$\alpha$ is a column vector with n entries, prove $\{\alpha, A\alpha , A^2\alpha,\cdots, A^{n-1}\alpha\}$ linearly independent

Question

the question is that

Suppose $\alpha $ is a column vector with n entries and $\alpha$ is not zero vector, then there exists square matrix $A$ (n by n) such that {$\alpha, A\alpha , A^2\alpha,\cdots, A^{n-1}\alpha$} are linearly independent.

What I have known so far
Whether we can construct a special basis for $\mathbb{R}^n$ which looks like {$\alpha, b_1, b_2, \cdots, b_{n-1}$} and a mapping T from $\mathbb{R}^n$ to $\mathbb{R}^n$ such that, $T(\alpha) = b_1, \ \ T(b_1) = b_2, \ \ T(b_2) = b_3, \ \ \cdots, \ \ T(b_{n-2}) = b_{n-1}$
By construction this mapping, we can say each result of the mapping is linear independent.

But the problem is how to transfer this mapping to a particular matrix?

Answer 1

This answer is just the extension of the LinearAlgebruh's comment.
I describe it for $4 \times 4$ matrix, similar procedure can be made for other dimensions.

Let $\alpha=\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}$,

At least one of components of $\alpha$ is non-zero.
WLOG let it be $a$.

We can extend $\alpha$ to the set of $4$ vectors and compose from them full rank matrix $B$

$B=\begin{bmatrix} a & 0 & 0 & 0 \\ b & 1 & 0 & 0 \\ c & 0 & 1 & 0\\ d & 0 & 0 & 1 \end{bmatrix}$,

We have ${B^{-1}B=I}$ and $B^{-1}\alpha=e_1$ .

Now we can take permutation matrix $P$

$P=\begin{bmatrix} 0& 0& 0 & 1 \\ 1 & 0 & 0& 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{bmatrix}$,

With such matrix we have:

$Pe_1=e_2$,
$Pe_2=e_3$,
$Pe_3=e_4$,
$Pe_4=e_1$

So the searched matrix $A=BPB^{-1}$.

Let us see what we have now:
$\alpha$,

$(BPB^{-1})\alpha=BPe_1=Be_2=e_2$,

$(BPB^{-1})^2\alpha=BP^2B^{-1}\alpha=BPPe_1=BPe_2=Be_3=e_3$,

$(BPB^{-1})^3\alpha=BP^3B^{-1}\alpha=BPPPe_1=BP^2e_2=BPe_3=Be_4=e_4$

and we have $\alpha , e_2, e_3, e_4 $ which are linearly independent.

Notice that any full rank matrix $B$ which has the first column as the vector $\alpha$ does the same job with permutation matrix $P$ because then $B^{-1}$ sends the vector $\alpha$ to column vector of the identity matrix, but undoubtedly presented matrix $B$ is one of the easiest to construct.