Showing something is an inner product 
Let us consider the quadratic form $q(t, x, y, z)=-t^{2}+x^{2}+y^{2}+z^{2}$ on $\mathbb{R}^{4} .$ Find the corresponding symmetric bilinear form $f$. The space $\mathbb{R}^{4}$ with the bilinear form $f$ is called the Minkowski space and it is the mathematical setting in which Einstein's theory of special relativity is most conveniently formulated. A vector $v \in \mathbb{R}^{4}$ is called timelike if $f(v, v)<0 .$ The timelike vectors correspond to directions to which matter can move in the spacetime $\mathbb{R}^{4} .$ Let $v \in \mathbb{R}^{4}$ be timelike. Then the orthogonal complement $\{v\}^{\perp}$ is called the local rest space of an observer moving to the direction $v$. Show that $f$ restricted to $\{v\}^{\perp}$ is an inner product, that is, show that
$$
g:\{v\}^{\perp} \times\{v\}^{\perp} \rightarrow \mathbb{R}, \quad g(u, w)=f(u, w), u, w \in\{v\}^{\perp},
$$
is positive definite.

I'm trying to do the question above.
I have found $f(v,w) = v^TAw$ where $$A = \begin{pmatrix} -1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \end{pmatrix}$$ but I'm really not show how to proceed - I mean, what does $f$ restricted to $\{ v \}^\bot$ mean for starters?
any help please! (this is for an elementary algebra course so keep it basic please).
 A: First, is easy to see that $f: \mathbb{R}^4 \times \mathbb{R}^4 \to \mathbb{R} $ is given explicit by the bilinear form:
$$f((t_1,x_1,y_1,z_1),(t_1,x_1,y_1,z_1))=-t_1t_2 + x_1x_2+ y_1y_2 + z_1z_2 .$$
Now, consider $v \in \mathbb{R}^4$ a timelike vector, then $V=\{v\}^{\perp}$ is a 3-dimensional subspace of $\mathbb{R}^4$, so $f$ restricted to $V$ means that we are restricting the function $f$ to the subset $V\times V \subset \mathbb{R}^4 \times \mathbb{R}^4$. Lets call the restriction $g: V \times V \to \mathbb{R}$. Is not hard to see that $g$ is a symmetric, bilinear form defined on the vector space $V$, what we want to show is that $g$ defines a inner product, since $g$ is bilinear and symmetric the only thing we realy need to proof, is that $g$ is definite positive ($g(u,u) \geq 0, \forall u \in V$ and $g(u,u)=0$, iff $u=0$).
To fix the ideas, supose firt that $\{e_j\}_{j=1}^{4}$ is the canonical base of $\mathbb{R}^4$ and $v=e_1$, so $f(v,v)=-1 <0$ and:
$$V=\{(0,x,y,z) \in \mathbb{R^4} : x,y,z \in \mathbb{R}\}, $$
and just using the definition of $f$ we have that:
$$f((0,x_1,y_1,z_1),(0,x_1,y_1,z_1))= x_1x_2+ y_1y_2 + z_1z_2 .$$
Then $g$ is the canonical inner product of $V =\mathbb{R^3} \subset \mathbb{R}^4$. As you notice $f(u,v)=u^TAv$, but now see that $A$ restricted to $V$ is just the identity matrix.
Now lets get back to the general case where $v \in \mathbb{R}^4$ is any vector, with $f(v,v)<0$. Take $u \in V$, lets show that $g(u,u) \geq 0$, to see that we make:
$$v= \sum_{j=1}^4v_je_j, \; \; u= \sum_{j=1}^4u_je_j,$$
so we have that $v_1 \neq 0$ (why ?) and:
$$ -v_1^2+\sum_{j=2}^4 v_j^2 < 0 \implies 1- \sum_{j=2}^4 \frac{v_j^2}{v_1^2} >0.$$
On the other hand, using that $u \in V$, we know that:
$$ \sum_{j=1}^4 u_jv_j=0.$$
Finally using the Cauchy-Schwarz inequality we have:
\begin{align}
g(u,u) & = -u_1^2 + \sum_{j=2}^4 u_j^2 \\
 & = - \frac{(v_2u_2 + v_3u_3 + v_4u_4)^2 }{v_1^2} + \sum_{j=2}^4 u_j^2 \\ 
 & \geq -\left(\sum_{j=2}^4 u_j^2 \right)\left(\sum_{j=2}^4 \left(\frac{v_j}{v_1}\right)^2 \right) + \sum_{j=2}^4 u_j^2\\
 & = \left(\sum_{j=2}^4 u_j^2 \right)\left(1-\sum_{j=2}^4 \left(\frac{v_j}{v_1}\right)^2\right) \\
 & \geq 0.
\end{align}
That is, for every $u \in V$, $g(u,u) \geq 0$. Now try to show that if $g(u,u)=0$, then $u=0$, once you done that $g$ is a inner product in $V$.
