- #1

gtfitzpatrick

- 380

- 0

\begin{pmatrix}\alpha & \beta \\ 1 & 0 \end{pmatrix}

[/tex] has distinct eigenvalues [tex]\lambda1[/tex] and [tex]\lambda2[/tex].

If P=[tex]

\begin{pmatrix}\lambda1 & \lambda2 \\ 1 & 0 \end{pmatrix}

[/tex]

proove that P[tex]^{}^-^1[/tex]AP = D =diag{[tex]\lambda1[/tex] , [tex]\lambda2[/tex]}.

deduce that, for every positive integer m, A[tex]^{}m[/tex] = PD[tex]^{}m[/tex]P[tex]^{}^-^1)[/tex]

so i just tryed to multiply the whole lot out, (p^-1 is easy to find, just swap,change signs)

and i got

[tex]

\begin{pmatrix}\lambda1(\alpha - \lambda2) + \beta & \lambda2(\alpha - \lambda2) + \beta \\ \lambda1(-\alpha + \lambda1) - \beta & \lambda2(-\alpha + \lambda1) - \beta \end{pmatrix}

[/tex]

am i going the right road with this or should i be approaching it differently?