How to Show Linearity of a Function?

[PhysicsRock]

Homework Statement

Take the Vector space of functions from a set ##\{ 0,1,...,n \}## to a field ##K##, ##V_n = \text{Fun}(\{ 0,1,...,n \},K)##. The notation ##j_K## refers to ##\underbrace{1_K + ... + 1_K}_{j \text{-times}}##. Let ##f \in V_{n+1}##. For a function ##\partial : V_{n+1} \rightarrow V_n## and ##f \in V_{n+1}##, show that ##\partial## is a linear function for ##\partial : i \rightarrow (i+1)_K \cdot f(i+1)##.

Relevant Equations

/

I don't really know how I am supposed to approach that. In general, I know how to show that a function is linear, which is to show that ##f(\alpha \cdot x) = \alpha \cdot f(x)## and ##f(x_1 + x_2) = f(x_1) + f(x_2)##. However, for this specific function, I have no idea, since there is nothing provided about ##f##, so if I wanted to show the multiplicative property, I couldn't just drag anything out of ##f## without loss of generality. So I could really use some help to figure this out.

Thank you in advance.

 

[pasmith]

Homework Statement:: Take the Vector space of functions from a set ##\{ 0,1,...,n \}## to a field ##K##, ##V_n = \text{Fun}(\{ 0,1,...,n \},K)##. The notation ##j_K## refers to ##\underbrace{1_K + ... + 1_K}_{j \text{-times}}##. Let ##f \in V_{n+1}##. For a function ##\partial : V_{n+1} \rightarrow V_n## and ##f \in V_{n+1}##, show that ##\partial## is a linear function for ##\partial : i \rightarrow (i+1)_K \cdot f(i+1)##.
Relevant Equations::

Should this not be [tex]
\partial(f) : i \mapsto (i+1)_K \cdot f(i+1)[/tex]

EDIT: Also, I think we need [itex]\partial: V_{n} \to V_{n+1}[/itex] for [itex]\partial(f)(n)[/itex] to be defined.

I don't really know how I am supposed to approach that. In general, I know how to show that a function is linear, which is to show that ##f(\alpha \cdot x) = \alpha \cdot f(x)## and ##f(x_1 + x_2) = f(x_1) + f(x_2)##. However, for this specific function, I have no idea, since there is nothing provided about ##f##, so if I wanted to show the multiplicative property, I couldn't just drag anything out of ##f## without loss of generality. So I could really use some help to figure this out.

You are trying to show that [itex]\partial : V_{n+1} \to V_n[/itex] is linear, ie. [itex]\partial(f_1 + f_2) = \partial (f_1) + \partial (f_2)[/itex] for all [itex]f_1, f_2 \in V_{n+1}[/itex] and [itex]\partial(\alpha f) = \alpha \partial (f)[/itex] for every [itex]\alpha \in K[/itex] and every [itex]f \in V_{n+1}[/itex].

The operations on [itex]V_{n+1}[/itex] and [itex]V_n[/itex] are pointwise: [itex](f_1 + f_2)(x) = f_1(x) + f_2(x)[/itex], [itex](\alpha f)(x) = \alpha f(x)[/itex].

 

[PhysicsRock]

Should this not be [tex]
\partial(f) : i \mapsto (i+1)_K \cdot f(i+1)[/tex]
You are trying to show that [itex]\partial : V_{n+1} \to V_n[/itex] is linear, ie. [itex]\partial(f_1 + f_2) = \partial (f_1) + \partial (f_2)[/itex] for all [itex]f_1, f_2 \in V_{n+1}[/itex] and [itex]\partial(\alpha f) = \alpha \partial (f)[/itex] for every [itex]\alpha \in K[/itex] and every [itex]f \in V_{n+1}[/itex].

The operations on [itex]V_{n+1}[/itex] and [itex]V_n[/itex] are pointwise: [itex](f_1 + f_2)(x) = f_1(x) + f_2(x)[/itex], [itex](\alpha f)(x) = \alpha f(x)[/itex].

Yes, it's supposed to be ##\partial(f)##. Thank you, I forgot to state that. This helps.

 

[PhysicsRock]

Should this not be [tex]
\partial(f) : i \mapsto (i+1)_K \cdot f(i+1)[/tex]

EDIT: Also, I think we need [itex]\partial: V_{n} \to V_{n+1}[/itex] for [itex]\partial(f)(n)[/itex] to be defined.
You are trying to show that [itex]\partial : V_{n+1} \to V_n[/itex] is linear, ie. [itex]\partial(f_1 + f_2) = \partial (f_1) + \partial (f_2)[/itex] for all [itex]f_1, f_2 \in V_{n+1}[/itex] and [itex]\partial(\alpha f) = \alpha \partial (f)[/itex] for every [itex]\alpha \in K[/itex] and every [itex]f \in V_{n+1}[/itex].

The operations on [itex]V_{n+1}[/itex] and [itex]V_n[/itex] are pointwise: [itex](f_1 + f_2)(x) = f_1(x) + f_2(x)[/itex], [itex](\alpha f)(x) = \alpha f(x)[/itex].

But one thing I don't understand: how do you know that ##f## obeys the requirements of a linear function?

 

[pasmith]

But one thing I don't understand: how do you know that ##f## obeys the requirements of a linear function?

[itex]f[/itex] doesn't have to be linear; its domain is not even a vector space!

 

[PhysicsRock]

EDIT: Also, I think we need ##\partial : V_n \rightarrow V_{n+1}## for ##\partial(f)(n)## to be defined.

I thought about that too. The assignment though says clearly what I stated in the problem description.

 

[pasmith]

The problem statement asks you to show that [itex]\partial[/itex] is linear; it says nothing about [itex]f[/itex], other than that it is an arbitrary vector in [itex]V_n[/itex]. Perhaps the fact that you used it in your statement of the definition of a linear function is confusing you.

 

[PhysicsRock]

The problem statement asks you to show that [itex]\partial[/itex] is linear; it says nothing about [itex]f[/itex], other than that it is an arbitrary vector in [itex]V_n[/itex]. Perhaps the fact that you used it in your statement of the definition of a linear function is confusing you.

I pretty much only translated the assignment. I felt like including everything was helpful to receive the needed aid.

 

[PeroK]

I pretty much only translated the assignment. I felt like including everything was helpful to receive the needed aid.

When you have something as abstract as this, you need to get a grip on what everything is.

I would have called ##\partial## an operator that maps functions in ##V_{n+1}## to functions in ##V_n##.

For ##f \in V_{n+1}##, what does ##d = \partial(f)## look like?

We know that ##d \in V_n## must be defined for ##i = 0, 1 \dots n##. Working from the definition:
$$d(0) = 1_Kf(1)$$$$d(1) = 2_Kf(2)$$$$\dots d(n) = (n+1)_Kf(n+1)$$Now that we see how ##\partial## transforms functions the fun can begin!

Let ##f, g \in V_{n+1}## and ##a \in K##. We need to show that:
$$\partial(f +g) = \partial(f) + \partial(g)$$$$\partial(af) = a\partial(f)$$Perhaps that's more help than I should have given you. It seems to me that the material you're studying exceeds your current capability to think in terms of abstract mathematics.