Eigenvalues

[gtfitzpatrick]

The real matrix A=

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


has distinct eigenvalues \lambda1 and \lambda2.
If P=

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




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

deduce that, for every positive integer m, A^{}m = PD^{}mP^{}^-^1)


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


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




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

[HallsofIvy]

Just doing the calculation should show that, but I don't get that for the calculation.

If
P= \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 0\end{bmatrix}
then
P^{-1}= \begin{bmatrix}0 & 1 \\ \frac{1}{\lambda_1} & -\frac{\lambda_1}{\lambda_2}\end{bmatrix}

Is that what you got?

You will also want to use the fact that the characteristic equation for A is x^2- \alpha x- \beta= 0 so \lambda_1^2- \alpha \lambda_1- \beta= 0 and \lambda_2^2- \alpha \lambda_2- \beta= 0.

[Horse]

Just doing the calculation should show that, but I don't get that for the calculation.

If
P= \begin{bmatrix}\lambda_1 & \lambda_2 \\ 1 & 0\end{bmatrix}
then
P^{-1}= \begin{bmatrix}0 & 1 \\ \frac{1}{\lambda_1} & -\frac{\lambda_1}{\lambda_2}\end{bmatrix}

Is that what you got?


Nice reply!


You will also want to use the fact that the characteristic equation for A is x^2- \alpha x- \beta= 0 so \lambda_1^2- \alpha \lambda_1- \beta= 0 and \lambda_2^2- \alpha \lambda_2- \beta= 0.


Perhaps, something like:

\lambda_{1} = \frac{\alpha \pm \sqrt{\alpha^{2}+4\beta}}{2}

\lambda_{2} = \frac{\alpha \pm \sqrt{\alpha^{2}+4\beta}}{2}

not sure though how it will solve the problem.

[Random Variable]

Couldn't you prove it by showing that the columns of P are the eigenvectors of A?

[HallsofIvy]

Yes, if you already have that theorem. But the obvious way to do it is to just do the product P^{-1}AP.

[gtfitzpatrick]

Thanks for all the replies. i had it wrong when getting p^-1 i only went and forgot 1/detA!!!

so for P^1AP i get
= \begin{bmatrix}\lambda_1 & \lambda_1 \\ \alpha-\lambda_1^2/\lambda_2+\alpha/\lambda_1 & \alpha - \lambda_1^2/\lambda_2\end{bmatrix}


I thought diag(\lambda_1,\lambda_2) =
D= \begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}


i still think im doing something wrong
confussed!

[gtfitzpatrick]

After repeatedly trying and not getting anywhere im after realising i had the question wrong P should read


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





So im off the try this new version.

But if i wanted to prove it like you said random variable how would i go about that?
take

\lambda_1^2- \alpha \lambda_1- \beta= 0 and
\lambda_2^2- \alpha \lambda_2- \beta= 0
and let them = columns of p?

[gtfitzpatrick]

right so now i got 1/(\lambda_1-\lambda_2) \begin{bmatrix}\lambda_1(\alpha - \lambda _2 ) + \beta & \lambda_2(\alpha - \lambda_2) + \beta \\ \lambda_1(-\alpha + \lambda_1) - \beta & \lambda_2(-\alpha + \lambda_1) - \beta\end{bmatrix}



not sure where to go from here...