Vector space, basis, linear operator

[boombaby]

Homework Statement

Let V be a vector space of dimension n. And the linear operators E=A^0, A^1, A^2, ... A^(n-1) are linearly independent. Prove that there exists a v in V such that V=<v, Av, A^2v, ..., A^(n-1)v>

Homework Equations

The Attempt at a Solution

Here are something that I tried.
the degree of the minimal polynomials p(t) such that p(A)=0 is larger than n-1. I wanted to start the proof from here but have no idea how to proceed.
assume V is over complex field C so that there is an eigenvector. However, it seems that this is also not the disired v to make them independent.
Any hint? Thanks a lot

 

[morphism]

Are you using <,> to denote span? If so, then it suffices to find a v such that {v, Av, ..., A^n-1v} is linearly independent. Stated differently, we want to make sure that whenever k₀, k₁, ..., k_n-1 are scalars that are not all zero, then

$$k_0 + k_1 Av + \cdots + k_{n-1} A^{n-1}v \neq 0$$.

Or, equivalently,

$$(k_0 + k_1 A + \cdots + k_{n-1} A^{n-1})v \neq 0 \iff v \not\in \ker(k_0 + k_1 A + \cdots + k_{n-1} A^{n-1})$$.

But what can you say about ker(k₀+ k₁ A + ... + k_n-1 A^n-1)?

 

[boombaby]

thanks for the hints! Here's what I think,

so k₀E+k₁A+...+k_n-1A^n-1 (called $$B_{k_{1},k_{2},..,k_{n-1}}$$)cannot be the 0 operator (since E, A, ..,A^n-1 are linearly independent), which implies that $$dim(Ker\! B_{k_{1},k_{2},..,k_{n-1}}) \leq n-1$$

(**) suppose we have V₁, V₂ as two proper sub space of V, then we can find a vector in V which is neither in V₁ nor V₂
(assume we cannot, find two vectors such that $$x\in V_{1} \: x\notin V_{2} \: y\in V_{2} \: y\notin V_{1}$$ x+y will give a contration)

If we have only finitely many $$B_{k_{1},k_{2},..,k_{n-1}}$$ I can find a v such that $$B_{k_{1},k_{2},..,k_{n-1}}v\neq 0 \: for\,all\: B_{k_{1},k_{2},..,k_{n-1}}$$ according to (**). Does this remain true if there are countably many or uncountably many subspace? I've no idea...

Am I thinking right?...

Thanks

 

[HallsofIvy]

What is your definition of "linearly independent" for linear operators?

 

[morphism]

Presumably linear independence refers to independence in the vector space L(V) of linear operators on V.

If we have only finitely many $$B_{k_{1},k_{2},..,k_{n-1}}$$ I can find a v such that $$B_{k_{1},k_{2},..,k_{n-1}}v\neq 0 \: for\,all\: B_{k_{1},k_{2},..,k_{n-1}}$$ according to (**). Does this remain true if there are countably many or uncountably many subspace?

No, but fortunately we only need to consider finitely many $B_{k_{1},k_{2},..,k_{n-1}}$. Consider the irreducible factors of the minimal polynomial of T.

 

[boombaby]

it's the same as other things I think...All the linear operators on V over field K forms a vector space, called L(V), and A_1,...,A_s in L(V) are linearly independent iff k_1*A_1+...k_s*A_s=0 implies k_1=...=k_s=0...

Edit:
I'll have a look at the irreducible factors...Thanks

 

[boombaby]

can you explain it a little more? I can not figure it out...
Thanks

 

[morphism]

You're going to want to use the rational canonical decomposition from here.

 

[boombaby]

Ah...it seems that there are lots of materials under the topic "rational canonical" that I cannot understand quite fast right now. I'll back to this question later. Anyway, Thanks a lot for your help!

 

[HallsofIvy]

it's the same as other things I think...All the linear operators on V over field K forms a vector space, called L(V), and A_1,...,A_s in L(V) are linearly independent iff k_1*A_1+...k_s*A_s=0 implies k_1=...=k_s=0...

Which is the same, then, as saying that if v is any non-zero vector, {A_1v, A_2v, ..., A_s v} are independent in V.

 

[boombaby]

Which is the same, then, as saying that if v is any non-zero vector, {A_1v, A_2v, ..., A_s v} are independent in V.

No, this is not true. if some A_i v=0, then they are not independent.