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

[victoranderson]

Please see attached question.
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)...

 

[Dick]

Please see attached question.
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)...

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.

 

[victoranderson]

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.

(b)
Thank you for your reply

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

 

[victoranderson]

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

Please see attached. Many thanks.

 

[Dick]

(b)
Thank you for your reply

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

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 $a_1$?

 

[Dick]

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

Please see attached. Many thanks.

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

 

[victoranderson]

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 $a_1$?

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 $$\phi^2 \neq 0$$

it leads to contradiction

 

[Dick]

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 $$\phi^2 \neq 0$$

it leads to contradiction

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 $\phi^4(v)$??

 

[victoranderson]

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 $\phi^4(v)$??

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 $a_1=0$

and it leads to contradiction??

 

[Dick]

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 $a_1=0$

and it leads to contradiction??

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?

 

[victoranderson]

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?

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

 

[Dick]

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

That's perfect.

 

[LCKurtz]

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

since $\phi^2\color\red{(v)} \neq 0$ 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

That's perfect.

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