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I am a physicist and very much used to the fact that any self-adjoint matrix ($H^{\dagger} =H$) in a finite-dimensional complex linear space can be uniquely specified by (a) the set of its (real) eigenvalues, and (b) the unitary matrix built from its (orthonormal) eigenvectors:

$$H = U^{\dagger} \cdot \rm{diag}\{ h_1, h_2, \ldots h_n \} \cdot U$$

where $(\cdot)^{\dagger} \equiv (\cdot)^{*T}$ denotes conjugate transpose.

I need a generalization of this for the classes of symmetric ($S^{T} =S$) and anti-symmetic ($A=-A^{T}$) complex matrices. The symmetric case seems easy:

$$S = U^{T}\cdot \rm{diag}\{ s_1, s_2, \ldots s_n \} \cdot U$$ where the singular values $s_1, s_2 \dots$ are non-negative reals and $U$ again is a generic the unitary matrix. (Here is a more precise statement, accounting for the extra choice of signs).

But I have a difficulty identifying the general singular value decomposition structure for an arbitrary anti-symmetirc matrix. I've observed numerically that the rank of $A$ is at most $n-1$ EDIT: when $n$ is odd ($n$ is the dimensionality of the linear space), so generalization needs to be 'clever'.

Can you help?

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Slava Kashcheyevs
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    $\pmatrix{0&-1\ 1&0}$ has full rank. Odd-dimensional complex skew-symmetric matrices, however, are known to have deficient ranks. – user1551 Jan 03 '14 at 16:35
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    One possible generalisation is that $A=U^TDU$ where $D$ is a direct sum of real multiples of $\pmatrix{0&-1\ 1&0}$ and the zero matrix, so that all nonzero singular values of $A$ occur in pairs. – user1551 Jan 03 '14 at 16:47
  • @user1551 Thank you! What you suggest looks to be equivalent to Youla decomposition which I've just found. – Slava Kashcheyevs Jan 03 '14 at 17:39
  • Here is more info on what @user1551 suggested http://en.wikipedia.org/wiki/Skew-symmetric_matrix#Spectral_theory – tom Jan 03 '14 at 17:51
  • @Slaviks Yes and no. Yes, the result can be derived from the Youla form. No, Youla form is a canonical form that can be formally written as $A=URU^T$, where $U$ is a unitary matrix and $R$ is some block triangular matrix whose diagonal blocks have some special strutures. It is something like the Schur decomposition, but using congruence instead of similarity in the decomposition. When $A$ is complex symmetric, the resulting Youla decomposition can be chosen to be something known as a Takagi decomposition. However, when $A$ is skew symmetric, the decomposition has no name, as far as I know. – user1551 Jan 03 '14 at 17:53
  • @user1551 In the introduction to Youla's 1964 paper (http://cms.math.ca/10.4153/CJM-1961-059-8) (form tom's wikipedia link), both Takagi decomposition and your suggested form are quoted "classical" and "canonical". Is there specific technical meaning in calling these forms "canonical"? – Slava Kashcheyevs Jan 03 '14 at 18:41
  • @Slaviks "Classical" refers to the result itself. The decomposition in the cases of symmetric matrices (due to Schur) and skew-symmetric matrices (due to L.K. Hua) were known long before Youla's paper. What wasn't known before Youla, however, is the result for a general complex matrix, and the condition under which the Youla decomposition is "canonic/canonical". – user1551 Jan 04 '14 at 12:07
  • A matrix form is called canonical if it is unique up to some sort of equivalence. E.g. the Schur form in Schur decomposition is not canonical: the strictly upper triangular part is not unique in general; in contrast, the Jordan form is canonical: the Jordan form of a complex matrix is unique, up to permutation of Jordan blocks along the main diagonal. – user1551 Jan 04 '14 at 12:07
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    I was mistaken in a previous comment that the Youla form is canonical. According to Youla, it is only canonical under certain conditions (as specified in the corollary on p.703). But these conditions are satisfied in particular when $A$ is complex skew-symmetric. So, if $A=-A^T$ and $A=U\Sigma U^T$ is a Youla decomposition, then positive multiplying factors for the 2x2 blocks $\pmatrix{0&-1\ 1&0}$ and the unit 1x1 blocks in $\Sigma$ are uniquely detemined. This is not surprising, as the multiplying factors are the singular values of $A$. – user1551 Jan 04 '14 at 12:08
  • Much useful! Can you put this as an answer? – Raffaele C Dec 21 '17 at 15:10

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