Are these vectors linearly independent or linearly dependent?

Question

I am really confused if $\alpha_1=(e^{\pi/2},1)$ and $\alpha_2=(\sqrt[3]{110},1)$ are linearly independent or linearly dependent in $\mathbb{R}^2$. (Hoffman and Kunze, page no. $48$.)

Consider $c_1\alpha_1+c_2\alpha_2=0$. Then $c_1(e^{\pi/2},1)+c_2(\sqrt[3]{110},1)=(0,0)$. This gives two equations: \begin{align*} c_1e^{\pi/2}+c_2\sqrt[3]{110}&=0\\ c_1+c_2&=0. \end{align*}

Then we get, $c_1=-c_2$ and $c_2(\sqrt[3]{110}-e^{\pi/2})=0$. Now, $e^{\pi/2}$ and $\sqrt[3]{110}$ are numerically very close. So for $0<c_2<1$, the smaller $c_2$ becomes, the closer the latter equation is to $0$.

Can we conclude that $\alpha_1$ and $\alpha_2$ are linearly dependent on basis of this?

But then if two vectors are linearly dependent, one of them is a scalar multiple of the other. I don't see how $e^{\pi/2}$ and $\sqrt[3]{110}$ are scalar multiple of each other. (I am only discussing the first components of each tuple because $1$ is clearly scalar multiple of $1$.)

Any idea if they are linearly independent or linearly dependent? Also, what about the linear independence/linear dependence of the set $\{e^{\pi/2},\sqrt[3]{110},1\}$ in $\mathbb{R}$?

This might be very simple and I am complicating things unnecessarily, nevertheless any explanation to clear my confusion would be appreciated. Thanks!

Edit: Thank you. I found each your answers helpful (except where I mentioned below). I wish I could accept more than one answers I found helpful :)

Answer 1

As noted in the other answers the key fact here is that $e^{\pi/2} \ne \sqrt[3]{110}$. And we can be sure of this fact also without an explicit calculation of the two numbers, because $\sqrt[3]{110}$ is an algebraic number, but $e^{\pi/2}=(-1)^{-i/2} $ is transcendental by the Gelfond–Schneider theorem.

This is important for the answer to your second question:

the three numbers (see as vectors) $\{\sqrt[3]{110},e^{\pi/2},1 \}$ are linearly dependent in the vector space $\mathbb{R}$ over the field $\mathbb{R}$ ( obviously), but are linearly independent in $\mathbb{R}$ over the field $\mathbb{Q}$, because there does not existst a rational number $q$ such that $\sqrt[3]{110}=qe^{\pi/2}$

Answer 2

The vectors are linearly independent. Otherwise, one of them would a multiple of the other one (as you wrote). But, since the second coefficient of both vectors is $1$, that would mean that $e^{\pi/2}=\sqrt[3]{110}$, that is, that $e^{3\pi/2}=110$. But they're different.

Answer 3

from the second equation we get $$c_2=-c_1$$ and then you will get $$c_1e^{\pi/2}-c_1\cdot \sqrt[3]{110}=0$$

Answer 4

"Close to zero" and "zero" are very much not the same thing when it comes to vectors (or, really, most things in math). The only actual solution to the equation $c_2(\sqrt[3]{110} - e^{\pi/2}) = 0$ is $c_2 = 0$; the math doesn't care if "small" $c_2$ gets you "close", it only cares about exact equality. In fact, if close enough were good enough, we could conclude by your argument that any two vectors were linearly independent - for sufficiently small $c_2$, for example, $c_2(2 - 1)$ is also very nearly zero.

Because the only true solution to your system of equations is $c_1 = c_2 = 0$, the vectors must be linearly independent.