# What is a necessary and sufficient condition for all the eigenvalues of a non-symmetric matrix to be positive?

Consider a real but not symmetric matrix $A$. To test if the matrix has positive eigenvalues, I've learnt from this forum that a symmetric matrix will be given by $B=A+A^T$. If all the eigenvalues of $B$ are positive, then it follows that A also has all the eigenvalues positive. So this is a sufficient condition.

For example, consider

$$A=\begin{bmatrix}1&4\\0&1\end{bmatrix}$$

It so happens that $B$ has one negative eigenvalue. Whereas $A$ has both positive eigenvalues. So what is a necessary (and sufficient) condition that $A$ has all positive eigenvalues?

Looking at Gershgorin's theorem , further rises the possibility of complex eigenvalues.

References:

1. Tests for positive definiteness for nonsymmetric matrices
2. p.322,Linear Algebra and its Applications, Gilbert Strang.
3. Necessary and sufficient condition for all the eigenvalues of a real matrix to be non-negative
• A non-symmetric matrix may have complex eigenvalues. Do you want their real parts to be positive? Aug 1, 2017 at 17:20
• A search for 'Lyapunov equation' might be helpful. Aug 1, 2017 at 17:39
• @RodrigodeAzevedo I want all the Eigen Values to be real Aug 2, 2017 at 7:06
• @TilakMallikarjun That is why symmetry is nice. It does ensure that the eigenvalues are real. Aug 2, 2017 at 9:03
• @RodrigodeAzevedo, I am having a non-symmetric matrix, whose sign of Eigen values determine the nature of a phenomena. So I want to know if at all it is possible with the methods in Linear algebra to get an idea of the same without actually solving for Eigen values. Aug 2, 2017 at 10:08

Your result is false: take $A=\begin{pmatrix}1&-1\\1&1\end{pmatrix}$ then $A+ A^T=2I_2$ has his eigenvalues positive whereas eigenvalues of $A$ are $1 \pm i$ not even real...