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Let $T$ be a linear operator on ${\mathbb R}^3$, which is represented by standard ordered basis, as follows: $$T(e_1) = 2 e_1,$$

$$T(e_2) = 2 e_2,$$

$$T( e_3) = -e_3.$$

I have to prove that $T$ has no cyclic vector. Also, to find what is the $T$-cyclic subspace generated by the $(1, -1, 3 )$ ?

Please suggest !

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If a matrix is brought into diagonal (or triangular) form, it's cyclic vectors are easy to detect: it is necessary and sufficient that for each distinct eigenvalue the projection of the vector on the generalised eigenspace be a cyclic vector for the subspace, where projecting in this case means extracting the appropriate coordinates. Your example is quite easy in that the matrix is already diagonal. Only multiple eigenvalues need attention. –  Marc van Leeuwen Feb 27 '12 at 11:23

1 Answer 1

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Minimal polynomial for $T$ is $(x-2)(x+1)$, but characteristic polynomial is $(x-2)^2(x+1)$. Both are not same. Hence $T$ doesn't have cyclic vector by following theorem:

Theorem: $T$ be a linear operator on vector space $V$ of $n$ dimensional. There exists a cyclic vector for T if and only if minimal polynomial and characteristic polynomial are same....

Proof:Suppose there exists a cyclic vector $v$ for $T$, that is we have $v\in V$ such that $\{v, Tv,..T^{n-1}v\}$ span $V$. Then matrix representation of $T$ will be some companion matrix , whose minimal and characteristic polynomial are same.

Now conversely, if minimal and charteristic polynomial are same, then we have minimal polynomial is of degree $n$. Take $v\neq 0 $, let minimal polynomial $p(x)= a_0+a_1x+...+a_{n-1}x^{n-1}$. Degree of $p(x)$ is equal to the cyclic subspace generate by $p(v)$ consider $\{v, Tv, T^2v,.. T^{n-1}v\}$, where $T$ is annihilator linear operator for $p(x), v$ (following Hoffman-Kunze). This is a cyclic base. Read section 7.1 in Linear algebra by Hoffman-Kunze.

Also we have $T(x,y,z)= (2x,2y,-z)$ Hence $T(1,-1,3)= (2,-2,-3)$ and $T^2(1,-1,3)= (4,-4,3)$. We have (4,-4,3) is linear combination of $(1,-1,3)$ and $(2,-2,3)$. Hence T-cyclic subspace generated by the $(1,−1,3)$ = Linear span of $\{(1,-1,3), (2,-2,-3)\}$

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Thanks a lot for the help. –  preeti Feb 27 '12 at 11:52
    
@zapkm $+1$, could u tell me which book did u get this theorem? –  Une Femme Douce May 10 '13 at 18:07

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