# Sign considering in question

1. Oct 28, 2011

### nhrock3

1)
$$q(x_{1,}x_{2,}x_{3})=$$
$$x_{1}^{2}+5x_{2}^{2}+26x_{3}^{2}+2x_{1}x_{2}+10x_{1}x_{3}+6x_{2}x_{3}$$

$$V=\{x=(x1,x2,x3)\in R^{3}:q(x)=0\}$$

check if V is a subspace of $$R^{3}$$ 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)
$$A=\left(\begin{array}{ccc}1 & 1 & 5\\1 & 5 & 3\\5 & 3 & 26\end{array}\right)=>A=\left(\begin{array}{ccc}1 & 0 & 0\\0 & 4 & 0\\0 & 0 & 0\end{array}\right)$$
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.
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
?

Last edited: Oct 28, 2011
2. Oct 29, 2011

### susskind_leon

I would say if the diagonal A' and A are related by the change of basis matrix T such that
T^-1 A' T= A, then you need to apply this transformation to the vector (0, 0, 1), so your basis would be {T^-1 (0, 0, 1) T}
That's why you diagonalized your matrix in the first place. Suppose you are acting your function on (x',y',z'), then your result is x'^2+4y'^2, so anything that has only z will be mapped to zero.