2
$\begingroup$

I'm wondering if these matrices have a name? (I'm somehow tempted to call them subunitary but it seems to be reserved for something else.)

The matrix $M \in \mathbb{C}^{n\times n}$ is called ..., if all the singular values $\sigma_1,\dots, \sigma_n$ are strictly smaller than 1.

Note that if all the singular values are 1 then $M$ is a unitary matrix. That is why I think it should be called subunitary.

The matrix $M \in \mathbb{C}^{n\times n}$ is called ..., if it fulfills $$M = \alpha U$$ with $|\alpha|<1, \alpha \in \mathbb{C}$ and $U \in U(n)$ ($U(n)$ is the group of unitary matrices). (maybe it has a name if one allows $|\alpha|=1$) It seems like this space is closed under multiplication. But obviously it is not a group as the $0$ matrix is not invertible.

  • 1
    I am confused about why the answer below was accepted, when it does not appear to answer the question. I don't know of a name other than (strictly) contractive scalar multiples of unitary matrices.2011-05-17
  • 0
    @Jonas Meyer: I accepted the answer, because it actually corresponds to the answer which I was searching for. In the beginning, I thought my matrices are of the form $\alpha U$. Closer inspection did show, that the eigenvalues in my application do in fact have different norms (in the mean time I figured out that my matrices might not be invertible at all). I will edit the question with the exact definition of the matrices whose name I'm searching for. I will inform Yuval of the change.2011-05-17

1 Answers 1

2

If you allow the eigenvalues to have different norms, all at most $1$, then it's a contraction.

  • 0
    I corrected the question. Are matrices which are not diagonalizable, but where all the singular values are below 1 called a contraction?2011-05-17
  • 1
    @Fabian: They're contractions in the sense of http://en.wikipedia.org/wiki/Contraction_%28operator_theory%29 if they act on $\mathbb{C}^n$ with Euclidean norm. They would be the set of operators with norm strictly less than one, sometimes called strict contractions.2011-05-17
  • 0
    @Jonas Meyer: thank you, Jonas, for clarifying. In fact, I have a Euclidean norm (the matrices appear in a scattering application in physics and the norm is the flux of particles -> the contraction then indicates that particles are lost). Sorry for the confusion and the trouble.2011-05-17
  • 1
    @Fabian: Thank you also, Fabian, for clarifying. It was no trouble.2011-05-17