I wonder how to proof that a given set of vectors is a frame and tight.
A nice explanation has been given here before: https://math.stackexchange.com/a/954403
However, I am puzzled about the following sentence:
"The frame is tight if there exists some $\alpha$ such that for any vector $v \in \Bbb{R}^4$ it happens that $\sum\limits_{k=1}^5 \big| \langle v, e_k\rangle \big|^2 = \alpha \|v\|^2$ where $\{e_k\}_{k=1}^5$ are the five vectors you gave."
I understand, that this is necessary to show, but how does one show this for any $v \in \Bbb{R}^4$?
The short answer is that you have to do the direct calculation since you are dealing with an equality. So just take $ v=( v_1, v_2, v_3, v_4)$ and start working.
Edit: Consider the following example where $w_1 = (1, 0), w_2 = (-1/2, \sqrt{3}/2)$ and $w_3 = (-1/2, -\sqrt{3}/2)$. Let us show $\{w_1, w_2, w_3\}$ is a tight frame. Observe \begin{align} \sum^3_{k=1}|\langle v, w_i\rangle|^2 =&\ |(v_1, v_2)\cdot (1, 0)|^2 + |(v_1, v_2)\cdot(-1/2, \sqrt{3}/2)|^2 + |(v_1, v_2)\cdot(-1/2, -\sqrt{3}/2)|^2\\ =&\ v_1^2+(-v_1/2+\sqrt{3}v_2/2)^2+(-v_1/2-\sqrt{3}v_2/2)^2\\ =&\ v_1^2+\frac{1}{4}v_1^2+\frac{3}{4}v_2^2-\frac{\sqrt{3}}{2}v_1v_2+\frac{1}{4}v_1^2+\frac{3}{4}v_2^2+\frac{\sqrt{3}}{2}v_1v_2\\ =&\ \frac{3}{2}v_1^2 + \frac{3}{2}v_2^2 = \frac{3}{2}\|v\|^2. \end{align}