# I Linear (in)dependency

Tags:
1. Oct 20, 2016

### Math_QED

Hello all.

I have a question about linear dependency.

Suppose we have a set $S$ of functions defined on $\mathbb{R}$.

$S = \{e^x, x^2\}$. It seems very intuitive that this set is linear independent. But, we did something in class I'm unsure about.

Proof:

Let $\alpha, \beta \in \mathbb{R}$.
Suppose $\alpha e^x + \beta x^2 = 0$
We need to show that $\alpha = \beta = 0$

(Here comes the part I'm unsure about)

Let $x = 0$, then $\alpha e^0 + \beta 0^2 = 0$
$\Rightarrow \alpha = 0$

But if $\alpha = 0$ then follows that $\beta = 0$.
So $S$ is linear independent.

My actual question:

Why can we conclude that the set is linear independent, just by saying that $x = 0$ makes it work? Shouldn't we show that it works for all $x \in \mathbb{R}$?

2. Oct 20, 2016

### Staff: Mentor

We can't. The conclusion is derived from $\alpha = 0$, not from $x=0$.
Yes. This is the crucial point. The equation $\alpha e^x + \beta x^2 = 0$ has to hold for all $x$, so especially for $x=0$.
And if already $x=0$ imply $\alpha = \beta = 0$, what chances are there for other values of $x$? The coefficients do not depend on $x$!

3. Oct 20, 2016

### Math_QED

So we can conclude this because the coefficients do not depend on $x$? From what I understood it mist hold for all x, so certainly for $x = 0$? I still don't fully understand I think.

To complicate things even further, let me suppose that we consider these functions on the domain $\mathbb{R}_0$, how do we show the linear dependency then?

4. Oct 20, 2016

### Staff: Mentor

Yes.
Yes.
True for all $x$ implies true for a certain $x$ as well, and everything derived from a single instance has to be true. It might not be sufficient to hold for all $x$, but it is necessary. And if something is wrong for one, it cannot be true for all.
What do you mean by $\mathbb{R}_0$? $\mathbb{R} - \{0\}$?
If we have a $0$, then the method above can be used.
If we don't have a $0$, we have to do some more work. E.g. by solving the system $\alpha e^x + \beta x^2 = 0$ for values $x \in \{1,2,-1,-2\}$. (I haven't done it, I simply listed enough values to be sure the system can only hold for $\alpha = \beta = 0$.)

The domain where the coefficients $\alpha \, , \, \beta$ are taken from is essential.
Until now we discussed linear independence over $\mathbb{Q}\, , \,\mathbb{R}$ or $\mathbb{C}$.
However, the two functions are not linear independent if we allowed the coefficients to be functions themselves.
We could get $\alpha(x) e^x + \beta (x) x^2 = 0$ with $\alpha(x) = -x^2 \neq 0$ and $\beta(x) = e^x \neq 0$.

Let me cheat here a little bit, because I don't want to think about the question, in which coefficient domain this could be done, that is also a field. So let us consider quotients of rational polynomials in one variable instead, which is a field. (The exponential function complicates things here.)
Let us further take $S=\{x,x^2\}$.
Then $\alpha x + \beta x^2 = 0 \Longrightarrow \alpha = \beta = 0$ if $\alpha \, , \, \beta \in \mathbb{Q}$.
But $\alpha x + \beta x^2 = 0 \nRightarrow \alpha = \beta = 0$ if $\alpha \, , \, \beta \in \mathbb{Q}(x)$.
In this case we have an equation $\alpha x + \beta x^2 = 0$ where we can choose $\alpha = -x \neq 0$ and $\beta = 1 \neq 0$.
So the elements of $S$ are linear independent over $\mathbb{Q}$, but linear dependent over $\mathbb{Q}(x)$.

Last edited: Oct 21, 2016
5. Oct 20, 2016

### Staff: Mentor

No, that's an incomplete summary of what you need to show. Suppose that your set is {x, 2x}.
Suppose $\alpha x + \beta 2x = 0$
Then $\alpha = 0$ and $\beta = 0$ clearly work.

From this one might mistakenly conclude that the functions x and 2x are linearly independent, which is not true.
What you left out from "We need to show that $\alpha = \beta = 0$" is that there can be no other solutions for these constants.