# Follow-up question regarding a cyclic basis with computations in linear algebra

This is a follow-up homework question(s) to a conceptual question I had asked earlier with regard to the cyclic basis of the vector space $V$ denoted by $\mathcal{B}= \{v,Av,…,A^{k-1}v\}$, where $A \in M_n(F)$, and $v \in F^n$. Also, there is a linear transformation $T: V \rightarrow V$ which is basically matrix multiplication by $A$, i.e. $T(v)= Av$. The question is as follows:

We are asked to find the explicit basis $\mathcal{B}$ for the module $V = F[x]v$, and the minimal polynomial $f= \min_{T}(x)$ for $T$ given specific choices for $A$ and $v$. We are also asked to find the coordinate matrix $[T]_{\mathcal{B}}$.

$1$. $A = \begin{bmatrix}12&2&3&-7\\ -5&2&-2&4\\6&1&5&-5\\13&3&4&-7\end{bmatrix}$, $F= \mathbb{Q}$, $v = \begin{bmatrix} 0\\ -1\\ -1\\ \ -1 \end{bmatrix}.$

$2$. $A= \begin{bmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{bmatrix}$, $0< \theta < 2\pi$, $F= \mathbb{R}$, $v= e_1$.

$3$. $A= \begin{bmatrix} 2&0&0\\ 0&3&0\\ 0&0&4 \end{bmatrix}$, $F= \mathbb{Q}$, $v= e_1+ e_2$.

$4$. $A= \begin{bmatrix} 1&1\\ 1&0 \end{bmatrix}$, $F= \mathbb{Z}_2$, $v= e_1$.

Well, for all of these, I computed the bases via matrix multiplication by $A, A^2,…, A^{k-1}$ with $v$ first. Is that the right approach? Does finding the minimal polynomial $f$ of the transformation $T$ require finding the characteristic polynomial of $A$ (which I already found), but what next? Finally, is the coordinate matrix $[T]_\mathcal{B}$ the companion matrix $C(f)$ of $f$?

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I think for the third problem, the minimal polynomial is just the product $(x-2)(x-3)= x^2 -5x +6$ since we're dealing in two dimensions. – Libertron Oct 5 '12 at 22:51

For the first part:

the matrix $A$ is an $n \times n$ matrix. As you said you form the vectors $\{ v,A\cdot v,\ldots,A^i\cdot v\}$ but you have to stop in that minimal $i\leq n$ such that $A^{i+1}\cdot v$ is a linear combination of $\{ v,A\cdot v,\ldots,A^i\cdot v\}$. Your $k$ is my $i-1$.

For the other parts:

yes the coordinate matrix $[T]_\mathcal{B}$ is the companion matrix $C(f)$ of $f$. We explained how to calculate the matrix $[T]_\mathcal{B}$ in your question Some really tricky concepts that need to be cleared up in linear algebra so, that and the first part of this question can help you find the minimal polynomial $f$ of $T$.

e.g. for $3.$ we have:

$A=\begin{pmatrix} 2&0&0\\ 0&3&0\\ 0&0&4 \end{pmatrix}$ and $v=\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}$ so

$$v=\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix} \ , Av=\begin{pmatrix} 2\\ 3\\ 0 \end{pmatrix} \ , A^2v=\begin{pmatrix} 4\\ 9\\ 0 \end{pmatrix}$$

and $A^2v=(-6)v+5Av \ .$

Thus $[T]_\mathcal{B}=\begin{pmatrix} 0&-6\\ 1&5 \end{pmatrix}$.

This is the companion matrix of $f$ so (as you correctly noticed) $$f(x)=x^2-5x+6=(x-3)(x-2) \ .$$

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