I split off this question from the thread here:(adsbygoogle = window.adsbygoogle || []).push({});

https://www.physicsforums.com/threads/error-in-landau-lifshitz-mechanics.901356/

In that thread, I was told that a symmetric matrix ##\mathbf{A}## with real positive definite eigenvalues ##\{\lambda_i\} \in \mathbb{R}^+## is always real. I feel that I must be overlooking something simple, as I can't seem to prove it. Clearly the determinant and trace are positive, (so the matrix is nonsingular) and if it's diagonalizable, then the matrix is similar to a real matrix (namely the diagonal eigenvalue matrix). But I'm not seeing how this implies that the original ##\mathbf{A}## is real.

I've seen various claims that a symmetric matrix can be written as ##\mathbf{A} = \mathbf{O}^T\mathbf{D}\mathbf{O}##, where ##\mathbf{O}## is an orthogonal matrix and ##\mathbf{D}## is the diagonal eigenvalue matrix, but sometimes (e.g., Wikipedia) the claim is explicitly for real symmetric matrices and sometimes (e.g., Mathworld) it is unspecified whether the claim is for real matrices or for all symmetric matrices. If it's true forallsymmetric matrices, then we're done, because orthogonal matrices are necessarily real. but if it's only true forrealsymmetric matrices, then we're back to square one.

I also tried working out the case for 2x2 matrices explicitly. The characteristic polynomial is

$$0 = \lambda^2-\lambda\mathrm{tr}(\mathbf{A})+\mathrm{det}(\mathbf{A})$$

Since ##\lambda^2 > 0##, we have that

$$\lambda(a_{11}+a_{22})-(a_{11}a_{22} - a_{12}^2) > 0$$

must be real. Writing this out explicitly as complex numbers gives:

$$\lambda((a+bi)+(c+di)) > [(a+bi)(c+di)-(e+fi)^2]$$

where ##a,b,c,d,e,f \in \mathbb{R}##. To get rid of the imaginary part, we need to satisfy:

$$0 = \lambda(b+d)-bc-ad+2ef$$

Since ##\mathrm{tr}(\mathbf{A}) > 0##, we get ##b+d = 0##. Since ##\mathrm{det}(\mathbf{A}) > 0##, we get ##bc+ad-2ef = 0##. So we have two conditions to satisfy:

$$b(c-a)+2ef=0$$

ensures reality, and:

$$\lambda(a+c)>ac+b^2-e^2+f^2$$

ensures positivity.

However, I don't see how in general, this implies ##b=d=f=0## such that the overall matrix is real. (EDIT: if one of ##b,d,f=0## then all of them must be zero by the first condition.)

It's a weird problem because I'm so used to dealing with either real symmetric matrices or complex Hermitian matrices, that I'm not sure what linear algebra rules apply to complex symmetric matrices. Thanks for any insight you can provide.

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# I Is a symmetric matrix with positive eigenvalues always real?

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