0
$\begingroup$

V is a vectorspace, $\sigma, \tau$ are two functions that are $ \mathbb{K}$-linear. Both functions map into their own Vectorspace $\sigma, \tau: V \rightarrow V.$ Proof $\sigma \circ $ $\tau $ is $\mathbb{K}$-linear under these conditions.

First of all, this exercise is a little bit weird, I guess im supposed to learn the definition of $\mathbb{K}$-linearity

Intuitively it is obvious: If two function map a vector into the same vectorspace and one function takes the output of the other function, the outcoming vector will stay the same. Now how do I proof this?

My attempt:

First the definitions:

$\sigma $ is $ \mathbb{K}$-linear: $\sigma(\lambda u+\mu v) = \lambda \sigma(u) + \mu \sigma(v)$

$\tau $ is $ \mathbb{K}$-linear: $\tau(\alpha w + \beta x) = \alpha \tau(w) + \beta \tau(x)$

Now with $\circ$ and respect, that the output of $\sigma$ is a vector, that serves as input for $\tau$ and $\tau$ being $\mathbb{K}$-linear:

$\tau(\alpha(\sigma(\lambda u+ \mu v)) + \beta (\sigma(\lambda u+ \mu v))) = \alpha \tau (\sigma(\lambda u + \mu v)) + \beta \tau (\sigma (\lambda u + \mu v))$

2 Answers 2

2

let $\xi=\sigma\circ\tau$

the linearity of $\xi$ requires: $$ \xi(\lambda u+\mu v) = \lambda \xi(u) + \mu \xi(v) $$ now $$ \xi(\lambda u+\mu v) = \sigma\circ\tau (\lambda u+\mu v)=\\ \sigma(\tau(\lambda u+\mu v))=\sigma(\lambda\tau(u)+\mu \tau(v))=\\ \lambda\sigma(\tau(u))+\mu\sigma(\tau(v)) =\\ \lambda\sigma\circ\tau(u)+\mu\sigma\circ\tau(v) =\\ \lambda\xi(u)+\mu\xi(v) $$

  • 0
    Ah, now I understand, taking $\xi$ as a function on it's own and proving $\mathbb{K}$-linearity for it, thanks! – 2017-01-27
  • 0
    yes, the result is so basic as to be almost a matter of "syntax", which is why it seemed "obvious" to you. introducing $\xi$ makes it a bit easier to see what is going on. i wrote it out as fully as possible for clarity, but in practice you could miss out the intermediate steps – 2017-01-27
1

All you need to prove is that $$\sigma \circ \tau (\alpha u + \beta v) = \alpha \, \sigma \circ \tau(u) + \beta \, \sigma \circ \tau(v) \quad \forall u,v \in V.$$ To show this, use the linearity of $\tau$ and $\sigma$: $$ \begin{align} \sigma \circ \tau (\alpha u + \beta v) &= \sigma(\alpha \, \tau(u) + \beta \, \tau(v)) \\ &= \alpha \, \sigma(\tau(u)) + \beta \, \sigma(\tau(v)). \end{align} $$