P is a symmetric positive definite matrix. I want to know under what condition of matrix A will the matrix A'P+PA be positive definite, assuming A,P are square matrices of same order.

A'P+PA is a symmetric matrix proof: let w=A'P then W'=(A'P)'=PA W=(W+W')/2 + (W-W')/2 where (W+W')/2 is symmetric and (W-W')/2 is skew symmetric therefore A'P+PA is symmetric (equals symmetric part w+w')

Basically I want to clarify if both A and P must be symmetric positive definite matrix for A'P+PA to be positive definite (or positive semi def). I am interested in knowing the properties of A which would ensure positive definiteness (or positive semi def) of A'P+PA.

  • $\begingroup$ A necessary condition for $A^\ast P+PA$ to be positive definite is that all eigenvalues of $A$ have positive real parts (Lyapunov's theorem). $\endgroup$
    – user1551
    Jan 15 '17 at 14:06

I think for positive definiteness of $A^T P + PA$, the matrix $A$ needs to be positive definite. Let us assume that $A$ has one negative eigenvalue such that $\exists q: \; A q = \lambda q$ for $\lambda<0$. Then using the same $q$, to compute the quadratic form of $A^T P + PA$, we have

$$ q^T (A^T P + PA) q = q^T A^T P q + q^T P A q = \lambda q^T P q + q^T P \lambda q = 2 \lambda q^T P q < 0,$$

where the $<0$ follows since $\lambda<0$ and $q^T P q>0$ (due to positive definiteness of $P$).

Note that this only shows that positive definiteness is necessary, not that it is sufficient (though I think it should be).

As for semidefiniteness, it is clear that if $A$ is semidefinite, it has at least one zero eigenvalue. Then, with the same argument as above, we can also show that $A^T P + P A $ can become zero for the corresponding eigenvector of $A$.

edit: As pointed out in the comments, from the argument above it is only necessary that the eigenvalues of $A$ are (non-negative) positive. Positive (semi)-definite matrices satisfy this but the converse is not true, matrices with strictly positive eigenvalues can be indefinite.

  • $\begingroup$ Positive semidefiniteness is both necessary and sufficient. $\endgroup$ Jan 15 '17 at 12:29
  • $\begingroup$ It is not clear from this answer or from your comment whether it is necessary that $A$ be a symmetric matrix. Perhaps it suffices for all eigenvalues of $A$ to be real and non-negative? $\endgroup$
    – pre-kidney
    Jan 15 '17 at 13:17
  • $\begingroup$ That is true. It is only necessary that all eigenvalues are non-negative. I was assuming that this implies (semi-)definiteness but it seems I was wrong. A counter-example is $A = \begin{bmatrix} 1 & -3 \\ 0 & 1 \end{bmatrix}$ which has the double eigenvalue 1 but is indefinite ($\begin{bmatrix} 1& 1\end{bmatrix} \cdot A \cdot \begin{bmatrix} 1\\1\end{bmatrix} = -1$). I've made an edit. $\endgroup$
    – Florian
    Jan 15 '17 at 15:28
  • $\begingroup$ Hi Florian, the point is that since A is not assumed to be symmetric, it may have complex eigenvalues. In fact, consider the matrix A = [0, 1; -1, 0]. It has no real eigenvalues. However, taking P to be the identity we see that A + A^T is the zero matrix which is psd. So it is definitely not the case that A must have any real eigenvalues at all, let alone be psd. $\endgroup$
    – pre-kidney
    Jan 15 '17 at 17:14
  • $\begingroup$ Hmm, I see your point. In the reals, the connection to eigenvalues will only work for diagonalizable matrices which is already too much to ask. Still, I would claim that an eigenvalue of $A$ that is negative (or has a negative real part) would lead to negative values in the QF of $A^TP+PA$. For sure this is true in the complex-valued case (which I'm more used to, that's the source of my wrong assertions). The OP hasn't made clear over which fields he's looking at the matrices. Thanks for your feedback though, this was instructive. $\endgroup$
    – Florian
    Jan 15 '17 at 18:25

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