# Find a cyclic vector of $\mathbb{R}^4$ such that...

Let $$T:\mathbb{R}^4 \rightarrow \mathbb{R}^4$$ be an operator with minimal polynomial $$m_T(x) = (x-2)^2(x+1)^2$$. Find a vector $$v \in \mathbb{R}^4$$ such that $$\mathbb{R}^4 = \langle \{v,T(v),T^2(v),T^3(v)\} \rangle$$.

I've been having a hard time with this one. The first thing that I can see is that since $$\deg m_T(x) = 4 = \dim \mathbb{R}^4$$ the minimal polynomial is the same as the characteristic one $$p_T(x)$$.

After a long time trying to get somewhere, I finally saw something... If we had a minimal polynomial like $$m_T(x)=(x-\lambda_1)(x-\lambda_2)(x-\lambda_3)(x-\lambda_4)$$, then $$p_T(x)=m_T(x)$$ and it would be possible to find such $$v$$. To see that, I'll show $$v = v_1+v_2+v_3+v_4$$ where $$v_i$$ is the eigenvector associated with $$\lambda_i$$ is such a vector. Let $$a,b,c,d \in \mathbb{K}$$ where $$\mathbb{K}$$ is the field in which the vector space is defined, therefore: \begin{align*} & 0= av+bT(v)+cT^2(v) +dT^3(v) =\\ &a(v_1+v_2+v_3+v_4) +\\ &b(\lambda_1v_1+\lambda_2v_2+\lambda_3v_3+\lambda_4v_4) + \\ &c(\lambda_1^2v_1+\lambda_2^2v_2+\lambda_3^2v3+\lambda_4^2v4)+\\ &d(\lambda_1^3v_1+\lambda_2^3v_2+\lambda_3v3+\lambda_4^3v4)\\ \end{align*} since $$\{v_1,v_2,v_3,v_4\}$$ is linearly independent we can rearrange what we've got in matrix notation: $$\left(\begin{matrix} 1&\lambda_1&\lambda_1^2&\lambda_1^3\\ 1&\lambda_2&\lambda_2^2&\lambda_2^3\\ 1&\lambda_3&\lambda_3^2&\lambda_3^3\\ 1&\lambda_4&\lambda_4^2&\lambda_4^3\\ \end{matrix}\right) \left(\begin{matrix} a\\b\\c\\d \end{matrix}\right) = \left(\begin{matrix} 0\\0\\0\\0 \end{matrix}\right)$$ and that Vandermonde matrix would be invertible since all eigenvalues are distrinct, therefore $$a=b=c=d=0$$ and the set $$\{v,T(v),T^2(v),T^3(v)\}$$ would be a basis for $$\mathbb{R}^4$$.

But unfortunately, that's not the case. I supposed that case because I had a belief that It'd give me some idea on how to procced in the case of the exercise where not all eigenvalues are distinct!

One other thing that I was able to see is that $$m_T(x) = x^4 -2x^3 -3x^2 +4x +4$$, since $$m_T(T) = 0$$ then $$T^4 = 2T^3+3T^2-4T-4T^0$$ and the matrix of $$T$$ in the basis $$B=\{v,T(v),T^2(v),T^3(v)\}$$ for such a $$v \in \mathbb{R}^4\setminus\{0\}$$ would be: $$[T]_B = \left(\begin{matrix} 0&0&0&-4\\ 1&0&0&-4\\ 0&1&0&3\\ 0&0&1&2 \end{matrix}\right)$$

but I also got stuck here, because with that matrix I have information about $$T$$ but it depends of the choice of $$v$$ (because that matrix is defined in a bases defined by the choice of $$v$$), and $$v$$ is what I want to find.

I'm completely lost at the moment. Any help would be highly appreciated.

Thanks!

I may be forgetting some details of the linear algebra, but I think that from your knowledge of the minimal polynomial you can deduce that $$T$$ in some basis $$\{v_1,v_2,v_3,v_4\}$$ has its Jordan canonical form with two blocks of eigenevalues $$2$$ and $$-1$$ as follows: $$\begin{pmatrix} 2 & 1 & & \\ & 2 & & \\ & & -1 & 1\\ & & & -1 \end{pmatrix}$$ Then we seek a vector $$v=av_1+bv_2+cv_3+dv_4$$ such that the matrix
$$\begin{pmatrix} v & Tv & T^2v & T^3 v \end{pmatrix} = \begin{pmatrix} a & 2a+b & 4a+4b & 8a+12b\\ b & 2b & 4b & 8b \\ c & -c+d & c-2d & -c+3d \\ d & -d & d & -d \end{pmatrix}$$
has nonzero determinant. A quick check with Mathematica says that the determinant of this is $$81b^2d^2$$, so for simplicity we could take $$v=v_2+v_4$$ to get our solution.
Geometrically, what's happening is that $$T$$ acts simultaneously by scaling and shearing on two separate copies of $$\mathbb{R}^2$$. The vectors $$v_1$$ and $$v_3$$ don't really go anywhere on applying $$T$$ (they're just scaled), whereas $$v_2$$ and $$v_4$$ really get "pushed around" by the shearing action. So maybe it makes intuitive sense that the combination of these two would wander around enough to give a basis for $$\mathbb{R}^4 = \mathbb{R}^2 \oplus \mathbb{R}^2$$.