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[tex]q(x_{1,}x_{2,}x_{3})=[/tex]

[tex]x_{1}^{2}+5x_{2}^{2}+26x_{3}^{2}+2x_{1}x_{2}+10x_{1}x_{3}+6x_{2}x_{3}[/tex]

[tex]V=\{x=(x1,x2,x3)\in R^{3}:q(x)=0\}[/tex]

check if V is a subspace of [tex]R^{3}[/tex] and

find the basis of V?

how i tried:

i diagonolized it the representative by rows and columns and i see

that

q is semi positive(having all posotive numbers except one zero)

[tex]A=\left(\begin{array}{ccc}1 & 1 & 5\\1 & 5 & 3\\5 & 3 & 26\end{array}\right)=>A=\left(\begin{array}{ccc}1 & 0 & 0\\0 & 4 & 0\\0 & 0 & 0\end{array}\right)[/tex]

my prof said that because q(x) is semi positive then V is subspace

but why??

and he didnt even looked that q(x)=0 it could be q(x)>=0

he made his desition without considering if q(x)=0 or q(x)>=0.

and when i asked him about the basis of V

i looked at it when q(x)=0 as kernel of A

but he said it worng and didnt say why

?

so how to find the basis

?

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