# Are these vectors linearly independent or linearly dependent?

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!

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.

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 $\sqrt[3]{110}=qe^{\pi/2}$

• Thank you so much. It clears away my confusions. – Kappa Mar 18 '18 at 4:58

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.

"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.

• Correct, but in numerical computation, its important to be aware of situations in which vectors are effectively linearly dependent (or matrices are effectively singular) to floating point precision even though they may be linearly independent in exact arithmetic. – Brian Borchers Mar 17 '18 at 17:20
• @BrianBorchers Yes, that's what I was wondering when I said they are numerically close. Would such a thing happen here? – Kappa Mar 18 '18 at 5:08

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

• I already have these equations in the above post. – Kappa Mar 18 '18 at 4:56