Isotropic subspace of $x_1^2-x_2^2+x_3^2$ I have the quadric form $$q(x_1,x_2,x_3,x_4) = x_1^2-x_2^2+x_3^2$$ on a 4-dimensional real vector space $V$, and I was asked to find a 3-dimensional isotropic subspace for that quadratic form.
I think it cannot exist and I tried to give a dimensional proof:
If such isotropic subspace $W$ exists, then $W\subset W^\bot$, so $\dim(W^\bot)\ge\dim(W)$. But then I would have $$4=\dim(V)=\dim(W)+\dim(W^\bot)\ge3+3$$ and this is a contraddiction.
However since the quadratic form is a degenerate form, I think that the equality $$\dim(V)=\dim(W)+\dim(W^\bot)$$ does not hold.
Any hint?
 A: Expanding on my comment. This is a bit too long to itself be a comment.
I was a little sloppy in what I said; it's not that $S$ can only be in $V_3$, it's just that that is the only case that matters.
All isotropic vectors are of the form $v_3 + v_0$ where $v_3 \in V_3$ is isotropic and $v_0 \in V^\perp$. Looking for non-degenerate isotropic vectors means looking for isotropic vectors of $V_3$.
If $S$ contains some $v_0$ then it is spanned by $v_3, v_0$. Completing $S$ to an isotropic 3-space requires another isotropic $v_3'$ orthogonal to $v_3$, so $v_3, v_3'$ span a totally isotropic plane in $V_3$.
If $S$ does not contain some $v_0$, then it is spanned by $v_3+v_0$ and $v_3'+\alpha v_0$ for some $\alpha$. If $v_3, v_3'$ are orthogonal then they span a totally isotropic plane in $V_3$, so assume otherwise. Clearly $w = v_3'-\alpha v_3 \in V_3\cap S$. The map $L(x) = x + \beta\langle x, w\rangle v_0$ fixes $w$ and $v_0$ and is an invertible linear isometry for each scalar $\beta \not= 0$. We can choose $\beta$ so that $L(v_3 + v_0) = v_3$, so the image of $S$ under $L$ is a totally isotropic plane in $V_3$.
