Let $F$ be a field , $A$ $\in$ $M_{n\times n} $($F$) (the set of all $n\times n$ matrices over field $F$ ) and $W$ = { $ B \in M_{n\times n } (F)$ $|$ $AB = BA$ }. Suppose there exists a column vector $v \in F^n$ such that { $v$ , $Av$ , $A^2 v$ , ... , $A^{n - 1} v$ } is a basis for $F^n$
a) Prove that $I$ , $A$ , $A^2$ ,...., $A^{n-1}$ are linearly independent vectors contained in $W$
b) Prove that {$I$ , $A$ , $A^2$ ,...., $A^{n-1}$} is a basis of W
I used proof by contradiction to solve part a. But I'm stuck at part b.
I know W is a subspace of $M_{n×n}(F)$. Also, since we already know the set {$I$ , $A$ , $A^2$ ,...., $A^{n-1}$} is linearly independent. I need to show that {$I$ , $A$ , $A^2$ ,...., $A^{n-1}$} spans W to prove that it is a basis. Its easy to prove that if $x \in$ span{$I$ , $A$ , $A^2$ ,...., $A^{n-1}$} then $x \in W$. But I am unable to prove the converse to complete the proof.
I have tried the following approach. Let $B \in W$
One can create the columns of B and solve the problem. Since { $v$ , $Av$ , $A^2 v$ , ... , $A^{n - 1} v$ } is a basis for $F^n$ , We know there exists linear combinations such that
$a_0v + a_1 A v + a_2 A^2 v + ... +a_{n-1} A^{n-1}v$ = $B e_1$ - (1st column of B)
$b_0v + b_1 A v + b_2 A^2 v + ... +b_{n-1} A^{n-1}v$ = $B e_2$ - (2nd column of B)
$c_0v + c_1 A v + c_2 A^2 v + ... +c_{n-1} A^{n-1}v$ = $B e_3$ - (3rd column of B)
so on... till n columns
where $e_i$ represents the $i$th the column of the identity Matrix
I am trying to create matrix B by [$Be_1$ $Be_2$ ... $Be_n$] and solve the problem. But I can't seem to be able to bring {$I$ , $A$ , $A^2$ ,...., $A^{n-1}$} out. Thank you for the help!
Since \begin{equation*}\tag{basis} v, Av, A^2 v, \dots , A^{n - 1} v \end{equation*} is a basis for $F^{n}$, any matrix is determined by its action on (basis).
Now let $B \in W$. Since $B$ commutes with $A$, we have $$B (A^{i} v) = A^{i} (B v),$$ that is, the action of $B$ on the whole (basis) is determined by its action on $v$ alone. The same clearly holds for any linear combination $C$ of powers of $A$, as such $C$ commute with $A$, and thus also lie in $W$.
So if \begin{equation*} B v = c_{0} v + c_{1} A v + \dots + c_{n-1} A^{n-1} v = (c_{0} I + c_{1} A + \dots + c_{n-1} A^{n-1}) v, \end{equation*} we have then \begin{equation*}\tag{equality} B = c_{0} I + c_{1} A + \dots + c_{n-1} A^{n-1}, \end{equation*} as LHS and RHS of (equality) agree on $v$, and both sides commute with $A$, so their action on $v$ alone determines them completely.