# 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? – Rodrigo de Azevedo Aug 1 '17 at 17:20
• A search for 'Lyapunov equation' might be helpful. – ekkilop Aug 1 '17 at 17:39
• @RodrigodeAzevedo I want all the Eigen Values to be real – Tilak Mallikarjun Aug 2 '17 at 7:06
• @TilakMallikarjun That is why symmetry is nice. It does ensure that the eigenvalues are real. – Rodrigo de Azevedo Aug 2 '17 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. – Tilak Mallikarjun Aug 2 '17 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...