Find a matrix $C \in M_{3\times 3} (\mathbb C)$ such that $A_4=C^{T}A_3C$ 
Let $$A_3=\begin{bmatrix} -1 & 2 & -1 \\ 2 & 0 & 0 \\ -1 & 0 & -1 \end{bmatrix},$$ $$A_4=\begin{bmatrix} 4 & 4 & 4 \\ 4 & 0 & 2 \\ 4 & 2 & 5 \end{bmatrix}.$$
  a) Show that $A_3$ and $A_4$ are congruent over $\mathbb C$, but not over $\mathbb R$.
  b) Find a matrix $C \in M_{3\times3} (\mathbb C)$ such that $A_4=C^{T}A_3C$.


My try:
a) $$\det A_3=4, \det A_4=-32 \Rightarrow \det A_3 \cdot \det A_4=-128$$ When the matrices are congruent over $K$ then the product of determinants is equal to $c^2$ for some $c\in K$. But $\sqrt{-128} \notin \mathbb R$, so $A_3$ and $A_4$ are not congruent over $\mathbb R$.$\det A_3 \neq 0$, $\det A_4 \neq 0$ so $\operatorname{rank} A_3= \operatorname{rank} A_4 =3$ and $A_3$ is congruent with $A_4$ over $\mathbb C$.I think this is a good way, but I have a problem with b):The matrix  $ C $ is determined by finding a base perpendicular to the bilinear form $ h $ where $A_3=G(h;st)$:$h(\epsilon_1, \epsilon_1)=-1, h(\epsilon_2, \epsilon_2)=0,h(\epsilon_3, \epsilon_3)=-1$ so let $\alpha_1=\epsilon_1$ because it is an isotropic vector.  We're looking for one $\alpha_2$ such that $h(\alpha_1,\alpha_2) = 0$ and $h(\alpha_2,\alpha_2)\neq0$$h(\alpha_1,\alpha_2) = 0 \Leftrightarrow x_{1}=2x_2-x_3$ so let $\alpha_2=(0,1,2)$ and then $h(\alpha_2, \alpha _2)=-4\neq 0$.  We're looking for one $\alpha_3$ such that $h(\alpha_1,\alpha_3) = 0,h(\alpha_2,\alpha_3) = 0$ and $h(\alpha_3,\alpha_3)\neq0$. 
$h(\alpha_1,\alpha_3) = 0,h(\alpha_2,\alpha_3) = 0 \Leftrightarrow -a+2b-c=0, -2c=0$ so let $\alpha_3=(2,1,0)$ then $h(\alpha_3,\alpha_3) = 4 \neq0$.  So we have $A=\left\{  \alpha _1, \alpha _2, \alpha _3\right\} $ and $C=M(id)_A^{st}$. 

Unfortunately it is not a correct answer because when I multiply $C^{T}, A_3, C$ I don't get $A_4$.  Can you tell me where did I make a mistake?

Note that if you write $A_{4}=C^{T}A_{3}C$, then you are essentially applying simultaneous row and column operations. $C^{T}$ represent some operations applied to the rows of $A_{3}$ and $C$ represent the same operations applied to the columns. Now you can start doing calculations
$$A_{3}=\left(\begin{matrix}-1&2&-1\\2&0&0\\-1&0&-1\end{matrix}\right)\stackrel{\sqrt{2}iR_{3}\rightarrow R_{3}\\\sqrt{2}iC_{3}\rightarrow C_{3}}{\longrightarrow}\left(\begin{matrix}-1&2&-\sqrt{2}i\\2&0&0\\-\sqrt{2}i&0&2\end{matrix}\right)\stackrel{R_{1}+\frac{1}{2}R_{2}\rightarrow R_{1}\\C_{1}+\frac{1}{2}C_{2}\rightarrow C_{1}}{\longrightarrow}\left(\begin{matrix}1&2&-\sqrt{2}i\\2&0&0\\-\sqrt{2}i&0&2\end{matrix}\right)\stackrel{R_{3}+\frac{1}{2}\left(1+\sqrt{2}i\right)R_{2}\rightarrow R_{3}\\C_{3}+\frac{1}{2}\left(1+\sqrt{2}i\right)C_{2}\rightarrow C_{3}}{\longrightarrow}\left(\begin{matrix}1&2&1\\2&0&0\\1&0&2\end{matrix}\right)\stackrel{R_{1}+R_{3}\rightarrow R_{3}\\C_{1}+C_{3}\rightarrow C_{3}}{\longrightarrow}\left(\begin{matrix}1&2&2\\2&0&2\\2&2&5\end{matrix}\right)\stackrel{2R_{1}\rightarrow R_{1}\\2C_{1}\rightarrow C_{1}}{\longrightarrow}\left(\begin{matrix}4&4&4\\4&0&2\\4&2&5\end{matrix}\right)$$
Applying the same operations to the identity matrix, we get
$$I=\left(\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}\right)\stackrel{\sqrt{2}iC_{3}\rightarrow C_{3}}{\longrightarrow}\left(\begin{matrix}1&0&0\\0&1&0\\0&0&\sqrt{2}i\end{matrix}\right)\stackrel{C_{1}+\frac{1}{2}C_{2}\rightarrow C_{1}}{\longrightarrow}\left(\begin{matrix}1&0&0\\\frac{1}{2}&1&0\\0&0&\sqrt{2}i\end{matrix}\right)\stackrel{C_{3}+\frac{1}{2}\left(1+\sqrt{2}i\right)C_{2}\rightarrow C_{3}}{\longrightarrow}\left(\begin{matrix}1&0&0\\\frac{1}{2}&1&\frac{1}{2}\left(1+\sqrt{2}i\right)\\0&0&\sqrt{2}i\end{matrix}\right)\stackrel{C_{1}+C_{3}\rightarrow C_{3}}{\longrightarrow}\left(\begin{matrix}1&0&1\\\frac{1}{2}&1&\frac{1}{2}\left(2+\sqrt{2}i\right)\\0&0&\sqrt{2}i\end{matrix}\right)\stackrel{2C_{1}\rightarrow C_{1}}{\longrightarrow}\left(\begin{matrix}2&0&1\\1&1&\frac{1}{2}\left(2+\sqrt{2}i\right)\\0&0&\sqrt{2}i\end{matrix}\right)$$
The end result is
$$C=\left(\begin{matrix}2&0&1\\1&1&\frac{1}{2}\left(2+\sqrt{2}i\right)\\0&0&\sqrt{2}i\end{matrix}\right)$$
Indeed, according to Wolfram Alpha this is correct.
