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*{{citation|title=Symplectic Geometry|first=Carl Ludwig|last= Siegel|journal= American Journal of Mathematics|volume= 65|issue=1|year=1943|pages=1–86|jstor= 2371774|doi=10.2307/2371774|id=Lemma 1, page 12}}
*{{citation|first=L.-K.|last= Hua|title= On the theory of automorphic functions of a matrix variable I–geometric basis|journal= Amer. J. Math.|volume= 66 |issue= 3|year=1944|pages= 470–488|doi=10.2307/2371910|jstor= 2371910}}
*{{citation|first=I.|last= Schur|title= Ein Satz über quadratische
*{{citation|first1=R.|last1= Benedetti|first2=P.|last2= Cragnolini|title=On simultaneous diagonalization of one Hermitian and one symmetric form|journal= Linear Algebra Appl. |volume=57 |year=1984| pages=215–226|doi=10.1016/0024-3795(84)90189-7|doi-access=free}}
</ref> In fact, the matrix <math>B=A^{\dagger} A</math> is Hermitian and [[Definiteness of a matrix|positive semi-definite]], so there is a unitary matrix <math>V</math> such that <math>V^{\dagger} B V</math> is diagonal with non-negative real entries. Thus <math>C=V^{\mathrm T} A V</math> is complex symmetric with <math>C^{\dagger}C</math> real. Writing <math>C=X+iY</math> with <math>X</math> and <math>Y</math> real symmetric matrices, <math>C^{\dagger}C=X^2+Y^2+i(XY-YX)</math>. Thus <math>XY=YX</math>. Since <math>X</math> and <math>Y</math> commute, there is a real orthogonal matrix <math>W</math> such that both <math>W X W^{\mathrm T}</math> and <math>W Y W^{\mathrm T}</math> are diagonal. Setting <math>U=W V^{\mathrm T}</math> (a unitary matrix), the matrix <math>UAU^{\mathrm T}</math> is complex diagonal. Pre-multiplying <math>U</math> by a suitable diagonal unitary matrix (which preserves unitarity of <math>U</math>), the diagonal entries of <math>UAU^{\mathrm T}</math> can be made to be real and non-negative as desired. To construct this matrix, we express the diagonal matrix as <math>UAU^\mathrm T = \operatorname{diag}(r_1 e^{i\theta_1},r_2 e^{i\theta_2}, \dots, r_n e^{i\theta_n})</math>. The matrix we seek is simply given by <math>D = \operatorname{diag}(e^{-i\theta_1/2},e^{-i\theta_2/2}, \dots, e^{-i\theta_n/2})</math>. Clearly <math>DUAU^\mathrm TD = \operatorname{diag}(r_1, r_2, \dots, r_n)</math> as desired, so we make the modification <math>U' = DU</math>. Since their squares are the eigenvalues of <math>A^{\dagger} A</math>, they coincide with the [[singular value]]s of <math>A</math>. (Note, about the eigen-decomposition of a complex symmetric matrix <math>A</math>, the Jordan normal form of <math>A</math> may not be diagonal, therefore <math>A</math> may not be diagonalized by any similarity transformation.)
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