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Show that $L:= [{ (x,y,z) \in R^3 : x^2 + y^2 + z^2 −2xy−2yz + 2xz =0 }] $ is a sub vector space of $R^3$

Proof: 1) show it is not empty (0,0,0) is inside thus not empty. 2) Scalar multiplication with $\lambda \in K$ multiplied $(x_1,x_2,x_3)\in L$ is closed. thus the product is also $\in L$

Addition is however not so clear. It seems to me I cannot proof the addition part. When I take two elements of $L$ add them at try to show that the sum is also in $L$ I reach a dead end.

Could someone please prove it?

lulu
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Mad
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  • Do you mean $x^2+y^2+z^2-2xy-2yz+2xz=0$? – lulu Nov 05 '19 at 21:45
  • I already edited it, yes that was what I meant. the equation must be fulfilled for the elements of $L$ exmples are $(0,0,0), (1,2,1) , (2,2,0) $ – Mad Nov 05 '19 at 21:46
  • What happens if you try brute force? If $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ both satisfy the relation, just test $(x_1+x_2,y_1+y_2,z_1+z_2)$. – lulu Nov 05 '19 at 21:53
  • if such is tried with (x1,x2,x3) and (y1,y2,y3) then you reach the following 0+0 + x1(y1-y2+y3) + x2(-y1+y2-y3) + x3(-y1-y2+y3) = 0 – Mad Nov 05 '19 at 21:54
  • Well, you should either get $0$ or you should reach a point where a good choice of coordinates gets you a counterexample. – lulu Nov 05 '19 at 21:55
  • Not with brute force as I mentioned the end result is not clear. Bjorns answer and reform of the equation is the right way to go. – Mad Nov 05 '19 at 21:59

2 Answers2

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The condition $$x^2+y^2+z^2-2xy-2yz+2xz=0$$ is equivalent to $$(x-y+z)^2=0 \Leftrightarrow x-y+z=0 $$ As for the addition, if $(x,y,z)\in L$ and $(x',y',z')\in L$, then $$(x+x')-(y+y')+(z+z')=(x-y+z)+(x'-y'+z')=0 $$ so $(x+x',y+y',z+z')\in L$ as well.

bjorn93
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  • Much thanks Bjorn! – Mad Nov 05 '19 at 21:58
  • if you would also be kind and give maybe a tip to a similar problem but with the equation $x^2-y^2-z^2 +2xy = 0$ – Mad Nov 05 '19 at 23:43
  • That wouldn't be a subspace. – bjorn93 Nov 05 '19 at 23:52
  • I took two elements and I am trying to show that their addition is in the subspace. But I am getting at the end $0+0 + (x_1 . x_2) -(y_1 . y_2) -(z_1 .z_2) + (x_1 . y_2)+(x_2 . y_1) = 0 $ How can I give a logical justification that it is not a subspace at this point? – Mad Nov 06 '19 at 00:01
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    If something isn't true, then you don't have to prove anything. You can just give a counterexample. In this case, two vectors which are in the set, but their sum isn't. Try with small numbers. – bjorn93 Nov 06 '19 at 00:08
  • Solved it. thank you – Mad Nov 06 '19 at 00:16
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Good, rank one, it really is just the single squared term

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ - 1 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 1 & 1 \\ - 1 & 1 & - 1 \\ 1 & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & - 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 1 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & - 1 & 1 \\ - 1 & 1 & - 1 \\ 1 & - 1 & 1 \\ \end{array} \right) $$

Will Jagy
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