- #1
member 587159
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}##?
Thanks in advance.
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}##?
Thanks in advance.