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I am a bit confused... Linear combination means

$$F(X)=af(x_1)+bf(x_2) + \cdots$$

and linearly independent means

$$af(x_1)+bf(x_2) + \cdots=0$$

where $a=b=\cdots=0$

My question: is a linear combination linearly independent or linearly dependent as $F(x)$ is not $0$?

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Both meanings you give look quite wrong. Can you try stating things again more carefully, and completely? –  Hurkyl Dec 29 '12 at 3:01
A linear comb. is always lin. independend. If you are working in a vector space we have $\alpha v = 0 \iff \alpha = 0$ if $v \neq 0$. But per definition is $0$ lin. independend. –  André Dec 29 '12 at 3:03
You are indeed a bit confused about what linearly independent means. A set of vectors $\{x_1,\ldots,x_n\}$ is linearly independent if $a_1x_1+\cdots+a_nx_n=0$ implies that $a_i=0$ for all $i$. Which makes your question nonsense. –  Alex Becker Dec 29 '12 at 3:05
ok, well, the only thing i could understand is linear combinations is always linearly independent, I am sorry as I am student of applied Maths and in 1999 I studied linear algebra for the last time... Thank you all for the patience... –  Intellectual_ Dec 29 '12 at 3:23

4 Answers 4

In linear algebra, a linear combination of vectors $v_1,v_2,\ldots, v_n$ is anything of the form $a_1v_1+\ldots +a_nv_n$, where $a_1,\ldots , a_n$ are scalars (referred to as the coefficients of the linear combination).

A set of vectors $\{v_1,\ldots, v_n\}$ is called linearly independent if $$a_1v_1+\ldots +a_nv_n=0 \Rightarrow a_1,\ldots ,a_n=0.$$

So, the connection between these definitions is that, for a set of linearly independent vectors, the only linear combination of $v_1,\ldots,v_n$ which equals $0$ is when all of the coefficients are $0$. If there is some linear combination of $v_1,\ldots,v_n$ equal to $0$ for which not all of the coefficients are $0$, then $v_1,\ldots,v_n$ are not linearly independent.

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A linear combination of a set of vectors $\bf{v}_1,\bf{v}_2,\ldots,\bf{v}_n$ is any sum $$\sum_i \alpha_i \bf{v}_i$$ for scalars $\alpha_i$.

The vectors are linearly independent if the only linear combination of them that's zero is the one with all $\alpha_i$ equal to 0.

It doesn't make sense to ask if a linear combination of a set of vectors (which is just a single vector) is linearly independent. Linear independence is a property of a set of vectors.

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Linear combination is one way to test independence. I think you mean to say $f_1$ and $f_2$. Putting the subscript inside the inputs seem to suggest you would get a number out of it and that would imply linear dependence

$f_1(x)$ and $f_2(x)$ are linearly independent if it happens that $a$ and $b$ are not scalar multiplies of each other.

If it happens that one of your function is $0$, then you can be sure that the other is a linear combination of the other, namely you just multiply $0$ to the other function to write it as the other.

I am also just going to assume neither are $0$ or are linearly dependent themselves already, so i.e. I don't look at the case where they could be constants already or multiplies of themselves. For a quick example, I assume the following can't happen $f_1 = 2e^x$ and $f_2 = e^x$

I am not sure this answers your question as your question doesn't seem to make much sense from the start

I hope this helps

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If we say "y is a linear combination of vectors $x_1,\ldots,x_m$" we mean that there are scalars $a_1,\ldots,a_n$ such that $y=a_1x_1+\cdots+a_mx_m$.

A set of vectors $z_1,\ldots,z_n$ is linearly dependent if there are scalars $b_1,\ldots,b_n$, not all zero, such that $b_1z_1+\cdots+b_nz_n=0$.

If you are confused, I think it's best to treat "linear combination" and "linearly dependent" as two almost unrelated ideas. Note that a linear combination of vectors will be a vector. Being "linearly independent" is a property of a set of vectors. (So a set of vectors might be large, or very large, or linearly dependent, or a subset of some subspace - these are all properties of sets.)

The short answer to your questions is "no".

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