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Solving linear system using the invariance of the trace and determinant

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0

$begingroup$

We have the system of equations
$$
v_k+1 = A v_k, quad forall k geq 1,
$$
where $mu leq 1$, $v_k = left( beginarrayc u_k-1 \ u_k endarray right)$, and $ A = left[ beginarraycc
0 & 1 \ 1 & i2mu endarray right]$.

Let $lambda_+$ and $lambda_-$ the eigenvalues of $A$. We want to prove that if $lambda_+ neq lambda_-$, then

$$ u_n = left( fraclambda_- lambda_+^n - lambda_+ lambda_-^nlambda_- - lambda_+ right) u_0 + left(fraclambda_-^n - lambda_+^nlambda_- - lambda_+ right) u_1.
$$

The given instructions say to do it making use of (*)"the invariance of the trace and determinant of $A$", but I don't see how to use these properties in this case.

I can nonetheless solve this problem by:

• Find the associatied eigenvectors $w_+$ and $w_-$ of $lambda_+$ and $lambda_-$ respectively, note that they are not orthogonal.


• Find $alpha, beta in mathbbC$ such that $v_1 = alpha w_+ + beta w_-$.


• By linearity, and eigenvectoricity, apply the matrix enough times since
  $$ A(alpha w_+ + beta w_- ) = alpha lambda_+ w_+ + beta lambda_- w_-. $$
  


• Rearrange terms.

My question is about an alternative (and possibly easier/faster way to solve this problem). How to use the instruction * ?

linear-algebra

share|cite|improve this question

edited 13 hours ago

pancho

asked 13 hours ago

panchopancho

531210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Mixing superscripts and exponents is not a good idea.
  $endgroup$
  – Yves Daoust
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I've fixed it.
  $endgroup$
  – pancho
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is the $i$ in font of $2mu$ intended ?
  $endgroup$
  – Yves Daoust
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes, else the matrix is diagonalizable.
  $endgroup$
  – pancho
  13 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

We have the system of equations
$$
v_k+1 = A v_k, quad forall k geq 1,
$$
where $mu leq 1$, $v_k = left( beginarrayc u_k-1 \ u_k endarray right)$, and $ A = left[ beginarraycc
0 & 1 \ 1 & i2mu endarray right]$.

Let $lambda_+$ and $lambda_-$ the eigenvalues of $A$. We want to prove that if $lambda_+ neq lambda_-$, then

$$ u_n = left( fraclambda_- lambda_+^n - lambda_+ lambda_-^nlambda_- - lambda_+ right) u_0 + left(fraclambda_-^n - lambda_+^nlambda_- - lambda_+ right) u_1.
$$

The given instructions say to do it making use of (*)"the invariance of the trace and determinant of $A$", but I don't see how to use these properties in this case.

I can nonetheless solve this problem by:

• Find the associatied eigenvectors $w_+$ and $w_-$ of $lambda_+$ and $lambda_-$ respectively, note that they are not orthogonal.


• Find $alpha, beta in mathbbC$ such that $v_1 = alpha w_+ + beta w_-$.


• By linearity, and eigenvectoricity, apply the matrix enough times since
  $$ A(alpha w_+ + beta w_- ) = alpha lambda_+ w_+ + beta lambda_- w_-. $$
  


• Rearrange terms.

My question is about an alternative (and possibly easier/faster way to solve this problem). How to use the instruction * ?

linear-algebra

share|cite|improve this question

edited 13 hours ago

pancho

asked 13 hours ago

panchopancho

531210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Mixing superscripts and exponents is not a good idea.
  $endgroup$
  – Yves Daoust
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I've fixed it.
  $endgroup$
  – pancho
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is the $i$ in font of $2mu$ intended ?
  $endgroup$
  – Yves Daoust
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes, else the matrix is diagonalizable.
  $endgroup$
  – pancho
  13 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

We have the system of equations
$$
v_k+1 = A v_k, quad forall k geq 1,
$$
where $mu leq 1$, $v_k = left( beginarrayc u_k-1 \ u_k endarray right)$, and $ A = left[ beginarraycc
0 & 1 \ 1 & i2mu endarray right]$.

Let $lambda_+$ and $lambda_-$ the eigenvalues of $A$. We want to prove that if $lambda_+ neq lambda_-$, then

$$ u_n = left( fraclambda_- lambda_+^n - lambda_+ lambda_-^nlambda_- - lambda_+ right) u_0 + left(fraclambda_-^n - lambda_+^nlambda_- - lambda_+ right) u_1.
$$

The given instructions say to do it making use of (*)"the invariance of the trace and determinant of $A$", but I don't see how to use these properties in this case.

I can nonetheless solve this problem by:

• Find the associatied eigenvectors $w_+$ and $w_-$ of $lambda_+$ and $lambda_-$ respectively, note that they are not orthogonal.


• Find $alpha, beta in mathbbC$ such that $v_1 = alpha w_+ + beta w_-$.


• By linearity, and eigenvectoricity, apply the matrix enough times since
  $$ A(alpha w_+ + beta w_- ) = alpha lambda_+ w_+ + beta lambda_- w_-. $$
  


• Rearrange terms.

My question is about an alternative (and possibly easier/faster way to solve this problem). How to use the instruction * ?

linear-algebra

share|cite|improve this question

edited 13 hours ago

pancho

asked 13 hours ago

panchopancho

531210

$endgroup$

share|cite|improve this question

edited 13 hours ago

pancho

asked 13 hours ago

panchopancho

531210

share|cite|improve this question

edited 13 hours ago

pancho

asked 13 hours ago

panchopancho

531210

asked 13 hours ago

panchopancho

531210

asked 13 hours ago

panchopancho

531210

1 Answer
1

active

oldest

votes

1

$begingroup$

The matrix equation boils down to

$$u_k+1=2imu,u_k+u_k-1,$$ which is a simple second order recurrence. The solution is a linear combination of the powers of the characteristic roots found from

$$u^2-2imu,u-1=0$$ and the coefficients are solution of

$$begincasesc_++c_-=u_0,\c_+u_++c_uu_-=u_1.endcases$$

---

Hint for using trace and determinant:

By induction,

$$v_n=A^nv_0.$$

The characteristic equation of $A$ is

$$lambda^2-textTr(A)lambda+textDet(A)=lambda^2-2imu,lambda-1=0,$$

so that the characteristic equation of $A^n$ is

$$lambda^2-(2imu)^n,lambda+(-1)^n=0.$$

Then the Eigenvectors of $A^n$ are the same as those of $A$.

share|cite|improve this answer

edited 11 hours ago

answered 13 hours ago

Yves DaoustYves Daoust

129k676227

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, thank you. Somehow it catches me out of guard, as recurrence solutions is not part of the course. But it does indeed answers the question.
  $endgroup$
  – pancho
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @pancho: still missing a way to find the Eigenvectors of $A^n$.
  $endgroup$
  – Yves Daoust
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Shouldn't them be the same ones as those of $A$ with eigenvalues $lambda_pm^n$?
  $endgroup$
  – pancho
  11 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @pancho: stupid me, that's right.
  $endgroup$
  – Yves Daoust
  11 hours ago
  

  
  
  
  

add a comment | 

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1

$begingroup$

The matrix equation boils down to

$$u_k+1=2imu,u_k+u_k-1,$$ which is a simple second order recurrence. The solution is a linear combination of the powers of the characteristic roots found from

$$u^2-2imu,u-1=0$$ and the coefficients are solution of

$$begincasesc_++c_-=u_0,\c_+u_++c_uu_-=u_1.endcases$$

---

Hint for using trace and determinant:

By induction,

$$v_n=A^nv_0.$$

The characteristic equation of $A$ is

$$lambda^2-textTr(A)lambda+textDet(A)=lambda^2-2imu,lambda-1=0,$$

so that the characteristic equation of $A^n$ is

$$lambda^2-(2imu)^n,lambda+(-1)^n=0.$$

Then the Eigenvectors of $A^n$ are the same as those of $A$.

share|cite|improve this answer

edited 11 hours ago

answered 13 hours ago

Yves DaoustYves Daoust

129k676227

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, thank you. Somehow it catches me out of guard, as recurrence solutions is not part of the course. But it does indeed answers the question.
  $endgroup$
  – pancho
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @pancho: still missing a way to find the Eigenvectors of $A^n$.
  $endgroup$
  – Yves Daoust
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Shouldn't them be the same ones as those of $A$ with eigenvalues $lambda_pm^n$?
  $endgroup$
  – pancho
  11 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @pancho: stupid me, that's right.
  $endgroup$
  – Yves Daoust
  11 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

The matrix equation boils down to

$$u_k+1=2imu,u_k+u_k-1,$$ which is a simple second order recurrence. The solution is a linear combination of the powers of the characteristic roots found from

$$u^2-2imu,u-1=0$$ and the coefficients are solution of

$$begincasesc_++c_-=u_0,\c_+u_++c_uu_-=u_1.endcases$$

---

Hint for using trace and determinant:

By induction,

$$v_n=A^nv_0.$$

The characteristic equation of $A$ is

$$lambda^2-textTr(A)lambda+textDet(A)=lambda^2-2imu,lambda-1=0,$$

so that the characteristic equation of $A^n$ is

$$lambda^2-(2imu)^n,lambda+(-1)^n=0.$$

Then the Eigenvectors of $A^n$ are the same as those of $A$.

share|cite|improve this answer

edited 11 hours ago

answered 13 hours ago

Yves DaoustYves Daoust

129k676227

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, thank you. Somehow it catches me out of guard, as recurrence solutions is not part of the course. But it does indeed answers the question.
  $endgroup$
  – pancho
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @pancho: still missing a way to find the Eigenvectors of $A^n$.
  $endgroup$
  – Yves Daoust
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Shouldn't them be the same ones as those of $A$ with eigenvalues $lambda_pm^n$?
  $endgroup$
  – pancho
  11 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @pancho: stupid me, that's right.
  $endgroup$
  – Yves Daoust
  11 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

The matrix equation boils down to

$$u_k+1=2imu,u_k+u_k-1,$$ which is a simple second order recurrence. The solution is a linear combination of the powers of the characteristic roots found from

$$u^2-2imu,u-1=0$$ and the coefficients are solution of

$$begincasesc_++c_-=u_0,\c_+u_++c_uu_-=u_1.endcases$$

---

Hint for using trace and determinant:

By induction,

$$v_n=A^nv_0.$$

The characteristic equation of $A$ is

$$lambda^2-textTr(A)lambda+textDet(A)=lambda^2-2imu,lambda-1=0,$$

so that the characteristic equation of $A^n$ is

$$lambda^2-(2imu)^n,lambda+(-1)^n=0.$$

Then the Eigenvectors of $A^n$ are the same as those of $A$.

share|cite|improve this answer

edited 11 hours ago

answered 13 hours ago

Yves DaoustYves Daoust

129k676227

$endgroup$

share|cite|improve this answer

edited 11 hours ago

answered 13 hours ago

Yves DaoustYves Daoust

129k676227

answered 13 hours ago

Yves DaoustYves Daoust

129k676227

answered 13 hours ago

Yves DaoustYves Daoust

129k676227