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$A \in \mathbb R^{4 \times 4}$ is matrix whose diagonal elements are zero. Can I say the following?

  1. If $A$ is neither skew-symmetric nor the zero matrix, then $A$ has at least one eigenvalue with positive real part.

  2. In my case I have all the diagonal elements equal to zero. What if in general $\mbox{tr}(A)=0$?

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Consider this matrix.

$\begin{bmatrix}0&1\\&0&1\\&&0&1\\&&&0\end{bmatrix}$

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Routh–Hurwitz (H) stability criterion:

This can also be used in general to find if a matrix $A\in \mathbb{R}^{n\times n}$ has eigenvalues with a positive real part.

Use RH criterion on the characteristic polynomial of $A$, that is, $det(sI-A)=0$.

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Routh–Hurwitz tells us that if the trace is zero, then either all eigenvalues have zero real parts, or there is a pair of eigenvalues whose real parts have opposite signs.

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