Does a combination of linearly independent vectors have a minimal value?

Question

My question is:

Let $v_1,v_2 \in \mathbb{R}^n$ be linearly independent. There exists $c > 0$ such that $$ \|v_1\cos(\beta t) - v_2\sin(\beta t)\| \geq c $$ for all $t \in \mathbb{R}$?

I know $\|v_1\cos(\beta t) - v_2\sin(\beta t)\| > 0$ for all $t \in \mathbb{R}$, because they are linearly independent. Also, there must be a $t_1 \in \mathbb{R}$

My geometric intuition tells me that the class of vectors $v_1\cos(\beta t) - v_2\sin(\beta t)$ with $t \in \mathbb{R}$ is an ellipsis, but I am not sure how to show it is true, if that is the case.

I also thought about assuming $v_1$ as the bigger vector of the two, and I thought about computing $t_1 \in \mathbb{R}$ such that $v_1\cos(\beta t_1) - v_2\sin(\beta t_1)$ is an orthogonal vector to $v_1$ to show that $v_1$ is in the direction of the biggest axis and $v_1\cos(\beta t_1) - v_2\sin(\beta t_1)$ is in the direction of the smallest axis.

However, this is not necessarilly true, because if the angle between $\theta$ is small, there would be a $t_2 \in \mathbb{R}$ such that $\|v_1\cos(\beta t_2) - v_2\sin(\beta t_2)\| > \max\{\|v_1\|,\|v_2\|\}$.

In summary, I am embarrassingly confused with this problem.

So, How can I prove my statement?

Note: I'm not asking for an actual proof, I'm asking for directions, what I might have missed and things like that.

Answer 1

Lets $v_1,v_2 \in \mathbb{R}^n$ be linearly independet vectors, $\beta \in \mathbb{R} \colon \beta \neq 0$. We are considering the function: $\begin{align} f \colon \mathbb{R} \rightarrow \mathbb{R}; t \mapsto \Vert v_1 \cos(\beta t) - v_2 \sin( \beta t) \Vert \end{align}$. Since $f$ is a composition of continous functions, it is continous, hence attains a minimum on the compact Invertvall $[0,\frac{2\pi}{\beta}]$. Since this function never vanishes, because $v_1,v_2$ are linearly independent and $\sin, \cos$ do not have a common root, we know that this minimum must be bigger than $0$. Now we mention that of course $f$ is periodic with period $\frac{2 \pi}{\beta}$, hence the minimum on $[0,\frac{2\pi}{\beta}]$ is also a global minimum, which prooves the statement.

The case $\beta=0$ is easy to see, since the function is constant.

Answer 2

As @Intelligenti pauca pointed out in his comment above, the vector

$ P(t) = v_1 \cos(\beta t) - v_2 \sin(\beta t) $

Describes an ellipse with conjugate semi-axes $v_1, v_2$. Conjugate axes can be transformed into the standard semi-axes of the ellipse, using the following trick.

Introduce a shift in the $t$ variable, by letting $t = t' - t_0 $, and

$ \beta t = \beta (t' - t_0) = \phi - \phi_0 $

Then

$P(\phi) = v_1 \cos(\phi - \phi_0) - v_2 \sin(\phi - \phi_0) $

Expand and collect terms,

$P(\phi) = \bigg(v_1 \cos(\phi_0) + v_2 \sin(\phi_0) \bigg) \cos(\phi) + \bigg( v_1 \sin(\phi_0) - v_2 \cos(\phi_0) \bigg) \sin(\phi) $

Now, we want the vector multiplying $\cos(\phi)$ to be perpendicular to the vector multiplying $\sin(\phi) $ so that they define the semi-minor and semi-major axes of the ellipse. Hence

$ \bigg(v_1 \cos(\phi_0) + v_2 \sin(\phi_0) \bigg) \cdot \bigg( v_1 \sin(\phi_0) - v_2 \cos(\phi_0) \bigg) = 0$

This reduces to

$ \dfrac{1}{2} \bigg( \|v_1 \|^2 - \| v_2 \|^2 \bigg) \sin(2 \phi_0) - \bigg( v_1 \cdot v_2 \bigg) \cos(2 \phi_0) = 0 \hspace{15pt} (*) $

Therefore,

$ \phi_0 = \dfrac{1}{2} \tan^{-1} \left( \dfrac{2 v_1 \cdot v_2}{\|v_1 \|^2 - \| v_2 \|^2 } \right) $

Now the semi axes are

$w_1 = v_1 \cos(\phi_0) + v_2 \sin(\phi_0) $

and

$w_2 = v_1 \sin(\phi_0) - v_2 \cos(\phi_0) $

So that the minimum required is the minimum of the two norms of $w_1$ and $w_2$

That is,

$ c = \text{Min}( \| w_1 \| , \| w_2 \| ) $