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?
1 Answer
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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 $T\widehat{A} = AT$.
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$\begingroup$ Realized after posting solution, it is just a matter of changing the basis. $\endgroup$ Commented Feb 6, 2013 at 2:41