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

Question

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!

Answer 1

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 $\mathbb{R}^4 = \mathbb{R}^2 \oplus \mathbb{R}^2$.