# Thread: Eigenvectors of recursive set of similarity transformations

1. ## Eigenvectors of recursive set of similarity transformations

Hi!

I need some help with the following:
Consider the sequence
$A_{1} = A \\
A_{2} = P ^{-1}_{1} A_{1} P_{1} \\
A_{3} = P ^{-1}_{2} A_{2} P_{2} \\
. \\
. \\
. ... \\
A_{k+1} = P ^{-1}_{k} A_{k} P_{k} .$

1) Write an expression for the eigenvectors of $A_{k+1}$ using the eigenvectors of $A_{k}$. Then use this to write an expression for the eigenvectors of $A_{k}$ using the matrices $P_{1},..., P_{k-1}$, and the eigenvectors of $A_{1} = A$.

I am completely lost and have no idea where to start. Any pointing me in the right direction would be greatly appreciated!

j

edit: I can't get line breaks to work in the above code...(?)

2. Originally Posted by Jodles
Hi!

I need some help with the following:
Consider the sequence
$A_{1} = A \\
A_{2} = P ^{-1}_{1} A_{1} P_{1} \\
A_{3} = P ^{-1}_{2} A_{2} P_{2} \\
. \\
. \\
. ... \\
A_{k+1} = P ^{-1}_{k} A_{k} P_{k} .$

1) Write an expression for the eigenvectors of $A_{k+1}$ using the eigenvectors of $A_{k}$. Then use this to write an expression for the eigenvectors of $A_{k}$ using the matrices $P_{1},..., P_{k-1}$, and the eigenvectors of $A_{1} = A$.

I am completely lost and have no idea where to start. Any pointing me in the right direction would be greatly appreciated!

j

edit: I can't get line breaks to work in the above code...(?)

$\mbox {If v is an eigenvector of } A_k \mbox{ with eigenvalue }\lambda \,,\,\, then\; A_kv = \lambda v$. $\mbox { As P is invertible } \exists u\in V\; s.t.\; Pu=v$, and so:

$A_{k+1}(u)=P_k^{-1}A_kP_k(u)=P_k^{-1}A_k(v)=P_k^{-1}(\lambda v)=\lambda P_k^{-1}v=\lambda u$

Try to take it from here now.

Tonio

3. Tonio, thank you! You saved my day!

Here's my take on the second part ("Write an expression for the eigenvectors of $A_{k}$ using the matrices $P_{1},..., P_{k-1}$, and the eigenvectors of $A_{1} = A$).

(I'm using $v_k$ and $v_{k+1}$ instead of $v$ and $u$, respectively.)

From the first part we now have: $v_{k+1} = P_k^{-1} v_k$. Obviously $v_k$ would be $v_k = P_{k-1}^{-1} v_{k-1}$ (1) ,
and further:
$v_{k-1} = P_{k-2}^{-1} v_{k-2}$
$v_{k-2} = P_{k-3}^{-1} v_{k-3}$
... and so on...

Inserted into (1), we get:
$v_k = P_{k-1}^{-1} P_{k-2}^{-1}P_{k-3}^{-1} v_{k-3}$

Thus a pattern evolves:
$v_k = P_{k-1}^{-1} \cdot P_{k-2}^{-1} \cdot ... \cdot P_{1}^{-1} v_{1}$ (2)
where $v_1$ is the eigenvector of $A_1 = A$, i.e. $Av_1 = \lambda v_1$

Does that look somewhat alright? Something I've missed? Or better ways to write it? (especially (2))...

J

4. Originally Posted by Jodles
Tonio, thank you! You saved my day!

Here's my take on the second part ("Write an expression for the eigenvectors of $A_{k}$ using the matrices $P_{1},..., P_{k-1}$, and the eigenvectors of $A_{1} = A$).

(I'm using $v_k$ and $v_{k+1}$ instead of $v$ and $u$, respectively.)

From the first part we now have: $v_{k+1} = P_k^{-1} v_k$. Obviously $v_k$ would be $v_k = P_{k-1}^{-1} v_{k-1}$ (1) ,
and further:
$v_{k-1} = P_{k-2}^{-1} v_{k-2}$
$v_{k-2} = P_{k-3}^{-1} v_{k-3}$
... and so on...

Inserted into (1), we get:
$v_k = P_{k-1}^{-1} P_{k-2}^{-1}P_{k-3}^{-1} v_{k-3}$

Thus a pattern evolves:
$v_k = P_{k-1}^{-1} \cdot P_{k-2}^{-1} \cdot ... \cdot P_{1}^{-1} v_{1}$ (2)
where $v_1$ is the eigenvector of $A_1 = A$, i.e. $Av_1 = \lambda v_1$

Does that look somewhat alright? Something I've missed? Or better ways to write it? (especially (2))...

J

Looks fine to me though all those indexes get me dizzy. I'd do a specific example with k= 2 or 3 and see if we get the correct pattern.

Tonio

5. Hehe, alright! Thanks again!

All the best,
Joachim