Considering the following set of vectors:
$w_1= \{1,a,1\}$, $w_2= \{1,0,-1\}$, $w_3= \{0,a,0\}$, $w_4= \{-2,0,2\}$, where: $a=\frac{0.408}{0.296}$
I am interested if they constitute a frame in $\mathbb{R}^3$.
This is only true if:
$A||v||^2 \leq \sum_{k=1}^{4} |<v,w_k>|^2 \leq B||v||^2$
So I started to calculate as follows:
$\sum_{k=1}^{4} |<v,w_k>|^2 \\= |(v_1,v_2,v_3)(1,a,1)|^2 +|(v_1,v_2,v_3)(1,0,-1)|^2+|(v_1,v_2,v_3)(0,a,0)|^2+|(v_1,v_2,v_3)(-2,0,2)|^2 \\ =2(3 v_1^2 + a v_1 v_2 + a^2 v_2^2 - 4 v_1 v_3 + a v_2 v_3 + 3 v_3^2)$
Now I am stuck. I have the feeling that there is something, as there is something like a norm there, but I don't know how to proceed. Can someone tell me how I can show that it is a frame, and if not, why not?
Further, once I possibly showed that it is a frame, is there a way to convert it into a tight frame?
As always, many thanks in advance!
Edit
One more step I didn't write out, is dividing by the norm $||v||^2=\sqrt{v_1^2+v_2^2+v_3^2}^2=v_1^2+v_2^2+v_3^2$
Thus:
$A \leq \frac{2(3 v_1^2 + a v_1 v_2 + a^2 v_2^2 - 4 v_1 v_3 + a v_2 v_3 + 3 v_3^2)}{v_1^2+v_2^2+v_3^2} \leq B$
Is the point I am stuck at.