Showing Linearity of $\varphi$ for $K(a)$

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Discussion Overview

The discussion revolves around demonstrating the linearity of the map $\varphi : K(a) \rightarrow K(a)$ defined by $\varphi(e) = ae$, where $K \leq K(a)$ is a field extension with $[K(a):K]=n$. Participants are also exploring the characteristic polynomial of $\varphi$ and its properties.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants propose verifying the linearity of $\varphi$ by checking the condition $\varphi(te + e') = t\varphi(e) + \varphi(e')$ for all $t \in K$ and $e, e' \in K(a)$.
  • One participant inquires about the characteristic polynomial of $\varphi$, suggesting that $a$ is a root of this polynomial.
  • Another participant states that the characteristic polynomial is given by $p(x) = \text{det}(x1_{K(a)} - \varphi)$, explaining that the map $x1_{K(a)} - \varphi$ sends $e \in K(a)$ to $xe - ae$.
  • A participant asks for clarification on what $x1$ means, leading to an explanation that $x1_{K(a)}$ sends $e \in K(a)$ to $xe$.

Areas of Agreement / Disagreement

Participants generally agree on the steps needed to show the linearity of $\varphi$ and the definition of the characteristic polynomial, but there is no consensus on the implications or further details regarding these concepts.

Contextual Notes

Unresolved aspects include the specific properties of the map $\varphi$ and the implications of $a$ being a root of the characteristic polynomial, as well as the notation used in the context of the discussion.

mathmari
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Hey! :o

Let $K \leq K(a)$ a field extension with $[K(a):K]=n$.

$K(a)$ is a vector space over $K$.

How can I show that the map $\varphi : K(a) \rightarrow K(a)$, with $\varphi(e)=ae$, is a $K-$linear map??
 
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mathmari said:
Hey! :o

Let $K \leq K(a)$ a field extension with $[K(a):K]=n$.

$K(a)$ is a vector space over $K$.

How can I show that the map $\varphi : K(a) \rightarrow K(a)$, with $\varphi(e)=ae$, is a $K-$linear map??

Hi mathmari,

To show that $\varphi$ is $K$-linear, you must verify

$$ \varphi(te + e') = t\varphi(e) + \varphi(e') \quad \text{for all} \quad t\in K, \, e, e'\in K(a).$$
 
Euge said:
Hi mathmari,

To show that $\varphi$ is $K$-linear, you must verify

$$ \varphi(te + e') = t\varphi(e) + \varphi(e') \quad \text{for all} \quad t\in K, \, e, e'\in K(a).$$

I understand! Thank you! (Sun)

I have also an other question...

I am also asked to show that $a$ is a root of the characteristic polynomial of $\varphi$.

Which is the characteristic polynomial of $\varphi$?? (Wondering)
 
mathmari said:
I understand! Thank you! (Sun)

I have also an other question...

I am also asked to show that $a$ is a root of the characteristic polynomial of $\varphi$.

Which is the characteristic polynomial of $\varphi$?? (Wondering)

It's $p(x) = \text{det}(x1_{K(a)} - \varphi)$. The map $x1_{K(a)} - \varphi$ sends $e \in K(a)$ to $xe - ae$.
 
Euge said:
It's $p(x) = \text{det}(x1_{K(a)} - \varphi)$. The map $x1_{K(a)} - \varphi$ sends $e \in K(a)$ to $xe - ae$.

Could you explain me what $x1$ is ?? (Wondering)
 
mathmari said:
Could you explain me what $x1$ is ?? (Wondering)

The map $x1_{K(a)}$ sends $e\in K(a)$ to $xe$.
 

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