Orthogonal Matrix: Properties & Conditions

[ali987]

Dear all,

I have a matrix, namely A. I calculate its eigenvalues by MATLAB and all of its eigenvalues lie on the unit circle(their amplitudes equal to 1). But A is not an orthogonal matrix (transpose(A) is not equal to inverse(A) ). What other condition or relationship may be correct for it?

 

[arkajad]

Is you matrix a real matrix or it has complex entries? Is it invertible?

 

[ali987]

It is definitely invertible. All entries are real, but eigenvalues can be complex.

 

[arkajad]

If your matrix is nxn, do you have n different eigenvalues or less than n?

 

[ali987]

I can't talk about it definitely, but, according to a numerical simulation results, I had several identical eigenvalues. so, there are less than "n" unique and different eigenvalues. But, as is mentioned, I can't extend this conclusion to general case, these are results of just a numerical example.

However, if I always have less than "n" different eigenvalues, what can I do?

 

[Fredrik]

A unitary matrix has eigenvalues on the unit circle. Not sure if "all eigenvalues on the unit circle" implies "unitary" though. I don't have time to think about that right now.

 

[ali987]

Unfortunately, that matrix is not unitary(or Orthogonal, since all entries are real).

A unitary matrix has eigenvalues on the unit circle, but if all eigenvalues of a matrix lie on the unit circle, that matrix is not definitely unitary. that's the problem.

 

[arkajad]

Take simple example

$$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$$

It has just one eigenvalue, namely 1, but it is not orthogonal.

 

[AlephZero]

Take simple example

$$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$$

It has just one eigenvalue, namely 1, but it is not orthogonal.

Or even better, take

$$A=\begin{pmatrix}1&1\\0&-1\end{pmatrix}$$

With distinct eigenvalues (1 and -1) and vectors.

It should be intuitively obvious that your hypothesis that "eigenvalues on the unit circle is sufficient" is probably wrong. There are only n eigenvalues, but orthogonality requires n² relationships to hold.

 

[arkajad]

Or even better, take

$$A=\begin{pmatrix}1&1\\0&-1\end{pmatrix}$$

With distinct eigenvalues (1 and -1) and vectors.

This matrix is nevertheless orthogonal with respect to the metric
$$\eta=\begin{pmatrix}2&1\\1&1\end{pmatrix}$$
in the sense that
$$A^T\eta A=\eta$$

 

[ali987]

Thanks every one for their useful comments, but, what about the main question?

We have a matrix (with real entries) whose all eigenvalues lie on the unit circle and that matrix is not orthogonal. What other conditions may be exist for this matrix? Or in other words, how can we proof that all eigenvalues of this matrix lie on the unit circle??