Positive definite matrix

Let A be a real nxn matrix.
What are the requirements of A for A+AT to be positive definite?
Is there a condition on eigenvalues of A, so that A+AT is positive definite?

Also I am not sure about the definition of a positive definite matrix. In some places it is written that the matrix must be symmetric, while in others it is defined for non-symmetric matrices. In many places I see theorems, which require a matrix to be positive definite and it is not clear to me if I can use it for non-symmetric matrices or not.

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Hello.

First, the definition:

Let's a matrix, A, is called definite positive if, $$vAv^t>0$$ for all v (v vector).

Here, I think you have the answer.

http://mathworld.wolfram.com/PositiveDefiniteMatrix.html

The problem is that this differs from how it is defined in Wikipedia. The Wikipedia definition is clear about symmetric matrices, but vague about non symmetric ones. So what do you do when you see some theorem, that speaks about positive definite matrices? Do you require the matrix to be symmetric or not?

I don't know if wikipedia is the ultimate source but definiteness is usually defined together with the partial ordering (actually Löewner partial ordering to sound very academic and sophisticated). Thus, non-hermitian definite matrices are of little use. Only gives some information about the sign of its eigenvalues. In other words $A-B \succ 0$ and thus $A \succ B$ makes no sense for nonhermitian matrices.

When the matrix entries are real $v^tAv = v^TA^Tv \in \mathbb{R}$, hence if $v^TAv > 0$ for all v , then $v^T(A + A^T)v = 2v^TAv> 0$ for all v. Note that, I did not use the term positive definite since A might be nonsymmetric. But obviously $A + A^T$ is symmetric hence positive definite.

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Positive definite matrix is automatically symmetric.

That was wrong in the real case. Thanks, D H.

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D H
Staff Emeritus
\begin{align*} A &= A_S + A_A \\[4pt] A_S &\equiv \frac 1 2 \left(A+A^T\right) \\[4pt] A_A &\equiv \frac 1 2 \left(A-A^T\right) \end{align*}
With this notation, the quadratic form $x^TAx$ becomes
$$x^TAx = x^TA_Sx + x^TA_Ax = x^TA_Sx$$
The final expression results because $x^TA_Ax \equiv 0$. The contribution of the skew symmetric part of a matrix to this quadratic form is identically zero.