Eigenvectors of tridiagonal matrix that is similar to a diagonal matrix

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An $n times n$ diagonal matrix $M$ and an $n times n$ tridiagonal matrix $J$ are given. Suppose there is an orthogonal matrix $Q$ such that $Q^TJQ = M$. It has been said that

The eigenvectors of $J$ are the columns of $Q$

I am new to linear algebra and I can't understand why this statement is true. I would be appreciated if you explain this.

---

Update

The symmetric matrix $A$ and orthogonal matrix $Q$ are given as follow
$$
A = beginpmatrix1 & 2 & 2 newline 2 & 1 & 2 newline 2 & 2 & 1 endpmatrix
$$
$$
Q = beginpmatrix-1/sqrt 2 & -1/sqrt 6 & 1/sqrt 3 newline 1/sqrt 2 & -1/sqrt 6 & 1/sqrt 3 newline 0 & sqrt 2/3 & 1/sqrt 3 endpmatrix
$$
The $Q^TAQ = $diag$(-1,-1,5) = M$. I expect the columns of $Q$ be the eigenvectors of $A$. The eigenvector matrix of $A$ is:

$$
E_vec = beginpmatrix0.6206 & 0.5306 & 0.5774 newline 0.1492 & -0.8027 & 0.5774 newline -0.7698 & 0.2722 & 0.5774 endpmatrix
$$
and $Q$ in double format is
$$
Q = beginpmatrix-0.7071 & -0.4082 & 0.5774 newline 0.7071 & -0.4082 & 0.5774 newline 0 & 0.8165 & 0.5774 endpmatrix
$$

what is wrong here? the $E_vec$ and $Q$ must be the same while they are not

linear-algebra eigenvalues-eigenvectors

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edited Jul 25 at 16:07

asked Jul 24 at 5:36

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up vote
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favorite

An $n times n$ diagonal matrix $M$ and an $n times n$ tridiagonal matrix $J$ are given. Suppose there is an orthogonal matrix $Q$ such that $Q^TJQ = M$. It has been said that

The eigenvectors of $J$ are the columns of $Q$

I am new to linear algebra and I can't understand why this statement is true. I would be appreciated if you explain this.

---

Update

The symmetric matrix $A$ and orthogonal matrix $Q$ are given as follow
$$
A = beginpmatrix1 & 2 & 2 newline 2 & 1 & 2 newline 2 & 2 & 1 endpmatrix
$$
$$
Q = beginpmatrix-1/sqrt 2 & -1/sqrt 6 & 1/sqrt 3 newline 1/sqrt 2 & -1/sqrt 6 & 1/sqrt 3 newline 0 & sqrt 2/3 & 1/sqrt 3 endpmatrix
$$
The $Q^TAQ = $diag$(-1,-1,5) = M$. I expect the columns of $Q$ be the eigenvectors of $A$. The eigenvector matrix of $A$ is:

$$
E_vec = beginpmatrix0.6206 & 0.5306 & 0.5774 newline 0.1492 & -0.8027 & 0.5774 newline -0.7698 & 0.2722 & 0.5774 endpmatrix
$$
and $Q$ in double format is
$$
Q = beginpmatrix-0.7071 & -0.4082 & 0.5774 newline 0.7071 & -0.4082 & 0.5774 newline 0 & 0.8165 & 0.5774 endpmatrix
$$

what is wrong here? the $E_vec$ and $Q$ must be the same while they are not

linear-algebra eigenvalues-eigenvectors

share|cite|improve this question

edited Jul 25 at 16:07

asked Jul 24 at 5:36

Drupalist

410218

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

An $n times n$ diagonal matrix $M$ and an $n times n$ tridiagonal matrix $J$ are given. Suppose there is an orthogonal matrix $Q$ such that $Q^TJQ = M$. It has been said that

The eigenvectors of $J$ are the columns of $Q$

I am new to linear algebra and I can't understand why this statement is true. I would be appreciated if you explain this.

---

Update

The symmetric matrix $A$ and orthogonal matrix $Q$ are given as follow
$$
A = beginpmatrix1 & 2 & 2 newline 2 & 1 & 2 newline 2 & 2 & 1 endpmatrix
$$
$$
Q = beginpmatrix-1/sqrt 2 & -1/sqrt 6 & 1/sqrt 3 newline 1/sqrt 2 & -1/sqrt 6 & 1/sqrt 3 newline 0 & sqrt 2/3 & 1/sqrt 3 endpmatrix
$$
The $Q^TAQ = $diag$(-1,-1,5) = M$. I expect the columns of $Q$ be the eigenvectors of $A$. The eigenvector matrix of $A$ is:

$$
E_vec = beginpmatrix0.6206 & 0.5306 & 0.5774 newline 0.1492 & -0.8027 & 0.5774 newline -0.7698 & 0.2722 & 0.5774 endpmatrix
$$
and $Q$ in double format is
$$
Q = beginpmatrix-0.7071 & -0.4082 & 0.5774 newline 0.7071 & -0.4082 & 0.5774 newline 0 & 0.8165 & 0.5774 endpmatrix
$$

what is wrong here? the $E_vec$ and $Q$ must be the same while they are not

linear-algebra eigenvalues-eigenvectors

share|cite|improve this question

edited Jul 25 at 16:07

asked Jul 24 at 5:36

Drupalist

410218

An $n times n$ diagonal matrix $M$ and an $n times n$ tridiagonal matrix $J$ are given. Suppose there is an orthogonal matrix $Q$ such that $Q^TJQ = M$. It has been said that

The eigenvectors of $J$ are the columns of $Q$

I am new to linear algebra and I can't understand why this statement is true. I would be appreciated if you explain this.

---

Update

The symmetric matrix $A$ and orthogonal matrix $Q$ are given as follow
$$
A = beginpmatrix1 & 2 & 2 newline 2 & 1 & 2 newline 2 & 2 & 1 endpmatrix
$$
$$
Q = beginpmatrix-1/sqrt 2 & -1/sqrt 6 & 1/sqrt 3 newline 1/sqrt 2 & -1/sqrt 6 & 1/sqrt 3 newline 0 & sqrt 2/3 & 1/sqrt 3 endpmatrix
$$
The $Q^TAQ = $diag$(-1,-1,5) = M$. I expect the columns of $Q$ be the eigenvectors of $A$. The eigenvector matrix of $A$ is:

$$
E_vec = beginpmatrix0.6206 & 0.5306 & 0.5774 newline 0.1492 & -0.8027 & 0.5774 newline -0.7698 & 0.2722 & 0.5774 endpmatrix
$$
and $Q$ in double format is
$$
Q = beginpmatrix-0.7071 & -0.4082 & 0.5774 newline 0.7071 & -0.4082 & 0.5774 newline 0 & 0.8165 & 0.5774 endpmatrix
$$

what is wrong here? the $E_vec$ and $Q$ must be the same while they are not

linear-algebra eigenvalues-eigenvectors

share|cite|improve this question

edited Jul 25 at 16:07

asked Jul 24 at 5:36

Drupalist

410218

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 16:07

edited Jul 25 at 16:07

edited Jul 25 at 16:07

asked Jul 24 at 5:36

Drupalist

410218

asked Jul 24 at 5:36

Drupalist

410218

asked Jul 24 at 5:36

Drupalist

410218

410218

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

There are some unnecessary conditions, the statement is much more general.
If $Q$ is orthogonal, then $Q^T=Q^-1$.
So in fact $Q^-1JQ=M$ is a diagonal matrix.

The point is that given ANY $ntimes n$ matrices $S, A, D$ such that $S$ is invertible and $D=S^-1AS$ is diagonal, then the eigenvectors of $A$ are the columns of $S$. To understand this, check out basis transformation:
https://en.wikipedia.org/wiki/Change_of_basis

share|cite|improve this answer

answered Jul 24 at 5:52

A. Pongrácz

1,829116

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I updated the question. Please take a look at it. Thanks
  – Drupalist
  Jul 25 at 16:07
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

There are some unnecessary conditions, the statement is much more general.
If $Q$ is orthogonal, then $Q^T=Q^-1$.
So in fact $Q^-1JQ=M$ is a diagonal matrix.

The point is that given ANY $ntimes n$ matrices $S, A, D$ such that $S$ is invertible and $D=S^-1AS$ is diagonal, then the eigenvectors of $A$ are the columns of $S$. To understand this, check out basis transformation:
https://en.wikipedia.org/wiki/Change_of_basis

share|cite|improve this answer

answered Jul 24 at 5:52

A. Pongrácz

1,829116

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I updated the question. Please take a look at it. Thanks
  – Drupalist
  Jul 25 at 16:07
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

There are some unnecessary conditions, the statement is much more general.
If $Q$ is orthogonal, then $Q^T=Q^-1$.
So in fact $Q^-1JQ=M$ is a diagonal matrix.

The point is that given ANY $ntimes n$ matrices $S, A, D$ such that $S$ is invertible and $D=S^-1AS$ is diagonal, then the eigenvectors of $A$ are the columns of $S$. To understand this, check out basis transformation:
https://en.wikipedia.org/wiki/Change_of_basis

share|cite|improve this answer

answered Jul 24 at 5:52

A. Pongrácz

1,829116

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I updated the question. Please take a look at it. Thanks
  – Drupalist
  Jul 25 at 16:07
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

There are some unnecessary conditions, the statement is much more general.
If $Q$ is orthogonal, then $Q^T=Q^-1$.
So in fact $Q^-1JQ=M$ is a diagonal matrix.

The point is that given ANY $ntimes n$ matrices $S, A, D$ such that $S$ is invertible and $D=S^-1AS$ is diagonal, then the eigenvectors of $A$ are the columns of $S$. To understand this, check out basis transformation:
https://en.wikipedia.org/wiki/Change_of_basis

share|cite|improve this answer

answered Jul 24 at 5:52

A. Pongrácz

1,829116

There are some unnecessary conditions, the statement is much more general.
If $Q$ is orthogonal, then $Q^T=Q^-1$.
So in fact $Q^-1JQ=M$ is a diagonal matrix.

The point is that given ANY $ntimes n$ matrices $S, A, D$ such that $S$ is invertible and $D=S^-1AS$ is diagonal, then the eigenvectors of $A$ are the columns of $S$. To understand this, check out basis transformation:
https://en.wikipedia.org/wiki/Change_of_basis

share|cite|improve this answer

answered Jul 24 at 5:52

A. Pongrácz

1,829116

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 5:52

A. Pongrácz

1,829116

answered Jul 24 at 5:52

A. Pongrácz

1,829116

answered Jul 24 at 5:52

A. Pongrácz

1,829116

1,829116

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I updated the question. Please take a look at it. Thanks
  – Drupalist
  Jul 25 at 16:07
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I updated the question. Please take a look at it. Thanks
  – Drupalist
  Jul 25 at 16:07
  

  
  
  
  

I updated the question. Please take a look at it. Thanks
– Drupalist
Jul 25 at 16:07

I updated the question. Please take a look at it. Thanks
– Drupalist
Jul 25 at 16:07

add a comment | 

 

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