Given the matrix of an R4 scalar product (b) in its standard basis, and a subspace of R4 (W), how to find the signature of the restriction of b to W? So, the scalar product's matrix in $R^4$ is the following:
\begin{pmatrix}
1 & 0 & 1 & 1 \\
0 & -1 & 2 & 0 \\
1 & 2 & 0 & 1 \\
1 & 0 & 1 & 0 \\
\end{pmatrix}
The subspace W of $R^4$ is defined by this equation: x1 + x2 + x3 + x4 = 0.
Well, I don't even know where to start because my teacher barely taught us restrictions, I know just a glimpse of them but never practiced on them. I found this in the exam but the majority of the students flunked this specific exercise. I know what the signature is by the way, however to get to it I must find the restriction, and I don't know how to do it.
Thank you for your help, I really want to know how to solve this.
 A: Let $A$ denote the given matrix, $A = \pmatrix{1 & 0 & 1 & 1 \\
0 & -1 & 2 & 0 \\
1 & 2 & 0 & 1 \\
1 & 0 & 1 & 0}$.
$A$ encodes the scalar product over $\Bbb R^4$ in the following sense: for any $x,y \in \Bbb R^4$, their scalar product is given by $f(x,y) = x^TAy$. The restriction  of this scalar product to $W$ refers to the same function with the restricted domain $W$. That is, the restriction of $f$ to $W$ is only defined for $x,y \in W$.
The matrix $A$ now gives us too much information; we need a matrix that characterizes a scalar product over the $3$ dimensional space $W$, which is to say that we need a $3 \times 3$ matrix. We can obtain such a matrix if we find a way to represent elements of $W$ with vectors of $3$ elements, i.e. with vectors from $\Bbb R^3$. The way to do this is to select a basis of $W$.
One basis of $W$ is given by
$$
\mathcal B = \{v_1,v_2,v_3\}; 
\quad v_1 = (1,-1,0,0),
v_2 = (1,0,-1,0),
v_3 = (1,0,0,-1).
$$
Any basis that you choose will give you the same final answer. For a vector $x \in W$, we define the coordinate vector $y = (y_1,y_2,y_3) = [x]_{\mathcal B}$ of $x$ relative to $\mathcal B$ so that
$$
x = y_1 v_1 + y_2 v_2 + y_3 v_3.
$$
Equivalently, if $B$ denotes the matrix whose columns are $v_1,v_2,v_3$, then $x = By$. Now, for two coordinate vectors $u,v \in \Bbb R^3$, the scalar product of the corresponding vectors is given by
$$
f(Bu,Bv) = (Bu)^TA(Bv) = u^T[B^TAB]v.
$$
Thus, the matrix of this scalar product is given by $B^TAB$.
In this case, the product $B^TAB$ comes out to
$$
B^TAB = \pmatrix{0 & 2 & 0\\2 & -1 & 0\\0 & 0 & -1}.
$$
You should find that the signature of this matrix (and hence the signature of the restricted bilinear form) is given by $n_+ = 1, n_- = 2, n_0 = 0$.
