To show that two quadratic forms are equivalent over $\mathbb{C}$ but not over $\mathbb{R}$ Consider the quadratic forms $p$ and $q$ given by $$q(x,y,z,w)=x^2+y^2+z^2+bw^2$$ and $$p(x,y,z,w)=x^2+y^2+czw.$$ 
Then we can write their respective matrices as 
$$
A = \pmatrix{1\\&1\\&&1\\&&&b}, \quad 
B = \pmatrix{1\\&1\\&&0&c/2\\&&c/2&0}
$$

To Prove: If $b$ and $c$ are non-zero than $p$ and $q$ are equivalent over $\mathbb{C}$ (need not be equivalent over $\mathbb{R}$). 
They are equivalent over $\mathbb{R}$ only if $b,c$ are non-zero and $b$ is negative.

Now that characteristic polynomial of $A$ and $B$ are $(x-1)^3(x-b)$ and $(x-1)^2(x^2-\frac{c^2}{4})$ respectively. So I can see from Sylvester's theorem why they are equivalent over $\mathbb{R}$ only if $b,c \neq 0 \ \& \ b<0$. But I can't prove the part over $\mathbb{C}$ I am trying to work with the inertia of matrices described here but I need help. 
 A: Write the coordinates in $q(x,y,z,w)$ as $\widetilde z:= \frac1{\sqrt c}\cdot (z - i\sqrt bw)$ and $\widetilde w:= \frac1{\sqrt c}\cdot (z+i\sqrt bw)$ (i. e. make a base change). Then you have
\begin{align*}
q(x,y,z,w) &= x^2+y^2+z^2+bw^2\\
&= x^2+y^2 + (z-i\sqrt bw)(z+i\sqrt bw)\\
&= x^2+y^2 + c\cdot \left(\frac{1}{\sqrt c}\cdot (z-i\sqrt bw)\right)\cdot \left(\frac1{\sqrt c}\cdot (z+i\sqrt bw)\right)\\
&= x^2+y^2 + c\widetilde z\widetilde w\\
&= p(x,y,\widetilde z,\widetilde w).
\end{align*}
Therefore, $p$ and $q$ are equivalent over $\mathbb C$.
If you want to calculate with matrices, notice that
$$\tag{$*$}
\begin{pmatrix}x\\y\\z\\w \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1
& 0 & 0\\ 0 & 0 & \frac1{\sqrt c} & -i\sqrt{\frac bc}\\
0 & 0 & \frac1{\sqrt c} & i\sqrt{\frac bc} \end{pmatrix}\cdot \begin{pmatrix}
x\\ y\\ \widetilde z\\ \widetilde w\end{pmatrix}.
$$
Therefore,
\begin{align*}
q(x,y,z,w) &= \begin{pmatrix}x & y & z & w\end{pmatrix}\cdot \begin{pmatrix} 1 &
0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & \frac c2\\ 0 & 0 & \frac c2 &
0\end{pmatrix}\cdot \begin{pmatrix}x\\y\\z\\w\end{pmatrix}\\
&=\begin{pmatrix}x & y & \widetilde z & \widetilde w\end{pmatrix} \cdot
\begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & \frac1{\sqrt c} &
\frac1{\sqrt c}\\ 0 & 0 & -i\sqrt{\frac bc} & i\sqrt{\frac bc} \end{pmatrix}
\cdot \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & \frac c2\\ 0
& 0 & \frac c2 & 0\end{pmatrix}\cdot \begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 &
0\\ 0 & 0 & \frac1{\sqrt c} & -i\sqrt{\frac bc}\\ 0 & 0 & \frac1{\sqrt c} &
i\sqrt{\frac bc}\end{pmatrix} \cdot \begin{pmatrix}x\\y\\ \widetilde z\\
\widetilde w\end{pmatrix}\\
&= \begin{pmatrix}x & y & \widetilde z & \widetilde w\end{pmatrix} \cdot
\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 &
b\end{pmatrix} \cdot \begin{pmatrix} x\\ y\\ \widetilde z\\ \widetilde
w\end{pmatrix}\\
&= p(x,y,\widetilde z, \widetilde w).
\end{align*}
To find the base change matrix, you could apply symmetric Gauss elimination to
the matrix of $q$: Use column operations (followed directly by the corresponding
row operation) to transform the matrix of $q$ into the matrix of $p$. Also, at
every step, keep track of the column operations (e. g. by applying them to the
identity matrix): the resulting matrix is your base change matrix (in the sense of ($*$)).
A: About Sylvester inertia, all you need is
$$
\left(
\begin{array}{rr}
1 & 1 \\
-1 & 1
\end{array}
\right)
\left(
\begin{array}{rr}
0 & 1 \\
1 & 0
\end{array}
\right)
\left(
\begin{array}{rr}
1 & -1 \\
1 & 1
\end{array}
\right) =
\left(
\begin{array}{rr}
2 & 0 \\
0 & -2
\end{array}
\right)
$$
where anything you require can be multiplied by a nonzero positive real number without penalty, should you wish to incorporate those $b,c$ numbers.
I should emphasize that inertia is about $P^T A P = D,$ where $P$ is real nonsingular. Different choices of $P$ give different diagonal entries in the diagonal matrix $D,$ but the count of negative, zero, positive diagonal entries does not change.
