Generating a vector space via a T-cyclic subspace

[Bipolarity]

I've been thinking about a problem I made up. The solution may be trivial or very difficult as I have not given too much thought to it, but I can't think of an answer of the top of my head.

Let $T:V → V$ be a linear operator on a finite-dimensional vector space $V$. Does there exist a vector $v \in V$ for which the T-cyclic subspace of $V$ generated by $v$ is $V$? This is certainly not true in general, since if $T$ is the zero transformation and $V$ has dimension greater than 1 then no T-cyclic subspace will equal $V$.

But what about for an arbitrary linear map?

BiP

 

[brmath]

Let's just look at a 2-dim space. Let T = $$\begin{pmatrix} 0 & 1 \\ 1 & 0\\ \end{pmatrix}$$
and v = (1,0). Then Tv = (0,1). Clearly v and Tv are a basis for the entire space.

 

[Bipolarity]

Let's just look at a 2-dim space. Let T = $$\begin{pmatrix} 0 & 1 \\ 1 & 0\\ \end{pmatrix}$$
and v = (1,0). Then Tv = (0,1). Clearly v and Tv are a basis for the entire space.

How can we generalize this for an arbitrary linear map?

BiP

 

[brmath]

Assuming we are in a finite dimensional vector space, of dimension n, you need to pick a T such that $$T^n= T$$ but if 1 < m < n $$T^m \neq T.$$
If you do have $$T^m = T$$ for 1 < m < n then the T-cyclic subspace has only dimension m, because that is all the independent vectors you can generate with powers of T.

Now there is bound to be some specific vector v such that $$T^mv = Tv$$ even though $$T^m \neq T.$$ That vector is in fact any eigenvector of $$T^m - T.$$ You can't generate the entire space V using that kind of vector v. However, there is bound to be a vector u which is not an eigenvector of $$T^m - T$$ for any 1 < m < n. You can use u to generate the entire space.

Does every n dimensional vector space have a linear operator which is cyclic of degree n? Yes. Start with my 2 dimensional example and see if you can find something similar in 3 dimensions. Once you get that far, you'll see how to generate such a T in n dimensions.

 

[blue_raver22]

olarBear,

This is an interesting question to consider. In general, the answer to whether there exists a vector $v \in V$ for which the T-cyclic subspace of $V$ generated by $v$ is $V$ will depend on the specific properties of the linear operator $T$ and the vector space $V$.

One approach to tackling this problem would be to consider the eigenvalues and eigenvectors of $T$. If $T$ has a full set of distinct eigenvalues, then it is possible to find a vector $v$ that generates a T-cyclic subspace that is equal to $V$. This is because the eigenvectors corresponding to distinct eigenvalues are linearly independent, and thus span the entire vector space $V$.

However, if $T$ has repeated eigenvalues or if the eigenvalues are not distinct, then it may not be possible to find a vector $v$ that generates a T-cyclic subspace that is equal to $V$. In this case, the T-cyclic subspace generated by $v$ will be a proper subspace of $V$.

Another approach to consider would be to examine the structure of the vector space $V$ itself. If $V$ has a basis consisting of eigenvectors of $T$, then it is possible to find a vector $v$ that generates a T-cyclic subspace that is equal to $V$. However, if $V$ does not have such a basis, then it may not be possible to find such a vector $v$.

Overall, the existence of a vector $v$ for which the T-cyclic subspace generated by $v$ is equal to $V$ will depend on the specific properties of the linear operator $T$ and the vector space $V$. Further investigation and analysis would be needed to determine the necessary and sufficient conditions for such a vector to exist.