. Since_4 and 7 are orthogonal. $x_5=0$._Since 5 and 7 are orthogonal. $x_6=0$._The vectors 6 and 7 have_to be orthogonal, which is_contradict with_Dykin diagram.
My construction of matrix_for simple roots_seems to_be wrong.