I have a question about linear dependency.

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

S={ex,x2}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 α,β∈R\alpha, \beta \in \mathbb{R}.

Suppose αex+βx2=0\alpha e^x + \beta x^2 = 0

We need to show that α=β=0\alpha = \beta = 0 is the only option to make sure that this linear combination equals 0.

(Here comes the part I’m unsure about)

Let x=0x = 0, then αe0+β02=0\alpha e^0 + \beta 0^2 = 0

⇒α=0\Rightarrow \alpha = 0

But if α=0\alpha = 0 then follows that β=0\beta = 0.

So SS is linear independent.

My actual question:

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

Can someone give a detailed explanation, as I didn’t quite understand the teacher’s explanation.

Thanks in advance.

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Actually you use two points; when you have that α=0\alpha=0, use x=1x=1 to get that β12=0\beta1^2=0, so also β=0\beta=0. This is just noting that x2x^2 is not the zero vector. Once you have proved α=β=0\alpha=\beta=0 for two points, what more do you need?

– egreg

2 days ago

Cause your first equality must be satisfied by All x∈ Rx \in \ \mathbb R and in particular for x=0x=0.

– Abdallah Hammam

2 days ago

Exactly, I only show that it holds for 1 point! It has to hold for all points in the domain! Could someone elaborate?

– Math_QED

2 days ago

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2 Answers

2

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The two functions are linearly dependent if exists (α,β)(\alpha, \beta) constants , different from (0,0)(0,0), such that αex+βx2=0\alpha e^x + \beta x^2=0 for all the xx in the considered domain.

So if for some specific values of xx the condition above does not subsists, it does not hold for all xx, so the functions are independent.

Your teacher attempted and find such a point at x=0x=0, for which (0,β)(0,\beta) results. This solution however does not hold for ,e.g., x=1x=1. For those two points no (α,β)≠(0,0)(\alpha,\beta) \ne (0,0) can be found, so the two functions are linearly independent.

We have ∀x∈R,αex+βx2=0\forall x \in \mathbb{R}, \alpha e^x + \beta x^2 = 0

Your goal is to determine all the possible values of α\alpha and β\beta. Clearly, α=β=0\alpha=\beta=0 is a solution, but can there be other solutions?

What are the properties that α\alpha and \beta\beta must satisfy?

In particular, we can derive some conditions by fixing certain values of xx.

By setting x=0x=0, we conclude that \alpha\alpha must be 00. We can recover \beta\beta by letting x=1x=1.

Clearly, we can get more conditions by fixing more values of xx, but just from these two conditions, we have concluded that the unique solution is \alpha=\beta=0\alpha=\beta=0.

Why is it unique?

– Math_QED

2 days ago

\alpha\alpha and \beta\beta must satisfies \alpha=0\alpha=0 and \alpha e + \beta = 0\alpha e + \beta = 0, We can solve the system and see that it is unique, that is both of them are zeroes.

– Siong Thye Goh

2 days ago