Given the frame $B=\{(1,1,0),(0,1,1),(1,1,1),(0,0,1),(0,1,-1)\}$, find (if possible) a (i) $(2,1)$ surgery, Given the frame $B=\{(1,1,0),(0,1,1),(1,1,1),(0,0,1),(0,1,-1)\}$, find (if possible) a (i) $(2,1)$ surgery, and a $(1,2)$ surgery that produce tight frames.
A frame is tight if and and only if the frame operator $B^TB$ is a multiple of identity. So, the off-diagonal entries must be zero and the diagonal entries must be equal to each other. 
$\textbf{Part i:}$
$\begin{pmatrix} 1 & 0 & 1 & a \\ 1 & 1 & 1 & b \\ 0 & 1 & 1 & c \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \\ a & b & c \end{pmatrix}=\begin{pmatrix} 2+a^2 & 2+ab & 1+ac \\ 2+ab & 3+b^2 & 2+bc \\ 1+ac & 2+bc & 2+c^2 \end{pmatrix}$
So to get matrix to produce a tight frame, we have the following equations:
\begin{cases}
2+a^2=3+b^2=2+c^2 \\
2+ab=0 \\
1+ac=0 \\
2+bc=0 \\
\end{cases}
From the first statement we have $$ab+2=bc+2 \iff ab=bc \iff a=c.$$ So $ac+1=0 \iff a^2=1=0$ but then there are no real values that will satisfy the equation. Hence there are no values of $(a,b,c)$ to produce a tight frame.
$\textbf{Part ii:}$
$\begin{pmatrix} 1 & 0 & 1 & 0 & a & d \\ 1 & 1 & 1 & 0 & b & e \\ 0 & 1 & 1 & 1 & c & f \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \\ a & b & c \\d & e & f \end{pmatrix}=\begin{pmatrix} 2+a^2+d^2 & 2+ab+de & 1+ac+df \\ 2+ab+de & 3+b^2+e^2 & 2+bc+ef \\ 1+ac+df & 2+bc+ef & 2+c^2+f^2 \end{pmatrix}$
So to get matrix to produce a tight frame, we have the following equations:
\begin{cases}
2+a^2+d^2=3+b^2+e^2=2+c^2+f^2 \\
2+ab+de=0 \\
1+ac+ef=0 \\
2+bc+df=0 \\
\end{cases}
What is a simple way of finding solutions for these equations?
 A: So to get matrix to produce a tight frame, we have the following equations:
\begin{cases}
1+a^2+d^2=3+b^2+e^2=3+c^2+f^2 \\
1+ab+de=0 \\
ac+df=0 \\
bc+ef=0 \\
\end{cases}
One way is to follow the mathematics of wishful thinking. Here, I wishfully think that I can choose many things to be zero. For instance, if $d = e = c = 0$, then the last two equations hold trivially and we are instead left with the system 
$$\begin{cases} 1 + a^2 = 3 + b^2 = 3 + f^2 \\
1 + ab = 0
\end{cases}$$
If $f = b$, then $f$ is covered. So really, we just have the pair $1 + ab = 0$ and $1 + a^2 = 3 + b^2$. Fortunately, the first means that $a = -1/b$. Plugging into the other gives 
$$1 + \frac{1}{b^2} = 3 + b^2.$$
Rearranging and calling $B = b^2$ gives
$$B^2 + 2B - 1 = 0,$$
with solutions $B = -1 \pm \sqrt 2$. Let's choose the positive square root, $B = 1 + \sqrt 2$. Then $b = \sqrt{1 + \sqrt 2}$, and $a = -1/b$.
So in total,
$$\begin{cases}
a = \frac{-1}{\sqrt{1 + \sqrt 2}}\\
b = \sqrt{1 + \sqrt{2}} \\
c = 0 \\
d = 0 \\
e = 0 \\
f = \sqrt{1 + \sqrt{2}}
\end{cases}$$
is a solution.
