Exercise 6.8.18

Answers

As what we did in the previous exercise, we set

K (t1 t2 t3 ) = 3t12+3t 22+3t 32−2t 1t3

and find

H ( (a1 a2 a3 ) , (b1 b2 b3 ) ) = 3a1b1+3a2b2+3a3b3−a1b3−a3b1

such that H(x,x) = K(x) and H is a bilinear form. This means that

K (t1 t2 t3 ) = (t1t2t3 ) ( 3 0−1 0 3 0 −10 3 ) (t1 t2 t3 )
= (t1t2t3 ) ( 1 2 1 20 0 0 1 − 1 2 1 20 ) (400 0 2 0 003 ) ( 1 20− 1 2 1 20 1 2 0 1 0 ) (t1 t2 t3 ) .

Note that here we may diagonalize it in sence of eigenvectors. Thus we pick

β = {( 1 2,0,− 1 2),( 1 2,0, 1 2),(0,1,0).}

And take

(t1′ t2′ t3′ ) = ( 1 20− 1 2 1 20 1 2 0 1 0 ) (t1 t2 t3 ) .

Thus we have

3t12+3t 22+3t 32−2t 1t= ( t1′t 2′t 3′ ) (400 0 2 0 003 ) (t1′ t2′ t3′ )
= 4(t1′)2 + 2(t 2′)2 + 3(t 3′)2.

Hence the original equality is that

4(t1′)2 + 2(t 2′)2 + 3(t 3′)2 + l.o.t = 0,

where l.o.t means some lower order terms. Hence S is a ellipsoid.

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2011-06-27 00:00
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