For a symmetric matrix M with complex entries, I want to diagonalize it using a matrix A, such that

$AMA^T = D$, where D is a diagonal matrix with real-positive entries.

Question 1: When can this be done?

Question 2: Is $A$ unitary, i.e., is $A^\dagger A = 1$ ?

Question 3: How do I construct $A$?

The question is motivated by Majorana masses of fermions, which are complex symmetric matrices, and need to be diagonalized as above to get the physical masses. Obviously masses need to be positive and the basis-rotation by $A$ must preserve probabilities, and needs to be unitary.

  • 2
    $\begingroup$ just to avoid confusion: the decomposition $AMA^T=D$ which you are seeking always exists, with real positive diagonal $D$ and unitary $A$, but this is not what is commonly called the "diagonalization" of $M$ (which would require $A^T=A^{-1}$). $\endgroup$ – Carlo Beenakker Mar 30 '13 at 14:51
  • $\begingroup$ How about this quiz - generalizing autonne takagi factorization: mathoverflow.net/q/302741/27004 $\endgroup$ – wonderich Jun 14 '18 at 22:26

You also ask how to construct the matrix $A$: it is the unitary matrix of eigenvectors of the Hermitian matrix $M\cdot M^{\dagger}$.

More explicitly: The masses $m_n$ can be obtained from the eigenvalues of the matrix product $H=M\cdot M^{\dagger}$, where $M^{\dagger}$ denotes the complex conjugate of the transpose of $M$. The eigenvalues $h_n$ of $H$ are real and nonnegative, so you obtain a nonnegative mass $m_n=\sqrt{h_n}$. The matrix $A$ is the matrix of eigenvectors of $H$, so that $H=A\cdot{\rm diag}(h_1,h_2,\ldots)\cdot A^{\dagger}$. It is a unitary matrix, $AA^{\dagger}=1$.

You'll note that the matrix $A$ is not unique, you can always multiply it by a diagonal matrix of phase factors $A\mapsto A\cdot{\rm diag}(e^{i\phi_1},e^{i\phi_1},\ldots)$. The easiest way to account for this, is to just take any $A$ and calculate

$A^{\dagger}\cdot M\cdot(A^{\dagger})^{T}={\rm diag}(e^{i\psi_1}m_1,e^{i\psi_2}m_2,\ldots)$.

Then the required phases $\phi_n$ are obtained by $\phi_{n}=-\psi_n/2$.

  • $\begingroup$ The keyword singular value decomposition (SVD) might be of interest to the OP $\endgroup$ – Aaron Hoffman Mar 30 '13 at 21:47
  • $\begingroup$ @CarloBeenakker I don't fully understand. Suppose $M$ is a unitary matrix, then $H = M M^\dagger = \mathbb I$, and so we can initially take $A = \mathbb I$ (note $h_n = 1$). But this is in conflict with your statement that $A^\dagger M (A^\dagger)^T$ should be diagonal? $\endgroup$ – Ruben Verresen Nov 22 '16 at 15:24
  • $\begingroup$ @RubenVerresen --- I assumed that the eigenvalues $h_n$ are all distinct, then the matrix of eigenvectors $A$ is unique up phase factors. $\endgroup$ – Carlo Beenakker Nov 22 '16 at 21:13
  • $\begingroup$ @CarloBeenakker Aha okay, thanks, so if I understand correctly the procedure you outlined above should work for sufficiently generic matrices, but as soon as we have extra properties for $M$ (such as unitarity) then we should look elsewhere. $\endgroup$ – Ruben Verresen Nov 22 '16 at 21:15
  • $\begingroup$ (Since I am interested in the case where $M$ is unitary I posted a new question, which I'm linking to here in case it may be useful for someone else in the future: math.stackexchange.com/questions/2026110/… ) $\endgroup$ – Ruben Verresen Nov 22 '16 at 21:17

I found the following theorem in Horn and Johnson's book:

(Takagi Factorization): If $A$ is symmetric, then there exists a unitary matrix $U$ and a real nonnegative diagonal matrix $\Sigma$ such that $A=U\Sigma U^T$, where the columns of $U$ are an orthonormal set of eigenvectors for $A\bar{A}$ and the corresponding diagonal entries of $\Sigma$ are the non-negative square roots of the corresponding eigenvalues of $A\bar{A}$.

Also, for example the matrix $A=\left(\begin{array}{cc}1& i\\\\ i& -1\end{array}\right)$ is an example of a complex symmetric matrix that is not diagonalizable.

  • $\begingroup$ Thanks newbie, this was useful too. But I marked Carlo's answer as "correct" because it gave an explicit constriction that I found helpful. $\endgroup$ – PoorPhysicist Apr 1 '13 at 6:08

I believe this theorem can be found in Horn and Johnson's "Matrix Analysis" book, though I don't have it at hand to check. At any rate, a complex symmetric matrix $M$ is diagonalizable if and only if its eigenvector matrix $A$ can be chosen so that $A^TMA = D$ and $A^TA=I$, where $D$ is the diagonal matrix of eigenvalues. (In other words there is a complex orthogonal, rather than unitary, matrix of eigenvectors). Based on this fact (or by a direct calculation) one can construct 2x2 complex symmetric matrices that are not diagonalizable.


Since no one said it, without the condition of question 2 this can always be done. Choose a vector $v$ such that $v^T M v\neq 0$. If this can't be done then the matrix is symmetric and symplectic, hence zero, hence diagonal. If it can, scale $v$ so the product is positive real, restrict to the kernel of $v^T M$, and apply induction.


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