All Eigenvalues Lie on the Unit Circle

[ali987]

Hi everyone

Consider a 2x2 partitioned matrix as follow:

A = [ B1 B2 ; B3 B4 ]

I'm sure that all eigenvalues of A are on the unit circle (i.e., abs
(all eig) = 1 ). but, I don't know how to prove it. Is there any
theorem?

 

[arkajad]

Is it, by chance, a unitary matrix: $$AA^\dag=I$$?

 

[Fredrik]

$$\begin{pmatrix}2 & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix}1 \\ 0\end{pmatrix}=2\begin{pmatrix}1\\ 0\end{pmatrix}$$

 

[HallsofIvy]

Hi everyone

Consider a 2x2 partitioned matrix as follow:

A = [ B1 B2 ; B3 B4 ]

I'm sure that all eigenvalues of A are on the unit circle (i.e., abs
(all eig) = 1 ). but, I don't know how to prove it. Is there any
theorem?

You can't prove it- it's not true! Fredrik gave an example in which the eigenvalue is 2, not on the unit circle.

Now, is there some condition on A, such as $$AA^\dag=I$$, as arkajad suggests, that you haven't told us?

 

[ali987]

Unfortunately, there is no special condition on A.
B1, B2, B3 and B4 are constructed from several matrices themselves, some of those matrices are symmetric and positive definite.
Is there any theorem which relate orthogonality of the A to orthogonality (or sth like that) of B1...B4?

 

[arkajad]

Is there any theorem which relate orthogonality of the A to orthogonality (or sth like that) of B1...B4?

No. What is important are the interrelations of your block matrices. Write $$AA^\dag$$ in the block matrix form and you will see. Is your matrix normal: $$AA^\dag=A^\dag A$$? If you do not know even this - then what do you know about your B matrices and relations between them? The key to your question is there.