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Is it possible to give a nice characterization of matrices $A \in R^{n \times n}$ which are self-adjoint with respect to some inner product?

These matrices include all symmetric matrices (of course) and some nonsymmetric ones: for example, the transition matrix of any (irreducible) reversible Markov chain will have this property.

Naturally, all such matrices must have real eigenvalues, though I do not expect that this is a sufficient condition (is it?).

About the only observation I have is that since any inner product can be represented as $\langle x,y \rangle = x^T M y$ for some positive definite matrix $M$, we are looking for matrices $A$ which satisfy $A^T M = M A$ or $M^{-1} A^T M = A$. In other words, we are looking for real matrices similar to their transpose with a positive definite similarity matrix.

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2 Answers 2

up vote 6 down vote accepted

In addition to having real eigenvalues, the matrix will have to be diagonalizable, i.e., there have to be enough eigenvectors to span $R^n$. Once these conditions are satisfied, you can take a basis consisting of $n$ eigenvectors and define an inner product by declaring these basis vectors to be orthonormal. That inner product will make your matrix self-adjoint, because there's an orthonormal basis with respect to which the matrix is diagonal with real entries on the diagonal.

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So, a subset of the set of normal matrices, then? –  J. M. Aug 15 '10 at 3:13
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@J. Mandalgan: No, there are plenty of examples of a matrix diagonalizable over the reals which does not commute with its transpose, such as $\left[\begin{array}{cc}0 & 1 \\\\ 0 & 1\end{array}\right]$. –  Tracy Hall Aug 15 '10 at 3:38
    
Thanks Tracy for the explanation; somehow I had remembered mistakenly that "normal" is synonymous with "diagonalizable". –  J. M. Aug 15 '10 at 6:20
    
J. Mangaldan: Normal matrices are precisely those that can be diagonalised via a unitary transformation. –  José Figueroa-O'Farrill Aug 15 '10 at 12:57
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Tracy Hall: normal matrices are not those which commute with their transpose, but those which commute with their adjoint relative to the inner product. If the inner product is the usual "dot" product, then this is indeed the transpose, but not necessarily otherwise. Your 2x2 matrix is not just normal, but indeed symmetric, relative to the inner product $\begin{bmatrix}\hphantom{-}1 & -1 \cr -1 & \hphantom{-}0\end{bmatrix}$. –  José Figueroa-O'Farrill Aug 15 '10 at 12:59

Take a look at: Indefinite linear algebra and applications --- Israel Gohberg,Peter Lancaster,L. Rodman, section 4.2.

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