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

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Linearity
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
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??
 
Physics news on Phys.org
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 $xe$.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 40 ·
2
Replies
40
Views
3K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K