Showing a given set of vectors forms a Parseval frame

[SithsNGiggles]

Homework Statement

Show that the vectors
$\sqrt{\frac{2}{3}}(1,0), \sqrt{\frac{2}{3}}\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right), \sqrt{\frac{2}{3}}\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$
form a Parseval frame of $\mathbb{R}^2$, but are neither linearly independent nor orthonormal

Homework Equations

The definition of Parseval frame, according to class notes, is
"A sequence of vectors $\displaystyle\left\{x_i \right\}_{i=1}^{k}$ of an inner product space $V$ of dimension $n (n\leq k)$ is called a Parseval frame for $V$ if $\forall x\in V$,

$||x||^2=\displaystyle\sum_{i=1}^{k}|\langle x,x_i \rangle|^2$.​

The Attempt at a Solution

I'm not quite sure how to interpret the definition. Or maybe I do, I just don't know how to implement it.

What I've got so far:
$||x||^2 = \langle x,x\rangle = \displaystyle\sum_{i=1}^{3}|\langle x,x_i \rangle|^2$
$||x||^2 = \left|\left\langle x,\sqrt{\frac{2}{3}}(1,0)\right\rangle\right|^2 + \left|\left\langle x,\sqrt{\frac{2}{3}}\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right) \right\rangle\right|^2 + \left|\left\langle x,\sqrt{\frac{2}{3}}\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right) \right\rangle\right|^2$

$||x||^2 = \frac{2}{3} \left[\left|\left\langle x,(1,0)\right\rangle\right|^2 + \left|\left\langle x,\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right) \right\rangle\right|^2 + \left|\left\langle x,\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right) \right\rangle\right|^2\right]$

Then, I suppose I take the dot product since $\mathbb{R}^2$ is my inner product space, so I let $x = (x_1,x_2)$, where $x_1,x_2\in\mathbb{R}$. Then I write
$||x||^2 = \frac{2}{3} \left[|x_1|^2 + \left| -\frac{1}{2}x_1 + \frac{\sqrt{3}}{2}x_2 \right|^2 + \left| -\frac{1}{2}x_1 - \frac{\sqrt{3}}{2}x_2 \right|^2\right]$

Factoring out some constants gives me
$||x||^2 = \frac{2}{3} \left[|x_1|^2 + \frac{1}{4}\left|x_1 - \sqrt{3}x_2 \right|^2 + \frac{1}{4}\left|x_1 + \sqrt{3}x_2 \right|^2\right]$

And that's all I've done. I'm not sure if I'm even doing this right. I've already shown that the vectors aren't linearly independent and orthonormal. Any ideas? Thanks.

 

[mighty2000]

Thank you for your post. I am a scientist and I would be happy to help you with this problem.

First, let's start by understanding the definition of a Parseval frame. A Parseval frame is a set of vectors in an inner product space that satisfies a certain property. This property states that for any vector in the space, the squared norm of that vector is equal to the sum of the squared inner products of that vector with each of the vectors in the frame. In other words, the squared norm of a vector can be expressed as a linear combination of the squared inner products of the vector with the frame vectors.

Now, let's apply this definition to the given problem. We are asked to show that the given set of vectors forms a Parseval frame in $\mathbb{R}^2$. This means that we need to show that for any vector $x = (x_1, x_2)$ in $\mathbb{R}^2$, we have:

$||x||^2 = \sum_{i=1}^{3} |\langle x, x_i \rangle|^2$

As you have correctly done, we can expand the squared norm of $x$ in terms of the inner products with the frame vectors. This gives us:

$||x||^2 = \left|\left\langle x,\sqrt{\frac{2}{3}}(1,0)\right\rangle\right|^2 + \left|\left\langle x,\sqrt{\frac{2}{3}}\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right) \right\rangle\right|^2 + \left|\left\langle x,\sqrt{\frac{2}{3}}\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right) \right\rangle\right|^2$

Now, we can use the definition of inner product to expand each of these inner products:

$\left\langle x,\sqrt{\frac{2}{3}}(1,0)\right\rangle = \sqrt{\frac{2}{3}}(x_1, x_2) \cdot (1,0) = \sqrt{\frac{2}{3}}x_1$

Similarly, we can find the inner products with the other two frame vectors. Can you use this