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Suppose a matrix $A\in\mathbb{R}^{n\times m}$ is given, $n>m$, with columns being subset to those of an rotation matrix (i.e., matrix with with orthonormal columns). Is it true that the sigular values of $A$ are all equal? Furthermore, does that imply that, in case the columns of $A$ are not orthonormal, the singular values are not equal?

A side note: Is every orthogonal matrix a rotation/reflection matrix? (suppose that a rotation matrix is the one that preserves isometry, i.e., distances.)

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up vote 1 down vote accepted

If $A$ has orthonormal columns, then $A^TA = I$ (the identity matrix). Since the singular values are the eigenvalues of $A^TA$, they are all equal to one.

On the other hand: Let $A$ have singular values equal to $c$, i.e. $$A = U\,cI\,V^T$$ with orthonormal $U$ and $V$ of the right size. Then $$A^TA = V\,cI\,U^T\,U\,cI\,V^T = c^2 I$$ which means that the rows of $A$ are orthogonal (but not normalized).

For the side note: An orthogonal matrix need not to preserve distances (e.g. $cI$ is orthogonal). An orthonormal matrix, however, does so.

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Thanks. Could you consider the small edit I've made. – user506901 Aug 28 '12 at 12:16
What is your definition of a rotation matrix? – Dirk Aug 28 '12 at 12:39
Preservation of isometry (i.e., the actual "character", without manipulating inter-point distances or angles) – user506901 Aug 28 '12 at 12:43

There are two cases where singular values are identical. The first case is for the deterministic scenerio where you have $B=AA^T$ is an identical matrix. I think if $B$ is all ones in the minor diagonal It also has identical singular values.$$\sigma_1=\sigma_2=\sigma_2,...,=\sigma_N$$

Basically from a given matrix; to get a matrix which has identical singular values, we need to manipulate the given matrix such that the minimum singular value is maximized. If we find a solution then


the general case will be satisfied with equality. Such a matrix can be as I said the identitiy matrix which has full rank. However it is not the only solution. To maximize the minimum singular value gives another solutions too. The idea is that we need to remove any dependency between rows and colums of this matrix! This implies that a fully random matrix will also satisfy


It is the same thing with taking the fourier transform of a dirac delta function and the result in the frequency domain is flat. Similarly when you take the fourier transform of the white noise which is random then you also get a flat spektrum.

As singular values can be seen as the frequency domain from another point of view. In this domain we have indentity matrix $\rightarrow$ dirac delta function and random matrix $\rightarrow$ white noise

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I do not get this. First: "Random matrix" is not precise. There are different probability distribution defined on the space of matrices. Second: Most "random matrices" will not have equal singular values (just check svd(randn(n,m)) in MATLAB). – Dirk Aug 28 '12 at 19:00
@Dirk then create random noise in MATLAB and plot the spectrum. It will also not be flat. It means I am just bullshitting. I am talking about a random process and you have a single realization. – Seyhmus Güngören Aug 28 '12 at 20:04

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