Consider the transform $\widehat{b}=T^{-1}b$, where $T=\begin{bmatrix}b & Ab & A^2b & \dots & A^{n-1}b \end{bmatrix}$ has full rank. Is it possible to find an explicit expression for 
\begin{equation}
\widehat{b}=\begin{bmatrix} \widehat{b}\_1 \\\ \vdots \\\ \widehat{b}\_n \end{bmatrix}
\end{equation}
in terms of the coefficients for $det(I\lambda-A)=det(I\lambda-\widehat{A})=a\_0+a\_1\lambda+\dots+a\_{n-1}\lambda^{n-1}+\lambda^n$. Further, is it possible to derive an explicit expresion for $\widehat{A}=T^{-1}AT$ in terms of these coefficients?



Linear algebra and Cayley Hamilton
Let $T^{(i)}=A^{i-1}b$ for $i=1,\cdots n$. 
You can solve $\widehat{b}$ from the equation $T\widehat{b}=b$. 
By Cramer's rule, 


$$
\widehat{b\_1}=\frac{det[b, T^{(2)}, \cdots, T^{(n)}]} {det T}
$$
$$
\widehat{b\_2}=\frac{det[b, b, T^{(3)}, \cdots, T^{(n)}]} {det T}
$$
$\cdots$
$$
\widehat{b\_n}=\frac{det[b, T^{(2)}, \cdots, T^{(n-1)},b]} {det T}
$$
Then you can see that 
$\widehat{b\_1}=1, \widehat{b\_i}=0$ for all $i=2,\cdots, n$. 


Further for $\widehat{A}$, use Cramer's rule to each column vector of $\widehat{A}$, in the equation $T\widehat{A} = AT$. 


