Sign considering in question

  • Thread starter nhrock3
  • Start date
  • #1
415
0
1)
[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
?
 
Last edited:

Answers and Replies

  • #2
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.
 

Related Threads on Sign considering in question

  • Last Post
Replies
6
Views
3K
Replies
4
Views
1K
  • Last Post
Replies
14
Views
1K
  • Last Post
Replies
20
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
830
Replies
4
Views
3K
  • Last Post
Replies
2
Views
982
Replies
1
Views
1K
Top