Prove that the polynomials $P_n$ has n distinct root for all n. $P_n$ are characteristic polynomials of a particular type of matrix.

$$
P_n=-xA_{n-1}-\frac{1}{2}A_{n-2}\
A_n = \left( \dfrac{-x+\sqrt{x^2-1}}{2}\right)^n + \left(\dfrac{-x-\sqrt{x^2-1}}{2}\right)^n
$$

$A_n$ satisfies the following recurrence relation:
$$
A_n=-x A_{n-1}-\frac{1}{4}A_{n-2}\
A_1=-x\
A_2=x^2-\frac{1}{2}
$$

Question

Prove that the polynomials $P_n$ has n distinct root for all n.  $P_n$ are characteristic polynomials of a particular type of matrix.

$$
P_n=-xA_{n-1}-\frac{1}{2}A_{n-2}\
A_n = \left(  \dfrac{-x+\sqrt{x^2-1}}{2}\right)^n +   \left(\dfrac{-x-\sqrt{x^2-1}}{2}\right)^n
$$

$A_n$ satisfies the following recurrence relation:
$$
A_n=-x A_{n-1}-\frac{1}{4}A_{n-2}\
A_1=-x\
A_2=x^2-\frac{1}{2}
$$

anonymous · Answer

im not sure if we need to find th characteristic  coefficient and show they are distinct

anonymous · Answer

@zzr0ck3r

zzr0ck3r · Answer

no idea

thomas5267 · Answer

I want to prove these matrices are diagonalisable.
$$
M_5=\begin{pmatrix}
0&\frac{1}{2}&0&0&0\
1&0&\frac{1}{2}&0&0\
0&\frac{1}{2}&0&\frac{1}{2}&0\
0&0&\frac{1}{2}&0&1\
0&0&0&\frac{1}{2}&0
\end{pmatrix}\

M_7=\begin{pmatrix}
0&\frac{1}{2}&0&0&0&0&0\
1&0&\frac{1}{2}&0&0&0&0\
0&\frac{1}{2}&0&\frac{1}{2}&0&0&0\
0&0&\frac{1}{2}&0&\frac{1}{2}&0&0\
0&0&0&\frac{1}{2}&0&\frac{1}{2}&0\
0&0&0&0&\frac{1}{2}&0&1\
0&0&0&0&0&\frac{1}{2}&0
\end{pmatrix}\
$$
In order words, $(M_n)_{i,i+1}=(M_n)_{i+1,i}=\frac{1}{2}$ except for $(M_n)_{2,1}=(M_n)_{n-1,n}=1$

thomas5267 · Answer

The old question has gotten so convoluted I figured that I will post a new one.

zzr0ck3r · Answer

This is just an algorithm...

anonymous · Answer

seems diagonalizable , but look at the diagonal zero which might have zero and  its determinant might be  zero in both and not being singular not sure though i can't help. maybe @Empty

zzr0ck3r · Answer

find the eigen values and the eigen vectors

thomas5267 · Answer

As far as I can check, from $M_3$ to $M_{15}$ the matrices all have distinct eigenvalues.

zzr0ck3r · Answer

ahh you are trying to show for the whole family?

thomas5267 · Answer

Yes.  Favard's theorem seems to apply in here.  Those polynomials should be orthogonal.

anonymous · Answer

well, there is a reason why i hate orthogonality. i'll watch and learn!

anonymous · Answer

@dan815  @Kainui @Empty  when ever your free just lets try on this!

thomas5267 · Answer

I know nothing about Favard's theorem but I found this seemingly useful theorem on google.

thomas5267 · Answer

I have no idea whether $A_n$ is orthogonal or not.  It does not seem like the case.

empty · Answer

Can you use the Cayley Hamilton theorem? It says that every matrix satisfies its characteristic equation.

thomas5267 · Answer

What I want to prove is that the matrices $M_n$ is always diagonalisable.  A sufficient but not necessary condition is that the characteristic polynomials of $M_n$ have n distinct roots.  Not sure how Cayley Hamilton Theorem will help though.