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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.

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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. – Jonas Meyer May 17 '11 at 4:27
@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. – Fabian May 17 '11 at 7:15
up vote 2 down vote accepted

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

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I corrected the question. Are matrices which are not diagonalizable, but where all the singular values are below 1 called a contraction? – Fabian May 17 '11 at 7:14
@Fabian: They're contractions in the sense of 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. – Jonas Meyer May 17 '11 at 7:59
@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. – Fabian May 17 '11 at 8:02
@Fabian: Thank you also, Fabian, for clarifying. It was no trouble. – Jonas Meyer May 17 '11 at 8:09

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