# Show that v, ø(v), ø(v)^2 are independent

1. Feb 8, 2014

### victoranderson

I can finish part (a)

For part b, how can I find ø(v) ?
Although I can find ø(v1) and ø(v2) but I think it is unrelated to ø(v)....

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2. Feb 8, 2014

### Dick

Part b isn't really related to the first part. If you assume $v, \phi(v), \phi^2(v)$ are linearly DEPENDENT what does that tell you about them? What's the definition of linear dependence? Try to use that to reach a contradiction to what you know about the vectors. If you can show one of them must be zero then you've shown they are linearly independent.

3. Feb 8, 2014

### victoranderson

(b)

This is my solution. I do not know if it is correct.
Assume v, ø(v) and ø(v)^2 are linearly dependent

$$=> a_{1}v+a_{2}\phi (v)+a_{3}\phi (v)^{2} = 0$$
for $$a_{1}, a_{2}, a_{3}$$ not all zero

since $$\phi (v)^{2} \neq 0$$ we have $$a_{3}=0$$

Contradiction, so v, ø(v) and ø(v)^2 are linearly dependent

Last edited: Feb 8, 2014
4. Feb 8, 2014

### victoranderson

I have another question about part c.
Why column 1 is M^2*v? How can we know?

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5. Feb 8, 2014

### Dick

That's not correct at all. You did get the definition of linear dependence right. So you have $a_{1}v+a_{2}\phi (v)+a_{3}\phi^2 (v) = 0$ (notice I wrote $\phi^2 (v)$ not $\phi (v)^2$ that's what you really want). It's worth trying to take it slowly and get this right. It's pretty basic. $\phi^2 (v) \ne 0$ doesn't imply anything about $a_3$. I'll give you a hint. What happens if you apply $\phi^2$ to that equation? What does that tell you about $a_1$?

Last edited: Feb 8, 2014
6. Feb 8, 2014

### Dick

It's much better to post new questions in new threads.

7. Feb 11, 2014

### victoranderson

I am a beginner in this topic...I know the trick is to add phi to the equation
is this a correct proof?

Assume v, ø(v) and ø(v)^2 are linearly dependent

$$=> a_{1}v+a_{2}\phi (v)+a_{3}\phi ^{2}(v) = 0$$
for $$a_{1}, a_{2}, a_{3}$$ not all zero

$$=> \phi^2( a_{1}v+a_{2}\phi (v)+a_{3}\phi ^{2}(v)) = 0$$

since $$\phi^2 \neq 0$$

8. Feb 11, 2014

### Dick

Nope, not correct. $\phi^2( a_{1}v+a_{2}\phi (v)+a_{3}\phi ^{2}(v)) = a_1 \phi^2(v) +a_2 \phi^3(v) + a_3 \phi^4(v)$, what can you say about $\phi^3(v)$ and $\phi^4(v)$??

9. Feb 11, 2014

### victoranderson

In my opinion, $\phi^3(v)$ = 0 => $\phi^4(v)$ = 0

so, $a_1 \phi^2(v) +a_2 \phi^3(v) + a_3 \phi^4(v)$=0 implies $a_1 \phi^2(v)$ = 0

which gives $a_1=0$

10. Feb 11, 2014

### Dick

It's not a contradiction yet. Now put $a_1=0$ into $a_{1}v+a_{2}\phi (v)+a_{3}\phi ^{2}(v) = 0$. Any ideas what to do next?

11. Feb 11, 2014

### victoranderson

Thanks dick, i know the trick now.
So this is my complete proof of part (b), please have a look.

Assume v, $\phi (v)$ and $\phi ^{2}(v)$ are linearly dependent

i.e $a_{1}v+a_{2}\phi (v)+a_{3}\phi^2 (v) = 0$ for $a_{1}, a_{2}, a_{3}$ not all zero

$\phi^2( a_{1}v+a_{2}\phi (v)+a_{3}\phi ^{2}(v)) = 0$

=> $a_1 \phi^2(v) +a_2 \phi^3(v) + a_3 \phi^4(v) = 0$

Since $\phi^3(v)$ = 0 => $\phi^4(v)$ = 0, we have $a_{1}=0$

So the equation $a_{1}v+a_{2}\phi (v)+a_{3}\phi ^{2}(v) = 0$ gives $a_{2}\phi (v)+a_{3}\phi ^{2}(v) = 0$

$\phi (a_{2}\phi (v)+a_{3}\phi ^{2}(v)) = 0$

=> $a_{2}\phi ^2(v)+a_{3}\phi ^{3}(v) = 0$

and once again, since $\phi^3(v)= 0$ and $\phi^2 \neq 0$

we have $a_{2}=0$

so the equation $a_{1}v+a_{2}\phi (v)+a_{3}\phi ^{2}(v) = 0$ gives $a_{3}\phi ^{2}(v) = 0$

since $\phi^2 \neq 0$ we have $a_{3}=0$

This leads to contradiction, since we assume $a_{1}, a_{2}, a_{3}$ not all zero

so v, $\phi (v)$ and $\phi ^{2}(v)$ are linearly INDEPENDENT

12. Feb 11, 2014

### Dick

That's perfect.

13. Feb 11, 2014

### LCKurtz

Almost. Correction in red. It isn't $\phi^2$ that isn't zero, it's $\phi^2(v)$ that isn't.